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CAPS Physical Sciences Grade 12
CAPS Physical Sciences Grade 12
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18 0 1 IA 1 2,1 2 Periodic Table of the Elements H 2 IIA 1,01 3 1,0 4 1,5 No He 13 IIIA 5 EN 14 IVA 2,0 6 15 VA 2,5 7 16 VIA 3,0 8 17 VIIA 3,5 9 Li Be Element B C N O F 6,94 9,01 AMU 10,8 12,0 14,0 16,0 19,0 11 0,9 12 1,2 Na Mg 23,0 24,3 19 0,8 20 13 3 IIIB 1,0 21 4 IVB 1,3 22 5 VB 1,5 23 6 VIB 1,6 24 7 VIIB 1,6 25 8 VII 1,5 26 9 VII 1,8 27 10 VII 1,8 28 11 IB 1,8 29 1,5 14 12 IIB 1,9 30 1,8 15 2,1 16 4,00 4,0 10 2,5 17 Ne 20,2 3,0 18 Al Si P S Cl Ar 27,0 28,1 31,0 32,1 35,45 39,9 1,6 31 1,6 32 1,8 33 2,0 34 2,4 35 2,8 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39,1 40,1 45,0 47,9 50,9 52,0 54,9 55,8 58,9 58,7 63,5 65,4 69,7 72,6 74,9 79,0 79,9 83,8 37 0,8 38 1,0 39 1,2 40 1,4 41 1,8 43 1,6 42 1,9 44 2,2 45 2,2 46 2,2 47 1,9 48 1,7 49 1,7 50 1,8 51 1,9 52 2,1 53 2,5 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85,5 87,6 88,9 91,2 92,9 95,9 (98) 101,1 102,9 106,4 107,9 112,4 114,8 118,7 121,8 127,6 126,9 131,3 55 0,7 56 0,9 5771 72 1,6 73 1,5 74 2,4 75 1,9 76 2,2 77 2,2 78 2,2 79 2,5 80 2,0 81 1,6 82 1,9 84 1,8 83 2,0 85 2,5 86 Cs Ba LaLu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 132,9 137,3 Lanthanides 178,5 180,9 183,8 186,2 190,2 192,2 195,1 197,0 200,6 204,4 207,2 209,0 (209) (210) (222) 87 0,7 88 0,9 89103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 Fr Ra AcLr Rf Db Sg Bh Hs Mt Ds Rg Cn Uut Uuq Uup Uuh Uus Uuo (223) 226,0 Actinides (261) (262) (263) (262) (265) (266) (269) (272) (277) (284) (289) (288) (293) (282) (282) Transition Metal 57 Metal 1,1 58 La Metalloid Nonmetal 1,1 59 Ce 138,9 140,1 1,1 61 1,1 60 Pr 140,9 62 Pm Nd (145) 144,2 1,1 63 Sm 64 Eu 150,4 1,2 65 Gd 152,0 66 Tb 157,3 1,2 67 Dy 158,9 1,2 68 1,2 69 Ho 162,5 164,9 Er 167,3 1,3 70 71 Tm 168,9 1,3 Yb 173,0 Lu 175,0 Noble Gas Lanthanide Actinide 89 1,1 90 1,3 91 1,5 92 1,3 94 1,4 93 1,3 95 1,3 96 1,3 97 1,3 98 1,3 99 1,3 100 1,3 101 1,3 102 1,3 103 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr 227,0 232,0 231,0 238,0 237,0 (244) (243) (247) (247) (251) (252) (257) (258) (258) (260) EVERYTHING SCIENCE GRADE 11 PHYSICAL SCIENCES VERSION 1 CAPS WRITTEN BY VOLUNTEERS COPYRIGHT NOTICE Your freedom to legally copy this book You are allowed and encouraged to copy any of the Everything Maths and Everything Science textbooks. You can legally photocopy any page or even the entire book. You can download it from www.everythingmaths.co.za and www.everythingscience.co.za, read it on your phone, tablet, iPad, or computer. You can burn it to CD, put on your flash drive, email it around or upload it to your website. The only restriction is that you have to keep this book, its cover, title, contents and shortcodes unchanged. This book was derived from the original Free High School Science Texts written by volunteer academics, educators and industry professionals. Everything Maths and Everything Science are trademarks of Siyavula Education. For more information about the Creative Commons AttributionNoDerivs 3.0 Unported (CC BYND 3.0) license see http://creativecommons.org/licenses/bynd/3.0/ AUTHORS AND CONTRIBUTORS Siyavula Education Siyavula Education is a social enterprise launched in 2012 with capital and support from the PSG Group Limited and the Shuttleworth Foundation. The Everything Maths and Science series is one of the titles developed and openly released by Siyavula. For more information about the writing and distribution of these or other openly licensed titles: www.siyavula.com info@siyavula.com 021 469 4771 Siyavula Authors Dr. Mark Horner; Heather Williams Siyavula and DBE team Jayanthi SK Maharaj (Veena); Marongwa Masemula; Ewald Zietsman; Bridget Nash; Prof. Gilberto Isquierdo; Karen Kornet; Dr. Kevin Reddy; Enoch Ndwamato Makhado; Clive Mhaka Siyavula and Free High School Science Text contributors Dr. Mark Horner; Dr. Samuel Halliday; Dr. Sarah Blyth; Dr. Rory Adams; Dr. Spencer Wheaton Iesrafeel Abbas; Sarah Abel; Dr. Rory Adams; Andrea Africa; Wiehan Agenbag; Matthew Amundsen; Ben Anhalt; Prashant Arora; Amos Baloyi; Bongani Baloyi; Raymond Barbour; CaroJoy Barendse; Richard Baxter; Tara Beckerling; Tim van Beek; Mariaan Bester; Jennifer de Beyer; Dr. Sarah Blyth; Sebastian Bodenstein; Martin Bongers; Thinus Booysen; Gareth Boxall; Stephan