engineeering science

MALAYSIAN AVIATION TRAINING ACADEMY

Apprentice Course For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved MIP/MECH/TN/003

INSTITUTE OF AEROSPACE TECHNOLOGY

AIRFORCE COLLEGE

PREPARATORY COURSE

ENGINEERING SCIENCE



RMAF TECHNICIAN ENHANCEMENT PROGRAMME

NAME : CLASS :



WARNING

This training note is intended for training purposes only. The information it contains is as accurate as

possible at the time of issue, and it is not subjected to amendment action. Where the information contained in this training note is at variance with official documents, the latter must be taken as the overriding authority. The contents in this training note shall not be reproduced in

any form without the expressed permission of MALAYSIAN AVIATION TRAINING ACADEMY

(MATA) SDN. BHD.

Preparatory Course Royal Malaysian Air Force Introduction Engineering Science 2.1- H.0 - 1 ________________________Apprentice Course – Mechanic_______________________

2.1 INTRODUCTION 2.1.1 What is Physics?

Physic is the most basic of the sciences. It deals with the behavior and structure of matter. The field of Physics is usually divided into the areas of motion, fluids, heat, sounds, light, electricity and magnetism and the modern topic of relativity, atomic structure, condensed-matter physics, nuclear physic, elementary particles, and astrophysics.

Physic, like other sciences, is a creative endeavor. It is not simply, a collection of facts. Important theories are created with the idea of explaining observations. To be accepted, theories are tested by comparing their predictions with the results of actual experiments. Note that, in general; a theory cannot be "proved" in an absolute sense.

Scientists often devise models of physical phenomena. A model is a kind of picture or analogy that seems to explain the phenomena. A theory often developed a model, usually deeper and more complex than a simple model.

A scientific law is a concise statement, often expressed in the form of an equation, which quantitatively describes a particular range of phenomena over a wide range of cases.

2.1.2 Measurement and Uncertainty.

In the quest to understand the world around us, scientists seek to find relationships among the various physical quantities they observe and measure. Measurements play a crucial role in physics, but can never be perfectly precise. It is important to specify the uncertainty of a measurement either by stating it directly using the ± notation, and/or by keeping only the correct number of significant figures. For example the width of a board might be write as 5.2 ± 0.1 cm. The ± 0.1 cm ("plus or minus 0.1 cm") represents the estimated uncertainty in measurement, so that the actual width most likely lies between 5.2 – 0.1 = 5.1 cm and 5.2 + 0.1 = 5.3 cm.

2.1.3 Units, Standards and the SI System.

Physical quantities are always specified relative to a particular standard or unit, and the unit used should always be started. The commonly accepted set of units today is the System International ( SI ), in which the standard unit of length, mass and time are the meter, kilogram and second. International System

The International system of units is called Sys’teme International d’ Unite’s (SI) and is essentially the same as we have come to know as the metric system. The International Committee on Weights and Measures has established seven fundamental quantities and has assigned official base units to each quantity. A summary of these quantities, their base units, and the symbols for the base units is given in table 1.1.

We can measure many quantities, such as volume, pressure, speed, and force, which are combinations of two or more fundamental quantities. However, no one has ever encountered a measurement that cannot be expressed in terms of length, mass, time current, temperature, luminous intensity, and amount of substance. Combinations of these quantities are referred to as derived quantities, and they are measured in derived units. Several common derived units are listed in Table 1.2. For training purposes only Rev.0 Issued 01 Dec 06 Malaysian Aviation Training Academy All right reserved MIP/MECH/TN/003


U.S. Customary System Units (USCS)

Unfortunately, the SI units are not fully implemented in many industrial applications. The United States is making progress toward the assimilation of SI units. However, wholesale conversions are costly, particularly in many mechanical and thermal applications, and total conversion to the International System will require some time. For this reason it is necessary to be familiar with older units for physical quantities. The USCS units for several important quantities are listed in table 1.3. Quantity Unit Symbol

Base Units Length meter m Mass kilogram kg Time second s Electric current ampere A Temperature kelvin K Luminous intensity candela cd Amount of substance mole mol

Table 1.1: Base units for seven fundamental quantities. Quantity SI unit USCS unit Length meter (m) Foot (ft) Mass kilogram (kg) Slug (slug) Time second (s) Second (s) Force (weight) Newton (N) Pound (lb)

R) Temperature Kelvin (K) degree Rankine (o Table 1.3: USCS units. For training purposes only Rev.0 Issued 01 Dec 06 Malaysian Aviation Training Academy All right reserved MIP/MECH/TN/003


Quantity Derived units Symbol Area Square meter m2

Volume Cubic meter m3

Frequency Hertz Hz s-1

Mass density (density) Kilogram per cubic meter Kg/m3

Speed, velocity Meter per second m/s Angular velocity Radian per second rad/s Acceleration Meter per second squared m/s2

Angular acceleration Radian per second squared rad/s2

Force Newton N kg.m/s2

Pressure (mechanical stress) Pascal Pa N/m2

Kinematic viscosity Square meter per second m /s 2

Dynamic viscosity Newton second per squared meter

N·s/m2

Work, energy, Quantity of heat Joule J N/m Power Watt W J/s Quantity of electricity Coulomb C Potential difference, Electromotive force Volt V J/C Electric field strength Volt per meter V/m Electric resistance, Ohm Ω V/A Capacitance Farad F C/V Magnetic flux Weber Wb V·s Inductance Henry H V·s/A Magnetic flux density Tesla T Wb/m2

Magnetic field strength Ampere per meter A/m Magnetomotive force Ampere A Luminous flux Lumen lm cd·sr Luminance Candela per square meter cd/m2

Illuminance lux lx lm/m2

Wave number 1 per meter m-1

Entropy joule per kelvin J/K Specific heat capacity joule per kilogram Kelvin J/(kg·K) Thermal conductivity watt per meter Kelvin W/(m·K) Radiant intensity watt per steradian W/sr Activity (of a radioactive source)

1 per second s-1

Table 1.2: Derived units for common physical quantities For training purposes only Rev.0 Issued 01 Dec 06 Malaysian Aviation Training Academy All right reserved MIP/MECH/TN/003


2.1.4 Converting Units.

Any quantity we measure, such as length, a speed, or an electric current, consists of number and unit. Often we are given quantity in one set of units, but we want it expressed in another set of units.

1 in = 2.54 cm 1 ft = 12 in. 1 yd = 36 in. = 3 ft 1 m = 39.4 in. = 3.28 ft 1 mi = 1.61 km = 5280 ft = 1760 yd

For example, suppose we measure that a table is 21.5 inches wide, and we want to express this in centimeters. We must use a conversion factor, which in this case is: - 1 in = 2.54 cm or written another way,

1 = 2.54 cm/in

Since multiplying by one does not change anything, the width in cm is:

21.5 inches = 21.5 in x 2.54 cm/in = 54.6 cm.

Note: the unit (inches in this case) cancelled out.

Example 1: What is the length of the table of the 100 m dash expressed in the yards?

Solution; Let us assume the distance is accurately known to four significant figures, 100.0 m. One yard (1 yd) is precisely 3 feet (36 inches), so we can write

1 yd = 3 ft = 36 in ( 1.0 ) = 36 in. (2.540 cm/in) = 91.44 cm

or; 1 yd = 0.9144 m

since ; 1 m = 100 cm. We can write this results as;

1 m = 1 yd = 1.094 yd 0.9144 then;

100 m = 100 m ( 1.094 yd/m) = 109.4 yd.

So a 100 m dash is 9.4 yard longer than a 100-yard dash. For training purposes only Rev.0 Issued 01 Dec 06 Malaysian Aviation Training Academy All right reserved MIP/MECH/TN/003


Example 2: Where the posted speed limit is 55 miles per hour ( mi/h or mph ), what is this speed: a) in meters per second (m/s) b) in kilometers per hour (km/h) Solution; a) we can write 1 miles as 1 mi = 1 mi x ( 5280 ft ) x( 12 in ) x ( 2.54 cm ) x ( 1 m ) 1 mi 1 ft 1 in 100 cm 1 mi = 1609 m ( 1.1 ) Note that each conversion factor is equal to one. We also know that 1 hour equals 60 min/h x (60 s/min) = 3600 s/h, so

55 mi/h = ( 55 mi) x ( 1609 m ) x ( 1 h ) h 1 mi 3600 s

= 25 m/s

b) now we use

1 mi = 1609 m = 1.609 km, then 55 mi/h = (55 mi/h) x (1.609 km) 1 mi

= 88 km/h Exponential notation

We can express measurements of length fairly easily if they fall in the range 0.001 m to 1000 m. When we try to talk about the size of an atom (0.000 000 000 1 m) or the distance to Pluto (6 000 000 000 000 m) we experience problems with the length of the number. It is common for scientists to use shorthand in which we write large numbers in a form using powers of ten. This is called exponential notation. For example:

• 10000 is written as 10 or 1 x 10 4 4

• 1/10000 or 0.0001 is written as 10 or 1 x 10 -4 -4

Sometimes instead of writing 104, we shorten the notation further so that

104 is written (pronouns or type in calculator) as 1E4. This final approach is often used in computer languages such as BASIC. It is common practice to adopt the following conventions:

Write numbers in exponential notation with one numeral in the digits place and others after the decimal point, that is, 3.43 x 10 not 34.3 x 10 . 4 3

Leave numbers between 0.1 and 100 as they are. Examples 56000000 = 5.6 x 10000000. It is written as 5.6 x 10 (or 5.6 E7). 7

0.00000056 = 5.6/10000000. It is written as 5.6 x 10- (or 5.6 E-7). 7.

For training purposes only Rev.0 Issued 01 Dec 06 Malaysian Aviation Training Academy All right reserved MIP/MECH/TN/003


Prefixes.

In the metric system we refer to multiples of the fundamental units by adding appropriate prefixes to the unit's name. We can use these prefixes not only for fundamental qualities of mass, length and time, but also for derived units of volume, velocity orally other unit. Commonly used prefixes and their meanings are listed in Table 1.4. Some derived quantities

Any measurement can be represented in terms of the fundamental units,

some of the more common units that can be derived are; area, volume, density, and velocity.

prefix Symbol Meaning Multiplier Common examples in use

Tera T 1012

1,000,000,000 10Giga gigawatt (GW) G 9

10 1,000,000 Mega megawatt (MW), megabyte (Mb) M 6

10 1,000 kilo kilometre (km), kilogram (kg) k 3

10 hecto hectopascal (hPa) h 2

10 deca da 1

10 0.1 deci d -1

10 0.01 centi centimetre (cm) c -2

10 0.001 milli millimetre (mm), millilitre (mL) m -3

10 0.000001 micro micrometre or micron (μm) μ -6

10 0.000000001 nano n nanometre (nm) -9

10- 0.000000000001 pico p picofarad (pF) 12

10 femto f femtosecond (fs) -15

Table 1.4: Commonly used prefixes

Area.

The amount of space covered by a body is called its area and is measured in square meter (m ). Area is the product of length and width, that is: 2

Area = length x width

A = l x w ( 1.2 )

As to the derivation of the square meter, symbol m2

Width (w)

Length (l)

Height (h)

Length (l)



Volume.