Brandt; Hannes Breytenbach; Alexander Briell; Wilbur Britz; Graeme Broster; Craig Brown; Michail Brynard; Deanne de Bude; Richard Burge; Bianca Bˆhmer; Jan Buys; George CalderPotts; Eleanor Cameron; Mark Carolissen; Shane Carollisson; Richard Case; Sithembile Cele; Alice Chang; Richard Cheng; Fanny Cherblanc; Dr. Christine Chung; Brett Cocks; RochÈ Compaan; Willem Conradie; Stefaan Conradie; Rocco Coppejans; Tim Craib; Andrew Craig; Tim Crombie; Dan Crytser; Jock Currie; Dr. Anne Dabrowski; Laura Daniels; Gareth Davies; Sandra Dickson; Sean Dobbs; Buhle Donga; William Donkin; Esmi Dreyer; Matthew Duddy; Christel Durie; Fernando Durrell; Dr. Dan Dwyer; Frans van Eeden; Alexander Ellis; Tom Ellis; Andrew Fisher; Giovanni Franzoni; Olivia Gillett; Ingrid von Glehn; Tamara von Glehn; Lindsay Glesener; Kevin Godby; Dr. Vanessa Godfrey; Terence Goldberg; Dr. Johan Gonzalez; Saaligha Gool; Hemant Gopal; Dr. Stephanie Gould; Umeshree Govender; Heather Gray; Lynn Greeff; Jaco Greyling; Martli Greyvenstein; Carine Grobbelaar; Suzanne GrovÈ; Dr. Tom Gutierrez; Brooke Haag; Kate Hadley; Alex Hall; Dr. Sam Halliday; Asheena Hanuman; Dr. Melanie Dymond Harper; Ebrahim Harris; Dr. Nicholas Harrison; Neil Hart; Nicholas Hatcher; Jason Hayden; Laura Hayward; Dr. William P. 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Well structured, impactful Corporate Social Investment (CSI) has the ability to contribute positively to nation building and drive positive change in the communities. MMI’s commitment to social investment means that we are constantly looking for ways in which we can assist some of South Africa’s most vulnerable citizens to expand their horizons and gain greater access to life’s opportunities. This means that we do not view social investment as a nice to have or as an exercise in marketing or sponsorship but rather as a critical part of our contribution to society. The merger between Metropolitan and Momentum was lauded for the complementary fit between two companies. This complementary fit is also evident in the focus areas of CSI programmes where Metropolitan and Momentum together cover and support the most important sectors and where the greatest need is in terms of social participation. HIV/AIDS is becoming a manageable disease in many developed countries but in a country such as ours, it remains a disease where people are still dying of this scourge unnecessarily. Metropolitan continues to make a difference in making sure that HIV AIDS moves away from being a death sentence to a manageable disease. Metropolitan’s other focus area is education which remains the key to economic prosperity for our country. Momentum’s focus on persons with disabilities ensures that this community is included and allowed to make their contribution to society. Orphaned and vulnerable children are another focus area for Momentum and projects supported ensure that children are allowed to grow up safely, to assume their role along with other children in inheriting a prosperous future. EVERYTHING MATHS AND SCIENCE The Everything Mathematics and Science series covers Mathematics, Physical Sciences, Life Sciences and Mathematical Literacy. The Siyavula Everything Science textbooks The Siyavula Everything Maths textbooks READ ON MOBILE MOBIWEB You can read all of the Everything Series textbooks on your mobile phone. Visit the Everything Maths and Everything Science mobi sites at: m.everythingmaths.co.za and m.everythingscience.co.za MXIT All Mxit users can read their Everything Series textbooks on Mxit Reach. Add Everything Maths and Everything Science as Mxit contacts or browse to the books on Mxit Reach. mxit>tradepost>reach>education>everything maths or everything science DIGITAL TEXTBOOKS READ ONLINE The online books feature videos, presentations, simulations and fully worked solutions to the questions and exercises found in the book. www.everythingmaths.co.za and www.everythingscience. DOWNLOAD FOR TABLETS For offline reading on your PC, tablet, iPad and Kindle you can download a digital copy of the Everything Series textbooks. Visit the Everything Maths and Everything Science websites and download the books. www.everythingmaths.co.za and www.everythingscience.co.za PRACTISE INTELLIGENTLY CHECK YOUR ANSWERS ON YOUR PHONE You can check your answer to any question in this textbook on your mobile phone by entering the shortcode found in the textbook