The amount of space taken up by a body is called its volume and is measured in cubic meter (m3). The volume of a rectangular body is the product of its length, width and height, that is:

Volume = length x width x height V = l x w x h ( 1.3 )

A cubic meter has side's 1 m long and is rather cumbersome for laboratory work. It is often more useful to measure volumes in terms of cubic centimeters (cm ). A cube with sides 1 cm (1 x 103 -2 m) long has a volume of 1 cm , or; 3

= 1 x 10 1 cm3 -2 m x 1 x 10 m x 1 x 10-2 -2 m = 1 x 10 1 cm3 -6 m3

Density The mass per unit volume of a substance is called its density, or;

mass Density = Volume m ( 1.4) ρ = v

Density is usually expressed in term of kg/m3 or g/cm3 thus ;

= 1 g 1 g/cm3 x (1 x 10-3 kg) x 1 cm3

cm3 1 g (1 x 10 m)-2 3

= 1 x 10-3 kg 1 x 10 m-6 3

= 1 x 10 kg/m3 3



Velocity The distance traveled in 1 second is called the velocity of a body, or;

distance Velocity = time (1.5)

s t

V =

Velocity is usually measured in m/s thus; 1 km/s = 103 m = 10 m/s 3

1 s 1 km/h = 103 m = 0.278 m/s

60 x 60 s For training purposes only Rev.0 Issued 01 Dec 06 Malaysian Aviation Training Academy All right reserved MIP/MECH/TN/003


Task 1 1. Convert the length given below to millimeters by extending the given numbers

without altering the full stop position. a) 3.7 m ……….mm c) 300 μm b) 32.4 cm………mm d) 3.5 x 104 nm

2. The derived, basic SI unit of area is the…………. Enter the unit factors in the

following list of most common multiples and sub-multiples, using full and decimal fraction number. a) 1 μm ……….m c) 1 mm ……….m 2 2 2 2

b) 1 cm ……….m 2 2

3. Enter the power of ten to obtain equivalent values:

a) 1 m …………..cm c) 1 km …………cm 2 2 2 2

b) 1 mm ……………cm 2 2

4. The derived basic SI unit for volume is the……… . Enter the power of ten to

obtain equivalent values: a) 1 cm ………..m3 3 c) 1 m …………cm 3 3

b) 1 mm ……….m 3 3

5. The basic SI unit of the mass is the …….. . Mark the decimal prefixes in the

following list and enter the basic unit factors, using full and decimal-fraction number. a) 1 tonne…………..kg c) 1 milligram ………….kg b) 1 gram …………..kg d) 1 microgram…………kg

6. Convert the masses given below to grams, using full and decimal fraction

numbers. a) 100000 mg d) 300 mg b) 0.267 µg e) 0.00000013 kg c) 0.000072 kg

Task 2 1. From Alor Star to Sungai Petani is 70 km. How far is this in miles? 2. Ahmad driving a 3-tone truck. How much actually the truck mass in

kilogram? 3. A lamppost height is 4.5 m. What is the lamppost height in cm? 4. 1 kg sugar equal to how many µg?


Preparatory Course Royal Malaysian Air Force Law Of Motion Physics 2.2 - H.0 - 1 ________________________ Apprentice Course – Mechanic _________________________

For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved MIP/MECH/TN/003

2.2 LAW OF MOTION (KINEMATICS).

Kinematics deals with the description of how object moves. The description of the motion of any object must always be given relative to some particular reference frame. The displacement of an object is the change in position of the object. 2.2.1 Average Velocity

The term “speed” refers to how far an object travels in a given time interval. If a car travels 240 kilometers (km) in 3 hours, we say its average speed was 80 km/h. In general, the average speed of an object is defined as the distance traveled along its path divided by the time it takes to travel this distance;

Average speed = distance traveled ( 2.1 )

time elapsed

Speed is simply a positive number, with units. Velocity on the other hand, is used to

signify both the magnitude (numerical value) of how fast an object is moving and the direction in which it is moving. (Velocity is therefore a vector quantity) The average velocity is defined in terms of displacement, rather than total distance traveled; Average velocity = displacement ( 2.2 ) time elapsed Average speed and average velocity often have the same magnitude, but sometimes they don’t. As an example, in Figure 2.1 when a person walked 70 m east and then 30 m west. The total distance traveled was 70 m + 30 m = 100 m, but the displacement was 40 m. suppose this walk took 70 s to complete.

West

70 m

0 40 m

Displacement

30 m East

Figure 2.1: Then the average speed was: Speed = distance = 100 m = 1.4 m/s time 70 s The magnitude of the average velocity, on the other hand, was: Velocity = displacement time = 40 m 70 s = 0.57 m/s



The average velocity, defined as the displacement divided by the elapsed time, can be written : v = x2 – x1 = Δx ( 2.3 ) t2 – t1 Δt The elapsed time is t2 – t1, and during this time interval the displacement of the object was Δx = x2 – x1 Example 1: Runner ‘ s average velocity. The position of a runner as a function of time is plotted as moving along the x-axis of a coordinate system. During a 3.0 s time interval; the runner’s position changes from x1 = 50.0 m to x2 = 30.5 m, as shown in Figure 2.2. What was the runner’s average velocity?

0 10 20 30 40 50 60

Δ x (X2) (X1)

finish start

Distance ( m)

Figure 2.2: A person run from x1 = 50.0 m to x2 = 30.5 m The displacement is -19.5 m Solution;

Average velocity is the displacement divided by the elapsed time. The displacement is Δx = x2 – x1 = 30.5 m – 50.0 m = -19.5 m. The time interval is Δt = 3.0 s. Therefore the average velocity is:- v = Δx Δt

= -19.5 m 3.0 s = - 6.50 m/s

The displacement and average velocity are negative, which tell us (if we didn’t already know it) that the runner is moving to the left along the x-axis, as indicated by the arrow in fig. 2.2. Thus we can say that the runner’s average velocity is 6.50 m/s to the left.



Example 2 : How far can a cyclist travel in 2.5 h along a straight road if her average speed is 18 km/h? Solution; We want to fine the distance traveled, so we use e.g. 2.3, letting Δx be the distance and v be the average speed and then rewrite it as :

v = Δx Δt Δx = v Δt = ( 18 km/h ) ( 2.5 h ) = 45 km 2.2.2 Displacement – Time Graphs

It is often useful to present records of motion in the form of a graph. The position of a moving object was recorded at different times and the results are shown in table 2.1. In the case of analyzing motion we can plot the position or displacement of the object the vertical axis (y-axis) and the elapsed time along the horizontal axis ( x-axis ), as shown figure 2.3.

Displacement (m) 1.0 2.0 3.0 4.0 4.0 4.0 4.0 4.0

Time (s) 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 Table 2.1: Displacement and time measurements. Displacement ( m) 5 3 1 Time ( s) 0 1 2 3 4 5 6 7 8 Figure 2.3: Displacement-time graph

Note that the plotted points are the direct observations. When we draw a line between these points we are assuming that the motion was uniform. That is called interpolation.



Example 3: The following table records the displacement of an object at different times. Draw a displacement – time graph for the object.

Displacement ( m ) 0 2 4 4 4 6 6 4 2 0

Time ( s ) 0 1 2 3 4 5 6 7 8 9

Displacement ( m ) 7 5 3 1 1 2 3 4 5 6 7 8 9 Time ( s ) 0

Figure 2.4 : A displacement- time graph for the motion of the object described in figure 2.5. Displacement ( m ) 4

3 2 1 A

B C

D

0 1 2 3 4 5 6 Time (s)



Figure 2.5:

the section AB represents change in displacement away from the reference point to the right ( in a positive direction )

At point B, the displacement and the distance travel is 4 m from A. the section BC has zero slope ( it is flat ). The object is stationary here.

At point C, the displacement and the distance travel is 4 m from A. the section CD represents change in displacement toward the reference point ( in a

negative direction ). At point D, the displacement is 0 m and distance travel is 8 m from A.

at point D the object is at same position as at point A. The displacement of the object at this point is zero.

2.2.3 Acceleration

An object whose velocity is changing is said to be accelerating. A car whose velocity increases in magnitude from zero to 80 km/h is accelerating. Acceleration specifies how rapidly the velocity of an object changing. Average acceleration is defined as the change in velocity divided by the time taken to make this change: Average acceleration = change of velocity ( 2.4 ) time elapsed In symbols, the average acceleration, a, over a time interval Δt = t2 – t1, during which velocity changes Δv = v2 – v1 is defined as: a = v2 – v1 = Δv . ( 2.5 ) t2 – t1 Δt Example 4: Average acceleration A car accelerates along a straight road from rest to 20 m/s in 5.0 s. What is the magnitude its average acceleration? Solution : The car starts from rest, so v1 = 0. The final velocity is v2 = 20 m/s. Then from e.g. 2.5, the average acceleration is;

a = v2 – v1 t2 – t1 = 20 m/s – 0 m/s 5.0 s = 4 m/s² # Careful !! : Do not confuse velocity with acceleration. Note : Acceleration tells us how fast the velocity changes, whereas velocity tells us

how fast the position changes.



Example 5: car slowing down. An automobile is moving to the right along a straight highway, which we choose to be the positive x-axis and the driver puts on the brakes. If the initial velocity is v1 = 15.0 m/s and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration? Solution ; a = v2 - v1 t = 5.0 m/s – 15.0 m/s 5.0 s = -2.0 m/s 2.2.4 Velocity – time graphs

We can use a table of position and time to produce the velocity of a moving body at a given time. We convert the displacement –time data using the following relationships: Velocity = Δs = s2 – s1 ( 2.6 ) Δt t2 – t1 at the time in the middle of the interval : t = t2 + t1

2 ( 2.7 )

During an experiment, the displacement of an object was recorded at regular time interval. The results are shown in table 2.2. The table includes derived velocity-time data. Displacement Time ( s ) Velocity ( m/s ) Time ( s )

0 0 2 0.5 2 1 6 1.5 8 2 10 2.5

18 3 14 3.5 32 4 18 4.5 50 5 22 5.5 72 6

Table 2.2 : Displacement-time and velocity-time measurements The following example shows how a typical set of data is derived. The displacement changes from 8 m to 18 m.



Now the relation gives the velocity:

Velocity = Δs = s2 – s1

Δt t2 – t1 = 18 m – 8 m 3 s – 2 s = 10 m/s

at time : t = t2 + t1 2 = 3 s + 2 s 2 = 2.5 s Displace

80 60 40 20

ment ( m)

0 1 2 3 4 5 6 Time ( s )

Figure 2.6 : A displacement-time graph of accelerated motion.

Just as we can display the changing position of a motion body on a position-time diagram (figure 2.6), so too can we display the variation in velocity using a velocity –time diagram (figure 2.7). In this case, the graph is linear. This indicates that the velocity is changing at a constant rate with time. This means that the object is accelerating at a constant rate. Acceleration is defined as the rate of change of velocity, so the slope of the velocity-time diagram gives the acceleration of an object: a = Δv = v2 – v1

Δt t2 –t1 = 20 m/s – 2 m/s 5.0 s – 0.5 s

= 4.0 m/s2



30 10 0 1 2 3 4 5 6

Velocity ( m/s )

Time ( s )

20

Figure 2.7 : The velocity-time graph derived from the displacement-time data.