into the search box on the mobisite. m.everythingmaths.co.za and m.everythingscience.co.za PRACTISE FOR TESTS AND EXAMS ON YOUR PHONE To do well in tests and exams you need practice. Practise the exercises from this textbook, additional exercises and questions from past exam papers on m.everythingmaths. co.za and m.everythingscience.co.za and Mxit Reach. m.everythingmaths.co.za and m.everythingscience.co.za MANAGE YOUR STUDIES YOUR DASHBOARD If you complete you practice homework and test questions at m.everythingmaths.co.za or m.everythingscience.co.za, you can track of your work. Your dashboard will show you your progress and mastery for every topic in the book and help you to manage your studies. You can use your dashboard to show your teachers, parents, universities or bursary institutions what you have done during the year. EVERYTHING SCIENCE When we look outside at everything in nature, look around us at everything manufactured or look up at everything in space we cannot but be struck by the incredible diversity and complexity of life; so many things, that look so different, operating in such unique ways. The physical universe really contains incredible complexity. Yet, what is even more remarkable than this seeming complexity is the fact that things in the physical universe are knowable. We can investigate them, analyse them and understand them. It is this ability to understand the physical universe that allows us to transform elements and make technological progress possible. If we look back at some of the things that developed over the last century ñ space travel, advances in medicine, wireless communication (from television to mobile phones) and materials a thousand times stronger than steel we see they are not the consequence of magic or some inexplicable phenomena. They were all developed through the study and systematic application of the physical sciences. So as we look forward at the 21st century and some of the problems of poverty, disease and pollution that face us, it is partly to the physical sciences we need to turn. For however great these challenges seem, we know that the physical universe is knowable and that the dedicated study thereof can lead to the most remarkable advances. There can hardly be a more exciting challenge than laying bare the seeming complexity of the physical universe and working with the incredible diversity therein to develop products and services that add real quality to peopleís lives. Physical sciences is far more wonderful, exciting and beautiful than magic! It is everywhere. Contents 1 Vectors in two dimensions 1.1 Introduction . . . . . . . . . . . . 1.2 Resultant of perpendicular vectors 1.3 Components of vectors . . . . . . 1.4 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 4 4 36 50 2 Newton’s laws 2.1 Introduction . . . . . . 2.2 Force . . . . . . . . . . 2.3 Newton’s laws . . . . . 2.4 Forces between masses 2.5 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 56 56 74 109 119 3 Atomic combinations 3.1 Chemical bonds . . 3.2 Molecular shape . 3.3 Electronegativity . . 3.4 Energy and bonding 3.5 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 136 148 153 159 160 4 Intermolecular forces 4.1 Intermolecular and interatomic forces . . . . . . . . . . . . . . . . . . 4.2 The chemistry of water . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 166 184 189 5 Geometrical optics 5.1 Summary of properties of light . . . . . . 5.2 Light rays . . . . . . . . . . . . . . . . . 5.3 Properties of light: revision . . . . . . . . 5.4 The speed of light . . . . . . . . . . . . . 5.5 Refraction . . . . . . . . . . . . . . . . . 5.6 Snell’s Law . . . . . . . . . . . . . . . . 5.7 Critical angles and total internal reflection 5.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 194 194 194 196 202 202 211 221 229 . . . . . . 234 234 234 235 237 238 248 . . . . . . . . . . 6 2D and 3D wavefronts 6.1 Introduction . . . . . . . . . . 6.2 Wavefronts . . . . . . . . . . . 6.3 Huygens principle . . . . . . . 6.4 Diffraction . . . . . . . . . . . 6.5 Diffraction through a single slit 6.6 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Ideal gases 252 7.1 Motion of particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 7.2 Ideal gas laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 7.3 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 8 Quantitative aspects of chemical change 8.1 Gases and solutions . . . . . . . . . . . . 