2.2.5 Displacement : the area under a velocity – time graph. Velocity ( m/s ) 4 3 2 1 0 1 2 3 4 5 6 7 8 time ( s )

uniform velocity

uniform acceleration Displacement (A)

Displacement (B)

Figure 2.8: Displacement is the area under a velocity –time graph Displacement from t = 0 s to t = 8 s is : s = Area = A + B s = ( 4 m/s x 4 s ) + ( ½ x 4 m/s x 4 s ) = 16 m + 8 m = 24 m



Example 6 : Determine the acceleration and displacement from the velocity-time graph shown in figure 2.9

Velocity (v)

40

30 20

10 0 2 4 6 8 10 12 14 16 18 20 22 Time (s) Figure 2.9: Velocity-time graph Solution : We can divide the graph into three sections. The first section is from time t = 0 s to time t = 6 s. The second section is from t = 6 s to t = 14 s, and the third is from t = 14 s to t = 22 s Section 1 Section 2 Section 3s = ½ x 40 m/s x 6 s s = 40 m/s x 8 s s = ½ x 40 m/s x 8 s = 120 m = 320 m = 160 m a = 40 m/s – 0 a = 0 m/s² a = 0 – 40 m/s 6 s – 0 s 8 s – 0 s = 6.7 m/s² = -5.0 m/s² The displacement from t = 0 s to t = 22 s is : s = 120 m + 320 m + 160 m = 600 m 2.2.6 Circular motion

To describe circular motion, we make use of angular quantities, such as angular displacement, angular velocity, and angular acceleration. These arcs defined in analogy to the corresponding quantities in linear motion:



Angular displacement (θ)

To understand what is meant by angular displacement, consider the rotating disk shown in figure 2.10. If point A rotates to point B as the disk turns on its axis, the angular displacement is donated by the angle θ.

To calculate an angular displacement in radians, we use the formula;

θ = λ ……………. ( 2.8 ) r

λ

A

B θ r

Figure 2.10 : Angular displacement θ is indicated by the shade portion of the disk.

In a complete circle there are 360°, which of course must correspond to an arc length equal to the circumference of the circle, λ = 2πr Thus θ = λ = 2πr r r = 2π rad ……………. in a complete circle. So: 360° = 2π rad = 1 rev. ( 2.9 ) One radian is therefore; 360˚ ≈ 360˚ ≈ 57.3˚

2π 6.28

1 rad = 57.3°



Example 7 : If the length is 6 cm and the radius is 10 cm, find the angular displacement in radians, degrees and revolutions. Solution; From the equation 2.8: Angular displacement in radian is; θ = λ r

= 6 cm 10 cm

= 0.6 rad Converting to degrees yields: θ = (0.6 rad ) 57.3° = 34.4 º

1 rad and since 1 rev = 360º θ = ( 34.4º ) 1 rev = 0.0956 rev 360 º Example 8 : A point on the edge of a rotating disk of radius 8 m moves through an angle of 37º. Compute the length of the arc described by the point. Solution; Convert 37º to radian units :- θ = ( 37º ) 1 rad = 0.646 rad. 57.3º The arc length is given by : λ = r θ = ( 8 m ) ( 0.646 rad ) = 5.17 m



Angular Velocity Thus the average angular velocity ( denoted by ω, the Greek lowercase letter omega ) is defined as : ω = Δθ (2.10) Δt Where Δθ is the angle through which the body has rotated in the time, Δt. Angular velocity is generally specified in radians per second ( rad/s ). We can relate the angular velocity ω to the frequency of rotation, ƒ where by frequency mean the number of complete revolution ( rev. ) per second. One revolution (of a wheel) corresponds to an angle of 2π rad/s. Hence, in general, the frequency, ƒ is related to the angular velocity, ω by: ƒ = ω 2π ω = 2πƒ ( 2.11) where f (frequency) is measured in revolutions per second ( rps ) or Hertz (Hz). Example 9 : The platter of the hard disk of a computer rotates at 5400 rpm (revolution per minute). What is the angular velocity, ω of the disk? Solution ; The angular velocity ω = 2πƒ = (2π rad/rev) 5400 rev/min 60 s/min = 570 rad/s Angular acceleration Angular acceleration, in analogy to ordinary linear acceleration, is defined as the change in angular velocity divided by the time required to make this change. The angular acceleration (denoted by α, the Greek lowercase letter alpha ) is defined as. α = ωf – ωi Δt = Δω ..................................... ( 2.12 ) Δt Angular acceleration α will be expressed as radians per second squared (rad/s²).



Example 10: A centrifuge rotor is accelerated from rest to 2000 rpm in 5.0 min. What is its average angular acceleration? Solution ; To calculate α = Δω/Δt we need the initial and final angular velocities. The initial angular velocity; Start from rest 0 rpm f = 0 ωi = 2πf = 0. The final angular velocity is; ωf = 2πf = ( 2π rad/rev ) 2000 rev/min 60s/min = 210 rad/s Then, since α = Δω/Δt and Δt = 5.0 min = 300 s, we have α = ωf – ωi Δt = 210 rad/s – 0 300 s = 0.7 rad/s² Centripetal forces The inward force necessary to maintain a uniform circular motion is defined as the centripetal force. Fc = m ac = mv² = 4Π2f2mr (2.13 ) r Where m is the mass of an object moving with a speed v in a circular path of radius R. The SI units for mv²/R are N. For problems in which the rotational speed is expressed in terms of the frequency, the centripetal force can be determine from :

ac = v2

r

Fc = mv² r Where; v = rω So; Fc = m r2 ω2

r Fc = m r ω2 ( 2.14 ) Where r is radius and ω is angular velocity.



2.2.7 Projectile Motion To examine the more general motion of objects moving through the air in two dimension near the earth’s surface such as a golf ball and speeding bullets. y vy = 0 at this point v vy vx vo vx vy vyo vx x 0 vxo a = g vy

Figure 2.11: Projective is fired with initial velocity Vo at angle θ to the horizontal. From the figure above a vector of Vo making an angle θ with the x axis has components. Vx0 = V0 cos θ ( 2.15 ) Vy0 = V0 sin θ ( 2.16 ) Kinematic Equations for Projective Motion. ( y-axis positive upward ax = 0, ay = -g = -9.80 m/s² ) Horizontal motion (x) Velocity, Vx = Vx0 ( 2.17 ) Range, R = Vx0 t ( 2.18 ) Vertical motion (y) Velocity, Vy = Vy0 – gt ( 2.19 ) Time travel, Ttravel = 2 x Vy0 ( 2.20 ) g Time at maximum height, Time max, Tmax = Vy0 ( 2.21) g



Velocity at particular time : V = √ Vx2 + Vy2 V = √ Vx2 + ( Vy0 – gt ) 2 ( 2.22 )

Example 11: A football is kicked at an angle θ = 37º with a velocity of 20 m/s as shown in figure below. Calculate: Solution; a) Component x and y of the initial velocity

37º

20 m/s

R

Vx0 = V0 cos θ = (20.0 m/s) (cos 37º) = (20.0 m/s) (0.799) = 16.0 m/s Vy0 = V0 sin θ = (20.0 m/s) (sin 37º) = (20.0 m/s) (0.799) = 12.0 m/s b) Time at maximum height tmax = Vyo g = 12.0 m/s 9.81 m/s² = 1.22 s c) The time travel before the football hit the ground. t travel = 2 x Vyo g = 2(12 m/s) 9.81 m/s² = 2.44 s d) How far away it hits the ground (Range) R = Vx0 t = (16.0 m/s) (2.44 s) = 39.14 m



Task 1 1. A car is observed at two different positions as indicated on the diagram below. The car is

observed to follow the path indicated. a) What is the distance travelled by the car? b) What is the car’s displacement?

0 1 2 3 4 5 6 7 8 km

2. A car travels from A to B by the path indicated in the diagram. The journey takes 25

minutes and the odometer indicates that the trip was 11.2 km.

A B 0 1 2 3 4 5 6 7 8 km

a) What is the distance travelled by the car? b) What is the car’s displacement going from A to B? c) Calculate the average speed of the car? d) Calculate the average velocity of the car?

Task 2 1. What must be your average speed in order to travel 230 km in 3.25 h? 2. A car travels 86 km. If the average speed was 8 m/s, how many hours were required

for the trip? 3. A bird flies at average speed of 25 km/h. how long does it take to fly 15 km? 4. Sound travels through the air at an average speed of 340 m/s. A boy drops a rock

from a bridge to the water 20 m below. How many second will be required for the sound of the splash to reach the boy’s ear.

5. A transfer truck traveled 640 miles on a run from Atlanta to New York. The entire trip

took 15 h, but the driver made two 30 min stops for meals. What was the average speed of the: - a) trip? b) driver’s driving speed?

6. A dump truck gets about 9 miles on a gallon of fuel, which is priced at a dollar a

gallon. What will be the cost of driving this truck for 2 h if its average speed is 30 mi/h?

7. An elevator is lifted at a constant speed of 40 ft/s. How much time is required for the

elevator to be lifted 200 ft?



8. Two cities are 2000 km apart. What must be the average speed of a light plane if it is

to make the trip in 10 h? Task 3 1. An object travelling in a straight line undergoes a change in velocity of 4 m/s in 2.0 s.

What is its acceleration? 2. A bicycle accelerates from rest and attains a velocity of 21 m/s in 3.5 s. Assuming that

the acceleration was constant and that motion is in a straight line, determine the magnitude of the bicycle’s acceleration.

3. A car moving at 30 m/s in a straight decelerates at a uniform rate of 2 m/s². How long

will it take to stop? 4. An object accelerates from rest in a straight line with a = 5 m/s² until its velocity is 60

m/s. What is the time taken to reach a velocity of 60 m/s? 5. The car decelerates from 100 km/h to zero in 20 s. Find the acceleration of the car

during this time interval. 6. An arrow leaves the bow 0.5 s after being released from a cocked position. If it has

reached a speed of 40 m/s in this time, what was the average acceleration? 7. An automobile accelerates from the speed of 50 km/h at the rate of 4 m/s2 for 3 s..

What is its speed at the end of the 3 s interval? 8. A car accelerates from rest to 12 m/s in 6.0 s. What was its acceleration in m/s2? 9. A car accelerates from rest at 5.0 m/s² for 4.0 seconds, and then accelerates at 2 m/s²

for a further 5.0 seconds. The car then maintains constant velocity for 3.0 seconds and then brake uniformly until stop in a further 10.0 seconds. The motion is in a straight line on a level surface.

a) Draw the velocity – time graph for the motion of the object over 22.0 second interval.

b) Determine the acceleration during the section when the brakes were applied. c) Find the displacement at times t = 4.0 s, 9.0 s, 12.0 s and 22.0 s respectively.



Task 4 1. What are the following angles expressed in radians:

(a) 30º (b) 57º (c) 90º (d) 360º (e) 420º

2. Convert the following radians into degrees:

(a) 6.28 rad (b) 0.03 rad (c) 2 x 10 ³ rad (d) 5.12 rad (e) 4π rad (f) π/4 rad

3. Find the following angular displacements in revolution. (a) 180º (b) 500º (c) 6.28 rad (d) 314.2 rad (e) 0.742 rad

4. (a) What angle in radians is subtended by an arc that has length 1.8 m and its part of a

circle of radius 1.2 m? (b) Express the same angle in degrees.

5. The angle between two radii of a circle is 0.62 rad. What are lengths is subtended if the

radius is 2.4 m.

Task 5

1. An electric drill makes 2000 revolutions per minute. What is its angular velocity? 2. A car is traveling at constant speed around a circular track. It makes one revolution in 45

s. What is its angular velocity? 3. A grinding wheel rotates at 1800 rpm. Calculate its angular velocity in rad/s. 4. Find angular velocity for each rotations:- i) 200 rpm ii) 2050 rpm iii) 2500 rpm



Task 6 1. A centrifuge accelerates from rest to 1500 rpm in 220 s. What was its angular

acceleration? 2. A wheel accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. Calculate its angular

acceleration. 3. A 4 kg ball is swung in a horizontal circle by a cord 2 m long. What is the tension in the

cord if the period is 0.5 s? 4. Two 4 kg object rotated about the center axis at 12 rev/s, as shown below, what is the

resultant force acting on each weight? 0.2 m 0.2 m 4.0 kg 4.0 kg

5. A drive shaft 60 mm in diameter rotates at 9 rev/s. What is the centripetal acceleration at

the surface of the shaft?