8.2 Stoichiometric calculations . . . . . . . . 8.3 Volume relationships in gaseous reactions 8.4 Chapter summary . . . . . . . . . . . . . 9 Electrostatics 9.1 Introduction . . . 9.2 Coulomb’s law . 9.3 Electric field . . . 9.4 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 290 298 310 312 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 318 318 328 341 10 Electromagnetism 10.1 Introduction . . . . . . . . . . . . . . . . . 10.2 Magnetic field associated with a current . . 10.3 Faraday’s law of electromagnetic induction 10.4 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 346 346 357 369 11 Electric circuits 11.1 Introduction . . . 11.2 Ohm’s Law . . . 11.3 Power and energy 11.4 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 372 372 399 413 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Energy and chemical change 12.1 Energy changes in chemical reactions . . . . 12.2 Exothermic and endothermic reactions . . . 12.3 Activation energy and the activated complex 12.4 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 418 425 429 432 13 Types of reactions 13.1 Acids and bases . . 13.2 Acidbase reactions 13.3 Redox reactions . . 13.4 Chapter summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 438 443 452 465 . . . . . 470 470 470 477 485 486 . . . . . . . . . . . . . . . . . . . . . . . . 14 The lithosphere 14.1 Introduction . . . . . . . . . . 14.2 The lithosphere . . . . . . . . 14.3 Mining and mineral processing 14.4 Energy resources . . . . . . . 14.5 Summary . . . . . . . . . . . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Quantities used in the book 490 Solutions to exercises 491 List of Definitions 508 Image Attribution 510 CONTENTS CHAPTER Vectors in two dimensions 1.1 Introduction 4 1.2 Resultant of perpendicular vectors 4 1.3 Components of vectors 36 1.4 Chapter summary 50 1 1 Vectors in two dimensions 1.1 Introduction ESBK2 In grade 10 you learnt about vectors in one dimension. Now we will take these concepts further and learn about vectors in two dimensions as well as components of vectors. As a very short recap, a vector has both a magnitude and a direction. There are many physical quantities, like forces, that are well described by vectors (called or known as vector quantities). We often use arrows to represent vectors visually because the length of the arrow can be related to the magnitude and the arrowhead can indicate the direction. We will talk about the head, tail and magnitude of a vector when using arrows to represent them. Below is a diagram showing a vector (the arrow). The magnitude is indicated by the length and the labels show the the tail and the head of the vector. The direction of the vector is indicated by the direction in which the arrow is pointing. Magnitude Tail Head When we write the symbol for a physical quantity represented by a vector we draw an arrow over it to signify that it is a vector. If the arrow is left out then we are referring only to the magnitude of the vector quantity. Key Mathematics Concepts • Theorem of Pythagoras — Mathematics, Grade 10, Analytical geometry • Units and unit conversions — Physical Sciences, Grade 10, Science skills • Equations — Mathematics, Grade 10, Equations and inequalities • Trigonometry — Mathematics, Grade 10, Trigonometry • Graphs — Mathematics, Grade 10, Functions and graphs 1.2 Resultant of perpendicular vectors ESBK3 In grade 10 you learnt about the resultant vector in one dimension, we are going to extend this to two dimensions. As a reminder, if you have a number of vectors (think forces for now) acting at the same time you can represent the result of all of them together with a single vector known as the resultant. The resultant vector will have the same effect as all the vectors adding together. 