6. An electron revolves in an orbit about the nucleus of an atom. If follows a circular path of radius 6 x 10-11 m. If the mass of the electron is 9.11 x 10-31 kg and its linear speed is 3.2 x 106 m/s, compute the centripetal acceleration and the centripetal force.

Task 7

1. A projectile is fired with an initial speed of 75.2 m/s at an angle of 34.5º above the horizontal on a long flat firing range. Determine:-

a) The total time in the air. b) The total horizontal distance covered. c) The velocity of the projectile 1.50 s after firing. 75.2 m/s 34.5° P



2. A projectile is shot from the ground level with an initial speed of 105 m/s at an angle of

37.0º with the horizontal as shown in the figure below.

a) Determine the time taken by the projectile to hit point P at the ground levels. b) Determine the range X of the projectile. c) Find vertical components of its velocity at time 7.0 s. d) Find magnitude of the velocity at time 7.0 s.

3. A football is kicked at ground level with a speed of 20.0 m/s and at an angle of 37.0º to

the horizontal. How much later does it hit the ground? 20 m/s 370

X

105 m/s

37º P

X

Preparatory Course Royal Malaysian Air Force Motion and Force Physics 2.3 - H.0 - 1 ________________________ Apprentice Course – Mechanic ________________________


2.3 MOTION AND FORCE 2.3.1 Force Defining force Newton’s second law be rearranged to express the relation between the observable quantities-mass and acceleration-and the less tangible idea of force:- Force = mass x acceleration F = m a ............................ (3.1)

In describing the magnitude of a force we therefore use SI units that are the result of combining the units of mass (kg) and acceleration (m/s²). The SI units of force are therefore kg m/s². The SI unit of force is called the Newton (N) and it is defined in terms of Newton’s second law: ☢ 1 Newton is the force necessary to give a mass of 1 kg an acceleration of 1 m/s²

(1 N = 1 kg. m/s2). 2.3.2 Vector Forces and Force. Vector A quantity that has direction as well as magnitude is a vector quantity. On a diagram, an arrow represents each vector. The arrow is always drawn so that it points in the direction of the vector represents. The length of the arrow is drawn proportional to the magnitude of the vector. Force is one of the examples of vector quantity. Addition of vector

Because vectors are quantities that have direction as well as magnitude, they must be added in a special way. There are four techniques of vectors addition. a) Adding vectors that are in same direction. For adding vectors that is in the same direction and same axis, we can use simple arithmetic. Example 1: x- axis F1 F2 F1 So, the Resultant Force is ; FR F2 Or,

F1 F2 F1

F2 So, the Resultant Force is ; FR



Example 2: y-axis F2

So, the Resultant Force is ;

F1 F2 F1 FR Or,

F1 F2 F1 FRSo, the Resultant Force is ;

F2 b) Adding vectors that are in opposite direction. Example 3: x-axis F1 F1 So, the Resultant Force is ; F2 F2

FR Or; F1 F1 F2

So, the Resultant Force is ; F2 FR



Example 4: y-axis F1 F1 F2 F2 FR So, the Resultant Force is ; Or; F1 F1 FR F2 F2 So, the Resultant Force is ; c) Adding vector that are not in same axis. Simple arithmetic cannot be used if the vectors are not along the same line. There are two methods to find the resultant vector ;

Calculation Draw vector diagram (in scales)

Calculation

When two vectors are given in different axis, we can calculate the resultant force.

Fy + = Fy FR Fx

θ Fx



By using Theorem Pythagoras;

FR = √ Fx² + Fy² (3.2) And its angle, θ; Tan θ = Fy Fx θ = tan-1 Fy (3.3) Fx

When a given vector is in angle, we can resolve the vector into two components,

which is in x-axis and y- axis. = Fy F FR

θ θ Fx Where, from Theorem Pythagoras; Fx = F cos θ (3.4) Fy = F sin θ So, when adding two or more vector which is in angle, we need to resolved each vector to its component each. Example 5: What is resultant Force, FR from the F1 and F2. FR F1 F2 F1 F2

θ + θ = θ θ



Solution; 1. First step; resolved each vector.

Resolved F1; Resolved F2;F1x = F1 cos θ F2x = F2 cos θ F1y = F1 sin θ F2y = F2 sin θ

2. Second step; add each vector component at the same axis.

FTx = F1x + F2x

= F1 cos θ + F2 cos θ FTy = F1y + F2y

= F1 sin θ + F2 sin θ 3. Resultant is calculated from FTx and FTy by using e.q 3.2;

FR = √ FTx ² + FTy²

4. The resultant angle , θ is calculated by using e.q 3.3; Tan θ = FTy FTx θ = tan-1 FTy FTx

Draw vector diagram (drawing must in scale) There are two methods to draw vector diagram; a) Tail- to-Tip F1 + F2 = FR F2 F1 b) Parallelogram (tip-to-tip) F1 + F2 = F2 FR F1



Force and Free Body Diagrams. Newton’s second law tells us that the acceleration of an object is proportional to the net force acting on the object. The net force, as mentioned earlier, is the vector sum of all forces acting on the object. For example, two forces of equal magnitude ( 100 N each ) are shown acting on an object at right angles to each other. We can see that the object will move at a 45° angle. The Pythagorean theorem tells us that the magnitude of the resultant force is (e.q 3.2) :- FR = √ Fx² + Fy²

FR = √ (100 N) ² + (100 N) ² = 141 N

(a) (b) Figure 3.1: a) Two forces, F1 and F2, act on an object. b) The sum or resultant of F1 and F2 is FR. Example 6: Calculate the sum of the two forces acting on the boat show in figure 3.2 (a)

F2 = 100 N

F1 = 100 N 45°

Figure 3.2(a)

45 °

37 °

F1 = 40 N

F2 = 30 N

FR = √ F1 2+ F22

F1

F2



Solution; y F1x F2x Figure 3.2(b) Figure3.2(c) These two forces are shown resolved in fig. 3.2(b). We add the forces using the method of compounds of F1 are: F1x = F1 cos 45° = (40 N)(0.707) = 28.3 N F1y = F1 sin 45° = (40 N)(0.707) = 28.3 N The components of F2 are: F2x = + F2 cos 37° = +(30 N)(0.799) = 24.0 N F2y = -F2 sin 37° = -(30 N)(0.602) = -18.1 N F2y, is negative because it points along the negative y axis. The components of the resultant force are fig 3.2(c): FTx = F1x + F2x = 28.3 N + 24 N = 52.3 N FTy = F1y + F2y = 28.3 N – 18.1 N = 10.2 N To find the magnitude of the resultant force, we use the Pythagorean theorem: FR = √ FTx ² + FTy²

= ( 52.3 N)² + ( 10.2 N)² = 53.3 N The only remaining question is the angle θ that the net force FR makes with the x-axis. We use: tan θ = Fy = 10.2 N = 0.195 Fx 52.3 N

and tan-1 (0.195) = 11°

F1y

F2y F2

F1

x.

Fx = F1x + F2x

θ

FR

y

x

Fy = F1y + F2y



2.3.3 Newton’s First Law of Motion Newton’s first law states : Every object continues in a state of rest or moving at constant speed in a straight line unless in compelled to change from that state by force acting upon it. Or, put more simple: If all external forces are balanced, an object will remain at rest or moving with constant velocity.

Since velocity is a vector, this means that neither the magnitude of the velocity nor its direction may change. 2.3.4 Inertia and mass.

All objects resist your attempt to change their velocity. You have to exert a force to start a stationary body moving. If the object is already moving you have to exert a force to increase the velocity or change the object’s direction. It is noticeable that it is harder to push a stationary car than a billiard ball. This resistance exists even in free space where there are no friction or drag forces. Sir Isaac Newton called this resistance to a change in velocity Inertia. Large massive bodies contain more matter than bodies of similar composition but smaller volume. Massive bodies resist a change in velocity more than light bodies and are said to have greater inertia.

Since this inertia is fundamental in understanding the motion of all things in our universe, we can describe objects by their inertia or inertial mass or, more simply, mass.

In what other situations do you describe thing or people as having inertia? During the 17th and 18th centuries, Newton summarized and developed the results of experiments performed over several centuries. His findings are known as ‘Newton’s laws of Motion’. 2.3.5 Newton’s Second Law of Motion. The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object. This is Newton’s second law of motion. As an equation, it can be written a = ΣF m where : a acceleration m mass ΣF net force (vector sum of all forces acting on the body).



We arrange this equation to obtain the familiar statement of Newton’s Second Law (e.q 3.1). ΣF = m a

Newton’s second law relates the description of motion to the cause of motion force. It is one of the most fundamental relationships in Physics. From Newton’s second law can make a more precise definition of force as an action capable of accelerating an object.

Example 7: If the force is given 2.0 N along the x axis and the mass is 0.5 kg and the acceleration will then automatically come our in m/s² when Newton’s second law is used : Solution; a = ΣF m

= 2.0 N 0.5 kg

= 4.0 m/s²

Example 8: Estimate the net force needed to accelerate a 1000 kg car at 5 m/s² Solution; ΣF = ma

= (1000 kg ) (5 m/s²)

= 5000 N

2 kg

2 N 5 N Figure 3.3:



Example 9: From the figure above find acceleration of the 2 kg object. Solution; a = ΣF m

= 5 N + (-2 N) 2 kg

= 3 N 2 kg

= 1.5 m/s²

2.3.6 Newton’s Third Law of Motion. Newton said, two bodies must be treated on an equal basis. The hammer exerts a force on the nail, and the nail exerts a force back on the hammer. This is the essence of Newton’s third Law of Motion: - Whenever one object exerts a force on a second object, the second exerts an equal and opposite force on the first. This law is sometimes paraphrased as “ to every action there is an equal and opposite reaction”. This is perfectly valid. But to avoid confusion, it is very important to remember that the “action” force and the “reaction” force are acting on different objects. Example 10:

What exerts the force on a car? What makes a car go forward? Solution; A common answer is that the engine makes the car move forward. But it is not so simple. The engine makes the wheels go around. But what good is that if they are on slick ice or mud? They just spin. A car moves forward due to the friction force exerted by the ground on the tires, and this force is the reaction to the force exerted on the ground by the tires. 2.3.7 Weight – The Force of Gravity and The Normal Force.

Galileo claimed that the objects dropped near the surface of the Earth would all fall with the same acceleration, g, if air resistance can be neglected. The force that gives rise to this acceleration is called the force of gravity. We now apply Newton’s second law to the gravitational force, and for the acceleration, a, we use the downward acceleration due to gravity, g. Thus the force of gravity on an object, Fg, whose magnitude is commonly called its weight, can be written as:- Fg = mg Weight = force of gravity ……… (3.5)

W = Fg



The direction of this force is down toward the center of the Earth. In SI units, g = 9.81

m/s2 = 9.81 N/kg, so the weight of 10 kg mass on Earth is 10 kg x 9.81 m/s2 = 98.1 N.

Fp ( 40N ) a

mg Figure 3.4 Example 11: What happens when a person pulls upward on the box with a force equal to, or greater than the box’s weight, say Fp = 100 N rather than the 40 N shown in figure 3.4. Solution;

The net force is now Σ Fy = FN – mg + Fp = FN – 98.1 N + 100 N and if we set this equal to zero, we would get FN = -1.9 N. This is nonsense, since the negative sign implies FN points downward, and the table surely cannot pull down on the box. The least FN can be is zero, which it will be in this case. The box accelerates upward since the net force is not zero; it is upward. Σ Fy = Fp – mg = 100 N – 98.1 N = 1.9 N So the box moves upward with an acceleration of magnitude:- ay = ΣFy m

= 2 N 10 kg = 0.19 m/s2



2.3.8 Friction Friction exists between two solids surfaces. There are two types of friction. 1. Static friction overcome to start motion. 2. Kinetic friction overcome to maintain motion. FN

Ffr

F

w = mg

Figure 3.5

Static friction V = 0

F (horizontal force acting on the object )

Figure 3.6

Ffr

1. The net force is zero on an object doesn’t move. 2. Force of static friction exerted by the surface keeping the object from moving. 3. No sliding and no moving (ΣF = 0 ). 4. The object is in equilibrium. 5. Force of static friction is given by :

F s = µ S FN (3.6) = µ S W

Where : Fs is static friction

F N is normal force W is weight µ S is coefficient of static friction



Kinetic friction

V, a = 0 ( constant velocity )

F

Ffr

rough surface

Figure 3.7

1. When the object starts to move the kinetic friction take over. 2. When a body is in motion along a rough surface, the force of kinetic acts opposite to the

direction of the body’s velocity. 3. The net force is zero on an object when its moving in constant velocity. The object is in

equilibrium (ΣF = 0 ) 4. Force of kinetic friction is given by

F K = µ K F N (3.7)

= µ K W Where; FK is kinetic friction

FN is normal force

W is weight

µK is coefficient of kinetic friction Applications involving Friction, Inclines

We now discuss some examples moving on an incline such as a hill or a ramp. Solving problems is usually easier if we choose the x - y coordinate system so that the x-axis points along the incline (either up the incline, or down the incline) and the y-axis perpendicular to the incline, as shown in figure below. (a)

θ

θ

W

Fy Fx

Ffr V, motion, a = 0

Figure 3.8: Forces on an object sliding down an incline plane with constant velocity.



b)

Fx

Ffr

θ

θ

Fy

W = mg

Fx = Ffr

Figure 3.9 From the diagram above weight force can be resolve into two component x and y. Fx = W sin θ (if a = 0) (3.8) Fy = W cos θ From the diagram also shown that the friction force is equal to force in component x if the object is in equilibrium. Ffr = Fx (3.9) And the normal force FN is equal to the component y.

FN = Fy So;



Example 12: The 2 kg object has just begun descending the 30º slopes. Figure 3.10

30º

a) Draw the free-body diagram

30º

W

FN

Fx

Ffr

30º Fy

b) Friction force exert by slope.

Ffr = Fx = mg sin θ

= ( 2 kg ) ( 9.81 m/s² ) sin 30º = 9.81 kg m/s² = 9.81 N

c) Calculate coefficient of static friction

Ffr = µs FN µs = Ffr = Fx FN Fy = W sin θ

W cos θ = sin 300_ cos 300

= 0.577



2.3.9 Moment Moment is defined as the product of force multiplied by moment arm. Moment = Force x moment arm M = F x d ..................................... (3.10)

d

F

Figure 3.11: examples of moment arms.

The perpendicular distance from the axis of rotation to the line of action of a force is called the moment arm (d) of that force. Example 13:

d = 0.8 m

F = 20 N

Figure 3.12

An ordinary situation such as the door in figure 3.12 (looking down from the above) you will find that the moment is : Moment = Force x moment arm

= 20 N x 0.8 m

= 16 Nm so moment produced on the door is 16 Nm.



Example 14:

A force of 20 N is exerted on a cable wrapped around a drum, which has a diameter of 0.12 m. What is the torque produce about the center of the drum?

0.12 m

20 N

The tangential force exerted by a cable wrapped around a drum. Figure 3.13

Solution;

Notice that the line of action of the 20 N force is perpendicular to the diameter of the drum. The moment arm is, therefore, equal to the radius of the drum (0.12 m/2). The moment is given by : M = F d = - ( 20 N ) ( 0.06 m ) = -1.2 Nm Example 15:

A mechanic exerts a 30 N force at the end of a 0.2 m wrench, as shown in figure 3.14. If this pull makes an angle of 60º with the handle, what is the torque produced on the nut?

30 N 60° Figure 3.14: 0.2 m



Solution;

The moment arm d would be calculated from sin 60º = r / 0.2 m, or : d = ( 0.2 m ) ( sin 60º ) = 0.17 m Moment = F d = ( 30 N ) ( 0.17 m ) = 5.20 Nm 2.3.10 Moment of couples

If we are to ensure that rotational effects are also balanced, we must require that there be no resultant moment. The second condition for equilibrium is : The algebraic sum of all the moment about any axis must be zero.

ΣM = 0 ΣM = M1 + M2 + M3 + ………….. = 0 (3.11)

= F1 d1 + F2 d2 + F3 d3 + ……. = 0

The second condition for equilibrium simply tells us that the clockwise moments are exactly equal in magnitude to the counter clockwise moment. Example 16: Determine the measure ‘λ’ to the balance figure below :

axis

ℓ 2 m

3 m

10 N 5 N 4 N

Figure 3.15:



Solution;

If the figure balances, the sums all the moment become zero. ΣM = 0 0 = M1 + M2 + M3 0 = F1 d1 + F2 d2 + F3 d3 0 = ( 10 N ) ( ℓ ) + ( -5 N) ( 2 m ) + ( - 4 N ) ( 3 m ) 0 = 10 ℓ N – 10 Nm – 12 Nm 10 λ N = 22 Nm λ = 22 Nm 10 N = 2.2 m Task 1 1. Determine the resultant force for the following forces below.

a) 10 N 20 N

b) 11.2 N 5 N c) 14 N 20 N d) 5 N 10 N 5 N e) 50 N 60 N 20 N 10 N



2. Determine the resultant force for force Fx and Fy as drawn below. Fy = 22 N Fx = 18 N 3. A force of 50 N acting on north and another force 120 N acting on East. What is the

combination force acted at the object? Task 2 1. Resolve the following forces into x and y component : i)

30º

60N

ii)

60º

30N

iii)

50º

100N iv)

47º 50N



2. Calculate the resultant force and direction of the two-combination force?

(refer to figure Q1) a. i + ii b. i + iii c. i + iv d. ii +iii e. ii +iv f. iii + iv

3. Calculate the resultant force and direction of the two forces below.

37º

42º

112N

144N

4. Calculate the sum of the forces acting on the object show in figure below.

30º

60º

15N

19N

5. From the figure above, find the direction for the resultant force. Task 3

F2

30º

10 N

F1

Figure 3.16:



1. A wood block weighted 10 N put as described in the above diagram. Determine the component forces of F1 and F2

F1

100 N

F2

25º

2. Figure above shown a student pulls the block with a rope. If the tension of the rope is

100 N, what is the force component of F1 (vertical) and F2 (horizontal)?

20º

m = 60 kg

Figure 3.17:

3. The object in figure: 3.17 have just begun descending the 20º slope.

a) Sketch the free-body diagram. b) If the object has mass of 60 kg, calculate the Fx and Fy .

●

70º

Figure 3.18: 4. A lawn – mover shown in the following figure 4 is pushed with a force of 200 N.

a) sketch a free – body diagram b) Find Fx c) Find Fy



Task 4

30 N

3 m

50º

Figure 3.19:

1. A 30 N force acts on the end of the 3 m lever a shown. Determine the moment of the force about 0. (Figure 3.19)

Figure 3.20: 2. A 100 N vertical force is applied to the end of the lever that is attached to a shaft at 0.

Determine (Figure 3.20) a) The moment of the 100 N force about 0 b) How far from the shaft a 240 N vertical force must act to create the same

moment about 0.

Figure 3.21:

O

A

100 N

24 m

60º O

d = 0.6 m

40º

F = 20 N

O



3. Determine the moment of the 20 N force about 0. (Figure 3.21)

0

0.12 m

50 N

Figure 3.22:

4. Find the moment about O which are used to tighten the bolts.(Figure 3.22) Task 5 1. Determine the measure ‘L’ to the balance shown in the following figure below: 8 N 12 N

2 m L

2. Determine the measure ‘d’ to the balance shown in the following figure below :

2 m 2 m d 16 N 6 N 4 N 3. Determine the weight ‘A’ to the balance shown in the following figure :

10 N A

4 m 2 m



4. a) Check whether the bar shown in the following figure is in equilibrium or not.

1 m 2 m

3 m

8 N 3 N 2 N

b) If not, determine the distance ‘x’ where the weight that exerts 8 N must be placed, in order to get balance.

Task 6 1. What is the mass of an object weighing 490 N? 2. A small stone of mass 129 g is falling with a constant speed of 19 ms-1 into water. What is

the force of the water on the stone?

Figure 3.23:

3. An object of mass 100 kg is suspended from a beam by a rope as shown in figure 3.23. Calculate the tension of the rope?

4. An astronaut of mass 75 kg in a spacesuit of mass 20 kg lands on Mars where its gravity

is 3.8 ms-2 . What would be the combined weight of the astronaut and spacesuit in mars? T1 8.0 kg T2 5.0 kg T3 1.0 kg

Figure 3.24:



5. Three objects of mass 8.0 kg, 5.0 kg and 1.0 kg connected to the ceiling by three pieces of light inelastic string, as shown in figure 3.24. The tensions in each string are respectively T1, T2 and T3. a. Calculate the value of the T1, T2 and T3. b. If the lower string supporting the 1.0 kg mass is cut. What will the new values of T1

and T3. Task 7 1. Discriminate two types of friction, for the case of sliding friction. 2. A force of 40.0 N is required to start at 5.0 kg box moving across a horizontal floor.

What is a coefficient of static friction between the box and the floor? 3. If the coefficient of kinetic friction between a 35 kg crate and floor is 0.30, what horizontal

force is required to move the crate at the steady speed across the floor? Task 8 1. Force of 150 N act on a 100 kg object as shown (figure 3.25 ). If we neglect friction, what

is the resulting acceleration of the mass? 150 N 100 kg Figure 3.25: 2. How much force is needed to get an acceleration 5 m/s2, act to the object with mass 100

kg? 3. Two forces act on a 200 kg mass as shown (figure 3.26). If we neglect friction, what is

the resulting acceleration of the mass?

20 N 60 N

200 kg

Figure 3.26:

4. A cart is pushed with a force of 40 N and pulled with a force of 60 N in the same direction, as shown in figure 3.27. If the cart mass is 200 kg, and we neglect friction, what is the resulting acceleration?

40 N 60 N

200 kg

Figure 3. 27:



Task 9 1. Two masses rest on frictionless table as shown in Figure 3.28. A 600 N forces push on

the 50 kg mass. a. What is acceleration of the 10.0 kg mass? b. What is the net force on the 50.0 kg mass? c. What force does the 50.0 kg mass exert in the 10.0 kg mass?

10.0 kg 50.0 kg 600 N Figure 3.28:

2. Three masses rest on a frictionless table. A 30 N force pushes on the 4.0 kg mass as shown in figure 3.29. a. What is the acceleration of the 1.0 kg mass? b. What is the net force on the 5.0 kg mass? c. What force does the 5.0 kg mass exert on the 4.0 kg mass?

1.0 kg 5.0 kg 4.0 kg 30 N

Figure 3.29:

3. Three blocks of steel rest on an ice rink. A 20 N force pushes on the 3.0 kg blocks as

shown in figure 3.30. a. What is the acceleration of the 3.0 kg block? b. What is the net force on the 5.0 kg block? c. What force does the 5.0 kg block exert on the 2.0 kg block?