4 1.1. Introduction We will focus on examples involving forces but it is very important to remember that this applies to all physical quantities that can be described by vectors, forces, displacements, accelerations, velocities and more. Vectors on the Cartesian plane ESBK4 The first thing to make a note of is that in Grade 10 we worked with vectors all acting in a line, on a single axis. We are now going to go further and start to deal with two dimensions. We can represent this by using the Cartesian plane which consists of two perpendicular (at a right angle) axes. The axes are a xaxis and a yaxis. We normally draw the xaxis from left to right (horizontally) and the yaxis up and down (vertically). We can draw vectors on the Cartesian plane. For example, if we have a force, F , of magnitude 2 N acting in the positive xdirection we can draw it as a vector on the Cartesian plane. y 2 1 F −2 −1 −1 1 2 x −2 Notice that the length of the vector as measured using the axes is 2, the magnitude specified. A vector doesn’t have to start at the origin but can be placed anywhere on the Cartesian plane. Where a vector starts on the plane doesn’t affect the physical quantity as long as the magnitude and direction remain the same. That means that all of the vectors in the diagram below can represent the same force. This property is know as equality of vectors. Chapter 1. Vectors in two dimensions 5 y 2 F In the diagram the vectors have the same magnitude because the arrows are the same length and they have the same direction. They are all parallel to the xdirection and parallel to each other. 1 F −2 −1 −1 F 1 x 2 −2 This applies equally in the ydirection. For example, if we have a force, F , of magnitude 2,5 N acting in the positive ydirection we can draw it as a vector on the Cartesian plane. Just as in the case of the xdirection, a vector doesn’t have to start at the origin but can be placed anywhere on the Cartesian plane. All of the vectors in the diagram below can represent the same force. y 2 F 1 −2 1 −1 −1 2 −2 y 2 F 1 F −2 −1 F −1 1 −2 6 1.2. Resultant of perpendicular vectors 2 x x The following diagram shows an example of four force vectors, two vectors that are parallel to each other and the yaxis as well as two that are parallel to each other and the xaxis. To emphasise that the vectors are perpendicular you can see in the figure below that when originating from the same point the vector are at right angles. y 2 y 2 F3 F1 1 1 F1 −2 −1 −1 F2 1 2 F3 x −2 F4 −1 −1 1 2 x F2 −2 F4 −2 Exercise 1 – 1: 1. Draw the following forces as vectors on the Cartesian plane originating at the origin: • • F1 = 1,5 N in the positive xdirection F2 = 2 N in the positive ydirection 2. Draw the following forces as vectors on the Cartesian plane: • F1 = 3 N in the positive xdirection • F2 = 1 N in the negative xdirection • F3 = 3 N in the positive ydirection 3. Draw the following forces as vectors on the Cartesian plane: • • • • F1 F2 F3 F4 = 3 N in the positive xdirection = 1 N in the positive xdirection = 2 N in the negative xdirection = 3 N in the positive ydirection 4. Draw the following forces as vectors on the Cartesian plane: • • • • F1 F2 F3 F4 = 2 N in the positive ydirection = 1,5 N in the negative ydirection = 2,5 N in the negative xdirection = 3 N in the positive ydirection Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 23F3 2. 23F4 3. 23F5 4. 23F6 www.everythingscience.co.za m.everythingscience.co.za Chapter 1. Vectors in two dimensions 7 y Vectors in two dimensions are not always parallel to an axis. We might know that a force acts at an angle to an axis so we still know the direction of the force and if we know the magnitude we can draw the force vector. For example, we can draw F1 = 2 N acting at 45◦ to the positive xdirection: 45◦1 F1 −1 −2 2 2 F1 1 45◦ 0 0 x 1 2 x We always specify the angle as being anticlockwise from the positive xaxis. So if we specified an negative angle we would measure it clockwise from the xaxis. For example, F1 = 2 N acting at −45◦ to the positive xdirection: y We can use many other ways of specifying the direction of a vector. The direction just needs to be unambiguous. We have used the Cartesian coordinate system and an angle with the xaxis so far but there are other common ways of specifying direction that you need to be aware of and comfortable to handle. Compass directions ESBK5 We can use compass directions when appropriate to specify the direction of a vector. For example, if we were describing the forces of tectonic plates (the sections of the earth’s crust that move) to talk about the forces involved in earthquakes we could talk the force that the moving plates exert on each other. Figure 1.1: A map of the 15 major tectonic plates that make up the Earth’s crust. 8 1.2. Resultant of perpendicular vectors The four cardinal directions are North, South, East and West when using a compass. They are shown in this figure: When specifying a direction of a vector using a compass directions are given by name, North or South. If the direction is directly between two directions we can combine the names, for example NorthEast is halfway between North and East. This can only happen for directions at right angles to each other, you cannot say NorthSouth as it is ambiguous. Figure 1.2: A sketch of the compass directions. Bearings ESBK6 Another way of using the compass to specify direction in a numerical way is to use bearings. A bearing is an angle, usually measured clockwise from North. Note that this is different to the Cartesian plane where angles are anti or counterclockwise from the positive xdirection. The resultant vector ESBK7 In grade 10 you learnt about adding vectors together in one dimension. The same principle can be applied for vectors in two dimensions. The following examples show addition of vectors. Vectors that are parallel can be shifted to fall on a line. Vectors falling on the same line are called colinear vectors. To add colinear vectors we use the tailtohead method you learnt in Grade 10. In the figure below we remind you of the approach of adding colinear vectors to get a resultant vector. y 4 3 2 1 y R F3 F4 F5 F2 F6 x R F1 1 2 3 4 5 6 7 x In the above figure the blue vectors are in the ydirection and the red vectors are in the xdirection. The two black vectors represent the resultants of the colinear vectors graphically. Chapter 1. Vectors in two dimensions 9 What we have done is implement the tailtohead method of vector addition for the vertical set of vectors and the horizontal set of vectors. Worked example 1: Revision: headtotail addition in one dimension QUESTION Use the graphical headtotail method to determine the resultant force on a rugby → player if two players on his team are pushing him forwards with forces of F1 = 600 N → and F2 = 900 N respectively and two players from the opposing team are pushing him → → backwards with forces of F3 = 1000 N and F4 = 650 N respectively. SOLUTION Step 1: Choose a scale and a reference direction Let’s choose a scale of 100 N: 0,5 cm and for our diagram we will define the positive direction as to the right. Step 2: Choose one of the vectors and draw it as an arrow of the correct length in the correct direction → We will start with drawing the vector F1 = 600 N pointing in the positive direction. Using our scale of 0,5 cm : 100 N, the length of the arrow must be 3 cm pointing to the right. y F1 = 600 N 1 0 0 1 2 3 4 5 6 7 8 x Step 3: Take the next vector and draw it starting at the arrowhead of the previous vector → → The next vector is F2 = 900 N in the same direction as F1 . Using the scale, the arrow should be 4,5 cm long and pointing to the right. y F1 = 600 N 1 F2 = 900 N 0 0 1 2 3 4 5 6 7 8 x Step 4: Take the next vector and draw it starting at the arrowhead of the previous vector 10 1.2. Resultant of perpendicular vectors → The next vector is F3 = 1000 N in the opposite direction. Using the scale, this arrow should be 5 cm long and point to the left. Note: We are working in one dimension so this arrow would be drawn on top of the first vectors to the left. This will get confusing so we’ll draw it next to the actual line as well to show you what it looks like. y F1 = 600 N 1 F2 = 900 N F3 = 1000 N 0 0 1 2 3 4 5 6 7 8 x Step 5: Take the next vector and draw it starting at the arrowhead of the previous vector → The fourth vector is F4 = 650 N in the opposite direction. Using the scale, this arrow must be 3,25 cm long and point to the left. y F1 = 600 N 1 F2 = 900 N F4 = 650 N F3 = 1000 N 0 0 1 2 3 4 5 6 7 8 x Step 6: Draw the resultant, measure its length and find its direction We have now drawn all the force vectors that are being applied to the player. The resultant vector is the arrow which starts at the tail of the first vector and ends at the head of the last drawn vector. yFR = 150 N F1 = 600 N 1 F2 = 900 N F4 = 650 N F3 = 1000 N 0 0 1 2 3 4 5 6 7 8 x The resultant vector measures 0,75 cm which, using our scale is equivalent to 150 N and points to the left (or the negative direction or the direction the opposing team members are pushing in). Chapter 1. Vectors in two dimensions 11 Exercise 1 – 2: 1. Find the resultant in the xdirection, Rx , and ydirection, Ry for the following forces: • F1 = 1,5 N in the positive xdirection • F2 = 1,5 N in the positive xdirection • F3 = 2 N in the negative