2.0 kg 5.0 kg 3.0 kg 20 N

Figure 3.30: Task 10 1. A force of 150 N at an angle 30 º from ground acts on a 100 kg object as shown (figure

3.31 ). If we neglect friction, and block accelerates horizontally. a. What is the horizontal component of Force? b. The resultant acceleration of the mass?

100 kg 250 N

30º Figure 3.31:



2. A 2.5 kg block is pulled along a smooth surface by a force at an angle of 50º to the

horizontal as shown figure 3.32. The block accelerates horizontally at 2.0 ms-2. a. What is the horizontal component of Force (F)? b. What is the magnitude of Force?

F

2.5 kg 50º Figure 3.32:

3. Two forces F1 and F2, act on a 200 kg mass. F2 act at an angle 45º from the horizontal as shown (figure 3.33). If we neglect friction. a. What is the horizontal component of Force (F2)? b. What is the resulting acceleration of the mass?

200 kg F2 = 100 N F1 = 20 N 45º Figure 3.33:

4. A cart is pushed with a force of 40 N and pulled with a force of 60 N act at an angle 60º

from the horizontal as shown (figure 3.34). If the cart mass is 200 kg, and we neglect friction, what is the resulting acceleration? a. What is the horizontal component of Force 60 N? b. What is the resulting acceleration of the mass?

200 kg 60 N 40 N 60º Figure 3.34:

5. Two masses rest on frictionless table as shown in Figure 3.35. A 200 N force pushes on

the 50 kg mass act at an angle 40º from the horizontal as shown (figure 3.35). a. What is the horizontal component of Force 200 N? b. What is acceleration of the 10.0 kg mass? c. What is the net force on the 50.0 kg mass? d. What force does the 50.0 kg mass exert in the 10.0 kg mass?

200 N

10.0 kg 50.0 kg 40º Figure 3.35:

Preparatory Course Royal Malaysian Air Force Bodies of Equilibrium Physics 2.4 - H.0 - 1 _______________________ Apprentice Course – Mechanic _________________________

2.4 BODIES IN EQUILIBRIUM 2.4.1 Static – The Study of Forces in Equilibrium

Objects within our experience have at least one force acting on them (gravity) and if

they are at rest then there must be other forces acting on them as well so that the net force is zero. An object at rest on a table, for example, has two forces acting on it, the downward force of gravity and the normal force the table exerts upward on it (figure 4.1). Since the net force is zero, the upward force exerted by the table must be equal in magnitude to the force of gravity acting downward. (Do not confuse these two forces with the equal and opposite forces of Newton’s third law which act on different bodies; here both forces act on the same body ) such a body is said to be in equilibrium ( Latin for “equal forces” or “balance” ).

Normal Force

Gravity Figure 4.1: The object in equilibrium, the net force on it is zero. Example 1 Calculate the force exerted on the leg by the traction apparatus shown in figure 4.2. Assume the pulleys are frictionless.

For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved

(a) (b) Figure 4.2: Traction apparatus exerts force on a leg.

3737 °

°

196.2 N

FR

20 kg

°

°

196.2 N

3737

MIP/MECH/TN/003


Solution Tension = 20 kg x 9.81 m/s2 = 196.2 N FR = 2 x 196.2 N cos 37o

= 313.4 N acting to the right

The leg is in equilibrium, so there must be another 313.4 N force acting on the leg to keep it at rest. What exerts this force?

2.4.2 The conditions for equilibrium For body to be at rest, the sum of the forces acting on it must add up to zero. Since force is a vector, the components of the net force must each be zero. Hence, a condition for equilibrium is that ΣFx = 0, ΣFy = 0, ΣFz = 0 (4.1)

We will mainly be dealing with forces that act in a plane, so we usually need only the x and y components. We must remember that if a particular force component points along the negative x or y-axis, it must have a negative sign. Equation 4.1 is called the first condition for equilibrium. Example 2

A 90 kg weakling cannot do even one pull-up. By standing on a scale (figure 4.3), he can determine how close he gets. His best effort results in a scale reading of 23 kg. What force is he exerting? Figure 4.3: (a) a person trying to do a pull up while standing on a scale.

(b) Simple free body diagram. For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved MIP/MECH/TN/003


Solution

There are three forces acting on our non-athlete, as shown in figure 4.3.Gravity, mg = ( 90 kg ) ( 9.81 m/s² ) downward. Two upward forces which are (1) the force the bar pulls upward on him, FB (equal and opposite the force he exerts on the bar)(2) the force the scale exerts on his feet, F

B

s. At best, Fs = ( 23 kg ) (g). The person doesn’t move so the sum of these forces is zero. ΣFy = 0 FB + FB s – mg = 0 We solve for FB: B

FB = mg – FB s = (90 kg x 9.81 m/s²) – (23 kg x 9.81 m/s²) = 657.3 N

That is, he could lift himself if his mass were only 67 kg. Although equation 4.1 must be true if an object is to be in equilibrium, they are not a

sufficient condition. Figure 4.4 shows an object on which the net force is zero. Although the two forces labeled F add up to give zero net force on the object, they do give rise to a net torque that will rotate the object. We see that if a body is to remain at rest, the net moment applied to it (calculated about any axis) must be zero. Thus we have the second condition for equilibrium: the sum of the torques acting on a body must be zero

ΣM = 0 (4.2)

F

F Figure 4.4: Although the net force on it is zero. The ruler will move (rotate). A pair of equal

forces acting in opposite directions but at different points on a body (as shown here) is referred to as a couple.

This will assure that the angular acceleration, αα, about any axis will be zero. If the

body is not rotating initially (ω = 0), it will not start rotating. Equation 4.1 and 4.2 are the only requirements for a body to be in equilibrium. For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved MIP/MECH/TN/003


2.4.3 Stability and Balance.

A body in static equilibrium, if left undisturbed, will undergo no translational or rotational acceleration since the sum of all the forces and the sum of all the moment acting on it are zero. However, if the object is displaced slightly, three different outcomes are possible: 1. The object returns to its original position, in which case it is said to be in stable

equilibrium. 2. The object moves even further from its original position, in which case it is said to be in

unstable equilibrium. 3. The object remains in its new position, in which case it said to be in neutral equilibrium.


Net force (a) (b) Figure 4.5: (a) stable equilibrium, and (b) unstable equilibrium

Consider the following examples. A ball suspended freely from a string is in stable

equilibrium, for if it is displaced to one side, it will quickly return to its original position (figure 4.5(a)). On the other hand, a pencil standing on its points is in unstable equilibrium. If its CG is directly over its (figure 4.5(b)), the net force and net moment on it will be zero. But if it is displace ever so slightly – say by a slight vibrations or tiny air current – there will be a moment on it, and it will continue to fall in the direction of the original displacement. Finally, an example of an object in neutral equilibrium is a sphere resting on a horizontal tabletop. If it is placed slightly to one side, it will remain in its new position.

(a) (b) (c) Figure 4.6: Equilibrium of a refrigerator resting on a surface.

MIP/MECH/TN/003


In most situations, such as in the design of structures and in working with human

body, we are interested in maintaining stable equilibrium or balance, as we sometimes say. In general, an object that CG is below its points of support, such as a ball on a string, will be in stable equilibrium. If the CG is above the base of support, we have a more complicated situation. Consider a standing refrigerator, figure 4.6(a). If it is tipped shown in figure 4.6(b). But if it is tipped too far, (figure 4.6(c)) it will fall over. The critical is reached when the CG is no longer above the base of support.

In general, a body whose CG is above its base of support will be stable if a vertical line projected downward from the CG falls within the base of support. This is because the normal force upward on the object (which balances out gravity) can be exerted only within the area of contact, so that if the force of gravity acts beyond this area, the net moment with widest face is more stable than a brick standing on its end, for it will take more of an effort to tip it over. In the extreme cases of the pencil in figure 4.5(b) the base is practically a point and the slightest disturbance will topple it. In general, the larger the base and the lower the CG, the more stable the object. In this sense, humans are much less stable than four – legged mammals, which not only have a larger base of support because of their four legs, but also have a lower centre of gravity. The human species has had to develop special apparatus, such as certain very strong muscles, in order to deal with the problem of keeping a person upright and at the same time stable. Because of their upright position, humans suffer from numerous ailments such as low back pain due to the large forces involved.

When walking and performing other kinds of movement, a person continually shifts the body so that its CG is over the feet, although in the normal adult requires no conscious thought. Even as simple a movement as bending over requires moving the hips backward so that the CG remains over the feet, and this repositioning is done without thinking about it. To see this, position yourself with your heels and back to a wall and try to touch your toes. You won’t be able to do it without falling. Persons carrying heavy loads automatically adjust their posture so that the CG of the total mass is over their feet (figure 4.7)

: Figure 4.7: Human adjusts their posture to achieve stability when carrying loads. For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved MIP/MECH/TN/003


2.4.4 Elasticity; Stress and Strain In this section we study the effects of these forces: any object changes shape under the action of applied forces. In section 2.4.5 we will see that if the forces are great enough, the object will break or fracture.


ΔL F = mg

Lo

Figure 4.8: Hooke's Law: ΔL ∝ applied force If a force is exerted on an object, such as the vertically suspended metal rod shown in figure 2.8, the length of the object changes. If the amount of elongation, ΔL. is small compared to the length of the object, experiment shows that ΔL is proportional to the weight or force exerted on the object. This proportionality can be written as an equation: F = k ΔL. (4.3) Where; F = represents the force (or weight) pulling on the object, ΔL = the change in length, and k = proportionality constant. Equation 4.3, which is sometimes called Hooke's law, after Robert Hooke (1635-1703) who first noted it, is found to be valid for almost any solid material from iron to bone, but it is valid only up to a point. If the force is too great, the object stretches excessively and eventually breaks. Figure 4.9, shows a typical graph of elongation versus applied force. Up to a point called the proportional limit, Eq. 4.3 is a good approximation for many common ma-terials, and the curve is a straight line.

MIP/MECH/TN/003


Beyond this point, the graph deviates from a straight line, and no simple relationship exists between F and ΔL. Nonetheless, up to a point farther along the curve called the elastic limit, the object will return to its original length if the applied force is removed. The region from the origin to the elastic limit is called the elastic region. If the object is stretched beyond the elastic limit, it enters the plastic region: it does not return to the original length upon removal of the external force, but remains permanently deformed (like bending a paper clip).


Elongation, ΔL

Figure 4.9: Applied force vs. elongation for a typical metal under tension. The maximum elongation is reached at the breaking point. The maximum force that can be applied without breaking is called the ultimate strength of the material. The amount of elongation of an object, such as the rod shown in figure 4.8, depends not only on the force applied to it, but also on the material from which it k made and on its dimensions. That is the constant k in Eq. 4.3 can be written in terms of these factors. If we compare rods made of the same material but of different lengths and cross-sectional areas it is found that for the same applied force, the amount of stretch (again assumed small compared to the total length) is proportional to the original length and inversely proportional to the cross-sectional area. That is the longer the object, the more it elongates for a given force and the thicker it is, the less it elongates. These findings can be combined with Eq. 4.3 to yield Δ L = 1 x F x Lo (4.4) E A Where;

Lo is the original length of the object. A is the cross-sectional area. ΔL is the change in length due to the applied force F is the force applied. E is a constant of proportionality, elastic modulus or Young's modulus.

MIP/MECH/TN/003


Elastic Modulus, E

E (Elastic modulus @ Young's modulus) and its value depend only on the material. The value of Young's modulus for various materials is given in Table 4.1 (note; shear and bulk modulus in this Table are discussed later in this section). Because E is a property only of the material and is independent of the object's size or shape, Eq. 4.4 is far more useful for practical calculation than Eq. 4.3. Material (Solid)

Elastic ModulusE (N/m2)

Shear ModulusG (N/m2)

Bulk Modulus B (N/m2)

Iron, cast 100 x 109 40 x 109 90 x 109

Steel 200 x 109 80 x 109 140 x 109

Brass 100 x 109 35 x 109 80 x 109

Aluminum 70 x 109 25 x 109 70 x 109

Concrete 20 x 109 Brick 14 x 109 Marble 50 x 109 70 x 109

Granite 45 x 109 45 x 109

Wood (pine) (Parallel to grain) (Perpendicular to grain)

10 x 109

1 x 109

Nylon 5 x 109 Bone (limb) 15 x 109 80 x 109 Liquids Water 2.0 x 109

Alcohol (ethyl) 1.0 x 109

Mercury 2.5 x 109

Gases Air, H2, He, CO2 1.01 x 105

At normal atmospheric pressure: no variation in temperature during process.

TABLE 4.1 - Elastic Modulus From Eq. 4.4, we see that the change in length of an object is directly proportional to the product of the object's length L0 and the force per unit area F/A applied to it. It is general practice to define the force per unit area as the stress. Stress = Force = F Area A Which has units of N/m2. Also the strain is defined to be ratio of the change in length to the original length:

strain = change in length = Δ L original length LO



And is dimensionless (no units). Strain is thus the fractional change in length of the object, and is a measure of how much the bar has been deformed. Stress is applied to the material by external agents, where as strain is the material's response to the stress. Equation 4.4 can be rewritten as F = E x Δ L A Lo E = F/A = stress (4.5) ΔL/ Lo strain Thus we see that the strain is directly proportional to the stress, in the linear (elastic) region of Fig. 4.9. Example 1 A 1.60-m-long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.30cm when tightened? Solution: We solve for F in Eq. 4.5 and note that the area: Given: Lo = 1.6 cm ΔL = 0.3 cm = 0.003 m d = 0.20 cm r = d/2 = 0.1cm = 0.001m E = 2.0 x 1011 N/m2 (obtained from Table 4.1) A = π r2 = (3.142)(0.0010 m)2

= 3.1 x 10-6 m2. Then F = E x Δ L x A Lo = ( 2.0 x 1011 N/m2) (0.0030 m ) (3.1 x 10-6 m2 ) 1.60 m = 1200 N Where we obtained the value for E from Table 4-1.The strong tensions in all the wires in a piano must be supported by a strong frame.



The rod shown in Fig. 4.8 is said to be under tension or tensile stress. For not only is

there a force pulling down on the rod at its lower end, but since the rod is in equilibrium we know that the support at the top is exerting an equal upward force on the rod at its upper.

F F (a) F (b) F

Figure 4.10: Stress exists within the material.

Fig. 4.10 (a). In fact, this tensile stress exists throughout the material. Consider, for example, the lower half of a suspended rod as shown in Fig. 4.10(b). This lower half is in equilibrium, so there must be an upward force on it to balance the downward force at its lower end. What exerts this upward force? It must be the upper part of the rod. Thus we see that external forces applied to an object give rise to internal forces, or stress, within the material itself. Figure 4.11: This Greek temple, in Agrigento, Sicily, built 2500 years ago, shows the post-

and-beam construction.



Strain or deformation due to tensile stress is but one type of stress to which materials

can be subjected. There are two other common types of stress: compressive and shear. Compressive stress is the exact opposite of tensile stress. Instead of being stretched, the material is compressed; the forces act inwardly on the body. Columns that support a weight, such as the columns of a Greek temple (Fig. 4.11), or those that support the beam subjected to compressive stress. Equations 4.4 and 4.5 apply equally well to compression and tension, and the values for the elastic modulus E are usually the same.


Lo

Figure 4.12: The three types of stress for rigid bodies. Shear Modulus, G

Figure 4.12 compares tensile and compressive stresses as well as the third type, shear stress. An object under shear stress has equal and opposite forces applied across its opposite faces. An example is a book or brick firmly attached to a tabletop, on which a force is exerted parallel to the top surface. The table exerts an equal and opposite force along the bottom surface. Although the dimensions of the object do not change significantly, the shape of the object does change as shown in the figure. An equation similar to Eq. 4.4 can be applied to calculate shear strain:

Δ L = 1 x F x Lo (4.6) G A

MIP/MECH/TN/003


But ΔL, Lo and A must be reinterpreted as indicated in figure 4.12(c). Note that A is the area of the surface parallel to the applied force (and not perpendicular as for tension and compression), and ΔL is perpendicular to L0. The constant of proportionality, 0, is called the shear modulus and is generally one half to one third the value of the elastic modulus, E (see Table 4-1). Figure 4.13 illustrates why ΔL and L0 for the fatter book shifts more for the same shearing force.

(a) (b)

Figure 4.13: The fatter book (a) shifts more than the thinner book (b) with the same applied shear force.

Figure 4.14: Balance of forces and torques for shear stress.

The rectangular object undergoing shear in Fig.4.12(c) would not actually be in equilibrium under the forces shown, for a net torque would exist. If the object is in fact in equilibrium, there must be two more forces acting on it, which balance out this torque. One acts vertically upward on the right, and the other acts vertically downward on the left, as shown in Fig. 4.14. This is generally true of shear forces. If the object is a brick or book lying on a table, these two additional forces can be exerted by the table and by whatever exerts the other horizontal force (such as a hand pushing across the top of a book). For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved MIP/MECH/TN/003


Bulk Modulus, B

If an object is subjected to inward forces from all sides, its volume will decrease. A

common situation is a body submerged in a fluid; for in this case, the fluid exerts a pressure on the object in all directions, as we shall see in FS1. Pressure is defined as force per area, and thus is the equivalent of stress. For this situation the change in volume, ΔV, is found to be proportional to the original volume, V0, and to the increase in the pressure, ΔP. We thus obtain a relation of the same form as Eq. 4.4 but with a proportionality constant called the bulk modulus, B: Δ V = - 1 Δ P Vo B B = - Δ P (4.7) ΔV/ Vo

The minus sign is included to indicate that the volume decreases with an increase in pressure. Values for the bulk modulus are given in table 4.1. Since liquid and gases do not have a fixed shape, only the bulk modulus applies to them. 2.4.5 Fracture If the stress on a solid object is too great the object fractures or breaks (Fig. 4.15). Table 4-2 lists the ultimate tensile strength, compressive strength. and shear strength for a variety of materials. These values give the maximum force per unit area that an object can withstand under each of these three types of stress. They are, however, representative values only, and the actual value for a given specimen can differ considerably. It is therefore necessary to maintain a "safety factor" of from 3 to perhaps 10 or more that is. The actual stresses on a structure should not exceed one tenth to one third of the values given in the table. You may encounter tables of the "allowable stresses" in which appropriate safety factors have already been included. Figure 4.15: Fracture as a result of three types of stress. For training purposes only Rev. 0 Issued 01 Dec 06 Malaysian Aviation Training Academy All rights reserved MIP/MECH/TN/003


Material Tensile Strength

(N/m2) Compressive Strength (N/m2)

Shear Strength (N/m2)

Iron , cast 170 x 106 550 x 106 170 x 106

Steel 500 x 106 500 x 106 250 x 106

Brass 250 x 106 250 x 106 200 x 106

Aluminum 200 x 106 200 x 106 200 x 106

Concrete 2 x 106 20 x 106 2 x 106

Brick 35 x 106 Marble 80 x 106 Granite 170 x 106 Wood (pine) (Parallel to grain) (perpendicular to grain)

40 x 106

35 x 106

10 x 106

5 x 106

Nylon 500 x 106 Bone (limb) 130 x 106 170 x 106

TABLE 4-2: Ultimate Strengths of Material (force/area) As can be seen in Table 4-2, concrete (like stone and brick) is reason-ably strong under compression but extremely weak under tension. Thus concrete can be used as a vertical column placed under compression but is of little value as a beam since it cannot withstand the tensile forces that arise (see Fig. 4.16). Reinforced concrete, in which iron rods are embedded in the concrete, is much stronger (Fig. 4.17). But the concrete on the lower edge of a loaded beam still tends to crack because of its weakness under tension. This problem is solved with pre-stressed concrete, which also contains iron rods or a wire mesh, but during the pouring of the concrete, the rods or wire are held under tension. After the concrete dries, the tension on the iron is released, putting the concrete under compression. The amount of compressive stress is carefully predetermined so that when the design loads are applied to the beam, they reduce the compression on the lower edge but never put the concrete into tension.


Figure 4.16: A beam sags, at least a title (but is exaggerated here), even under its own

weight. The beam thus changes shape so that the lower portion is compressed, and the upper portion is under tension (elongated). Shearing stress also occurs within the beam.

Tension

Compression

MIP/MECH/TN/003


Figure 4.17: Steel rods waiting for concrete to be poured around them to form a new

highway. Task 1 1. Three forces are applied to a tree sapling, as shown in figure below, to stabilize it. If F1 =

282 N and F2 = 355 N, find F3 in magnitude and direction.



2. The wire band shown in figure below has a tension FT of 2.0 N along it. It therefore exerts forces of 2.0 N on the tooth (to which it is attached) in the two directions shown. Calculate the resultant force on the tooth due to the wire, Fw.

FT FT

FT 70º 70º 70º FT

70º FW y

3. What should be the tension in the wire if the net force exerted on the tooth in figure above is to be 0.75 N? Assume that the angle between the two forces is 155o rather than the 140o in the figure.

Task 2 1. A nylon tennis string on a racket is under a tension of 250 N. if the diameter is 1.00 mm.

By how much is it lengthened from its untensioned length of 30.0 cm? 2. A marble column of cross-sectional area 2.0 m2 supports a mass of 25 000 kg.

(a) What is the stress within the column? (b) What is the strain?

Task 3 1. A vertical steel girder with a cross-sectional area of 0.15 m2 has a sign (mass 2000 kg)

hanging from its end (a) What is the stress within the girder? (b) What is the strain on the girder? (c) If the girder is 9.50 m long, how much is it lengthened? (Ignore the mass of the

girder itself.)

70 kg

20 cm 18 cm



2. A string made of by unknown material hang with a mass of 70 kg shown in above figure.

If the string diameter is 2 cm. Find…… a) Stress b. strain c. Elastic modulus for the material.


Preparatory Course Royal Malaysian Air Force Thermodynamics Physics 2.5 - H.0 - 1 _________________________ Apprentice Course – Mechanic ______________________

2.5 THERMODYNAMICS

When heat is transformed into any other form of energy, or when other forms of energy are transformed into heat, the total amount of energy (heat plus other forms) in the system is constant.

This is the first law of thermodynamics, the conservation of energy. To express it another way: it is in no way possible either by mechanical, thermal, chemical, or other means, to obtain a perpetual motion machine; i.e., one that creates its own energy.

A second statement may also be made about how machines operate. A steam engine uses a source of heat to produce work. Is it possible to completely convert the heat energy into work, making it a 100% efficient machine? The answer is to be found in the second law of thermodynamics:

No cyclic machine can convert heat energy wholly into other forms of energy. It is not possible to construct a cyclic machine that does nothing but withdraw heat energy and convert it into mechanical energy.

The second law of thermodynamics implies the irreversibility of certain processes that of converting all heat into mechanical energy, although it is possible to have a cyclic machine that does nothing but convert mechanical energy into heat.

The way that the gas temperature scale and the thermodynamic temperature scale are shown to be identical is based on the microscopic interpretation of temperature, which postulates that the macroscopic measurable quantity called temperature is a result of the random motions of the microscopic particles that make up a system.

2.5.1 Temperature.

In a qualitative manner, we can describe the temperature of an object as that which

determines the sensation of warmth or coldness felt from contact with it. It is easy to demonstrate that when two objects of the same material are placed

thermal contact, the object with the higher temperature cools while the cooler object becomes warmer until a point is reached after which no more change occurs, and to our senses, they feel the same. When the thermal changes have stopped, we say that the two objects are in thermal equilibrium. We can then define the temperature of the system by saying that the temperature is that quantity which is the same for both systems when they are in thermal equilibrium.

If we experiment further with more than two systems, we find that many systems can be brought into thermal equilibrium with each other; thermal equilibrium does not depend on the kind of object used. Put more precisely,

If two systems are separately in thermal equilibrium with a third, then they must also be in thermal equilibrium with each other, and they all have the same temperature regardless of the kind of systems they are. The statement in italics, called the zeroth law of thermodynamics restated as:

If three or more systems are in thermal contact with each other and all in equilibrium together, then any two taken separately are in equilibrium with one another.

Now one of the three systems could be an instrument calibrated to measure the temperature - i.e. a thermometer. When a calibrated thermometer is put in thermal contact with a system and reaches thermal equilibrium, we then have a quantitative measure of the temperature of the system. For example, a mercury-in-glass clinical thermometer is put under the tongue of a patient and allowed to reach thermal equilibrium in the patient's mouth - we then see by how much the silvery mercury has expanded in the stem and read the scale of the thermometer to find the patient's temperature.



Thermometer

A thermometer is an instrument that measures the temperature of a system in a

quantitative way. The easiest way to do this is to find a substance having a property that changes in a regular way with its temperature. The most direct 'regular' way is a linear one:

t(x) = ax + b,

Where t is the temperature of the substance and changes as the property x of the substance changes. The constants a and b depend on the substance used and may be evaluated by specifying two temperature points on the scale, such as 32° for the freezing point of water and 212° for its boiling point.

For example, the element mercury is liquid in the temperature range of -38.9° C to 356.7° C. As a liquid, mercury expands as it gets warmer, its expansion rate is linear and can be accurately calibrated.

The mercury-in-glass thermometer illustrated in the above figure contains a bulb filled with mercury that is allowed to expand into a capillary. Its rate of expansion is calibrated on the glass scale.

Figure 5.1: Thermometer Temperature Scales Scale based on 32° for the freezing point of water and 212° for the boiling point of water, the interval between the two being divided into 180 parts. The 18th-century German physicist Daniel Gabriel Fahrenheit originally took as the zero of his scale the temperature of an equal ice-salt mixture and selected the values of 30° and 90° for the freezing point of water and normal body temperature, respectively; these later were revised to 32 º and 96 º , but the final scale required an adjustment to 98.6° for the latter value.

Until the 1970 s the Fahrenheit temperature scale was in general common use in English-speaking countries; the Celsius, or centigrade, scale was employed in most other countries and for scientific purposes worldwide. Since that time, however, most English-speaking countries have officially adopted the Celsius scale. The conversion formula for a temperature that is expressed on the Celsius (°C) scale to its Fahrenheit (°F) representation is:

°F = (9/5 x °C) + 32. ………………… (5.1)


http://www.britannica.com/eb/article?idxref=266537



Celsius temperature scale also called Centigrade Temperature Scale, scale based on

0° for the freezing point of water and 100° for the boiling point of water. Invented in 1742 by the Swedish astronomer Anders Celsius, it is sometimes called the centigrade scale because of the 100-degree interval between the defined points. The following formula can be used to convert a temperature from its representation on the Fahrenheit (°F) scale to the Celsius (°C) value:

°C = 5/9 (°F - 32). ……………….. (5.2) The Celsius scale is in general use wherever metric units have become accepted, and it is used in scientific work everywhere.

The temperature at which water, ice, and water vapor coexist in equilibrium; its value is set as 273.16. The unit of temperature on this scale is called the Kelvin, and its symbol is K (no degree symbol used). To convert from Celsius to Kelvin, add 273.

K = ° C + 273. …………………… (5.3)

Thermodynamic temperature is the fundamental temperature; its unit is the Kelvin that is defined as the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.

2.5.2 Heat Definition

Energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred. i.e., heat flows from the hotter body to the colder. The effect of this transfer of energy usually, but not always, is an increase in the temperature of the colder body and a decrease in the temperature of the hotter body.

A substance may absorb heat without an increase in temperature by changing from one physical state (or phase) to another, as from a solid to a liquid (melting), from a solid to a vapor (sublimation), from a liquid to a vapor (boiling), or from one solid form to another (usually called a crystalline transition). The important distinction between heat and temperature (heat being a form of energy and temperature a measure of the amount of that energy present in a body) was clarified during the 18th and 19th centuries.





-273.15 0 100 °C

-459.7 32 212 °F

V

Figure 5.2: Temperature scale

Task 1 1. What is the definition of Thermal equilibrium? 2. What is heat? 3. What is the definition for temperature? Task 2 Convert the temperature scale below to the Fahrenheit scale:

a) 0 0C e) 80 0C b) 100 0C f) -800C c) 1000 0C g) 300 0C d) -100 0C h) -300 0C

0 491.7 671.7 °R

0 273.15 373.15 °K

0 80 °Re

0V

-218.5

MIP/MECH/TN/003


Task 3 Convert the temperature scale below to the Celsius scale:

a) 0 0F e) 800F b) 100 0F f) -800F c) 1000 0F g) 300 0F d) -100 0F h) -300 0F

Task 4 Convert the temperature scale below to the Kelvin scale:

a) 0 0C b) 100 0C c) -273 0C d) 273 0C e) -100 0C f) 300 0C g) 1000 0C h) -1 0C


Preparatory Course Royal Malaysian Air Force Solultions Physics H.0 -1 _____________________________ Apprentice Course – Mechanic _______________________


SOLUTIONS 2.1 Introduction Task 1 1

a) 3700 mm c) 0.3 mm b) 324 mm d) 3.5 x 10-2 mm

2

a) 1 x 10-12 m2 c) 1 x 10-6 m2 b) 1 x 10-4 m2

3

a) 1m2 = 1 x 104 m2 c) 1km2 = 1 x 1010 cm2 b) 1mm2 = 1 x 10-2 cm2

4 a) 1cm3 = 1 x 10-6 m3 b) 1mm3 = 1 x 10-9m3

5

a) 1 Tone = 1000 kg b) 1 gram = 0.001 kg c) 1 milligram = 0.000001kg d) 1 microgram = 0.000000001 kg

6

a) 100 g d) 300 g b) 0.000000267 g e) 0.00013 g c) 0.072 kg

Task 2 1. 43.5 mi 2. 3000 kg 3. 450 cm 4. 100000 μg

2.2 Law of Motion (kinematics) Task 1 1. a) 9 km

b) 7 km 2) a) 11.2 km c) 7.47 m/s

b) 8 km d) 5.33 m/s Task 2 1. 70.8 km/h 2. 2.9 hrs. 3. 0.6 hrs. 4. 59 ms 5. . a) 42.7 mi/h b) 45.7 mi/h 6. $ 6.67 7. 5 s 8. 200 km/h



Task 3 1. 2 m/s2 2. 6 m/s2 3. 15 s 4. 12 s 5. - 5 m/s2 6. 80 m/s2 7. 62 km/h 8. 2 m/s2 9. .a)

.b) -3 m/s2 c) 405 m. Task 4 1. a) 0.523 rad d) 6.283 rad

b) 0.995 rad e) 7.330 rad c) 1.57 rad

2. a) 360o d) 293.35o

b) 1.72o e) 720 º c) 114591.50 f) 45o

3. a) ½ rev d) 50 rev b) 1.389 rev e) 0.12 rev c) 1 rev

4. (a) 1.5 rad

(b) 86o

5. 1.49m Task 5 1. 209.44 rad/s 2. 0.14 rad/s 3. 188.50 rad/s 4. (i) 20.94 rad/s

(ii) 214.68 rad/s (iii) 261.80 rad/s

Task 6 1. 7.14 rad/s² 2. 1.93 rad/s² 3. Fc = 1260 N 4. Fc = 4548 N 5. ac = 95.9 m/s2 6. ac= 1.71 x 1023 m/s2 , Fc = 1.56 x 10-7 N



Task 7 1. (a) 8.68 s c) 67.95 m/s

(b) 538 m 2. (a) 12.88 s c) -5.48 m/s

(b) 1080 m d) 84.04 m/s 3. 3.25 s 2.3 Motion and Force Task 1 1. (a) 30 N d) 14.14 N

(b) 16.2 N e) 56.57 N (c) 24.41 N

2. 28.43 N 3. 130 N Task 2 1. i) Fx = 51.96 N, Fy = 30 N

ii) Fx = -15 N, Fy = 25.98 N iii) Fx = -64.28 N, Fy = -76.60 N iv) Fx = 34.09 N, Fy = -36.57 N 2. a. FR = 67.08 N θ = 56.57o

b. FR = 48.2 N θ = 75o

c. FR = 86.16 N θ = 2.73o

d. FR = 94.06 N θ = 32.56o

e. FR = 19 N θ = 20o

f. FR = 117.13 N θ = 75o

3. FR = 198.58 N θ = 8.38o 4. FR = 24.207 N 5. θ = 21.70o Task 3 1. F1 = 8.66 N

F2 = 5 N 2. F1 = 42.26 N

F2 = 90.63 N

3. (a) b) Fx = 201.31 N

Fy = 553.10 N

FyFx

θ

mg



4. (a) b) Fx = 68.40 N

c) Fy = -187.93 N

Fy F= 200 N

F700

x

Task 4 1. 57.85 Nm 2. (a) 1200 Nm b) 5 m 3. 9.192 Nm 4. 6 Nm Task 5 1. 3 m 2. 3 m 3. 20 N 4. (b) 1.125 m Task 6 1. 50 kg 2. 1.26 N 3. 981 N 4. 361 N 5. a. T1 = 137.3 N, T2 = 58.9 N, T3 = 9.81 N b. T1 = 127.5 N, T2 = 49 N Task 7 1. i) Static Friction

ii) Kinetic Friction 2. 0.81 3. 103 N Task 8 1. 1.5 m/s2 2. 500 N 3. 0.2 m/s2 4. 0.5 m/s2 2.4 Bodies in Equilibrium Task 1 1. 370 N, 116° 2. 1.36 N 3. 1.73 N Task 2 1. 19 mm 2. .a) 1.2 x 105 N/m2

.b) 2.4 x 10-6

Task 3 1. a) 1.3 x 105 N/m2 c) 0.0062 mm

c) 6.6 x 10-7 2. .a) 2.18 x 106 N/m2 c) 19.7 x 106 N/m2

b) 0.111



2.5 Thermodynamics Task 1 a) 32°F b) 212°F c) 1832°F d) -148°F e) 176°F f) -112°F g) 572°F h) -508°F Task 2 a) –17.8°C b) 37.8°C c) 537.8°C d) –73.3°C e) 26.7°C f) –62.2°C g) 148.9°C h) –184.4°C Task 3 a) 273 K b) 373 K c) 0 K d) 546 K e) 173 K f) 846 K g) 1273 K h) 272 K

engineeering science

Documents