xdirection x , and ydirection, R y for the following 2. Find the resultant in the xdirection, R forces: • F1 = 2,3 N in the positive xdirection • F2 = 1 N in the negative xdirection • F3 = 2 N in the positive ydirection • F4 = 3 N in the negative ydirection x , and ydirection, R y for the following 3. Find the resultant in the xdirection, R forces: • F1 = 3 N in the positive xdirection • F2 = 1 N in the positive xdirection • F3 = 2 N in the negative xdirection • F4 = 3 N in the positive ydirection x , and ydirection, R y for the following 4. Find the resultant in the xdirection, R forces: • F1 = 2 N in the positive ydirection • F2 = 1,5 N in the negative ydirection • F3 = 2,5 N in the negative xdirection • F4 = 3 N in the positive ydirection 5. Find a force in the xdirection, Fx , and ydirection, Fy , that you can add to the following forces to make the resultant in the xdirection, Rx , and ydirection, Ry zero: • F1 = 2,4 N in the positive ydirection • F2 = 0,7 N in the negative ydirection • F3 = 2,8 N in the negative xdirection • F4 = 3,3 N in the positive ydirection Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 23F7 2. 23F8 3. 23F9 4. 23FB www.everythingscience.co.za 12 5. 23FC m.everythingscience.co.za 1.2. Resultant of perpendicular vectors Magnitude of the resultant of vectors at right angles ESBK8 We apply the same principle to vectors that are at right angles or perpendicular to each other. Sketching tailtohead method The tail of the one vector is placed at the head of the other but in two dimensions the vectors may not be colinear. The approach is to draw all the vectors, one at a time. For the first vector begin at the origin of the Cartesian plane, for the second vector draw it from the head of the first vector. The third vector should be drawn from the head of the second and so on. Each vector is drawn from the head of the vector that preceded it. The order doesn’t matter as the resultant will be the same if the order is different. Let us apply this procedure to two vectors: y • F1 = 2 N in the positive ydirection 2 • F2 = 1,5 N in the positive xdirection 1 F1 First, we first draw a Cartesian plane with the first vector originating at the origin as shown on the right. 0 The second step is to take the second vector and draw it from the head of the first vector: y is the vector connectThe resultant, R, ing the tail of the first vector drawn to the head of the last vector drawn: y F2 2 0 1 F2 2 F1 F1 1 1 0 0 x 2 1 2 x R 0 0 1 2 x It is important to remember that the order in which we draw the vectors doesn’t matter. We If we had drawn them in the opposite order we would have the same resultant, R. can repeat the process to demonstrate this: Chapter 1. Vectors in two dimensions 13 We first draw a Cartesian plane with the second vector originating at the origin: The next step is to take the other vector and draw it from the head of the vector we have already drawn: y y 2 2 F1 1 1 F2 0 0 1 2 F2 0 x 0 1 x 2 y is the vector conThe resultant, R, necting the tail of the first vector drawn to the head of the last vector drawn (the vector from the start point to the end point): 2 F1 R 1 F2 0 0 1 2 x Worked example 2: Sketching vectors using tailtohead QUESTION Sketch the resultant of the following force vectors using the tailtohead method: • F1 = 2 N in the positive ydirection • F2 = 1,5 N in the positive xdirection • F3 = 1,3 N in the negative ydirection • F4 = 1 N in the negative xdirection 14 1.2. Resultant of perpendicular vectors SOLUTION Step 1: Draw the Cartesian plane and the first vector First draw the Cartesian plane and force, F1 starting at the origin: y Step 2: Draw the second vector Starting at the head of the first vector we draw the tail of the second vector: y F2 2 2 F1 F1 1 1 0 0 0 1 2 x 0 Step 3: Draw the third vector Starting at the head of the second vector we draw the tail of the third vector: y F1 0 0 1 2 x F2 2 F1 F3 1 2 Step 4: Draw the fourth vector Starting at the head of the third vector we draw the tail of the fourth vector: y F2 2 1 F4 1 F3 0 x 0 1 2 x Step 5: Draw the resultant vector Starting at the origin draw the resultant vector to the head of the fourth vector: y F2 2 F1 F4 1 F3 R 0 0 1 2 x Chapter 1. Vectors in two dimensions 15 Worked example 3: Sketching vectors using tailtohead QUESTION Sketch the resultant of the following force vectors using the tailtohead method by first determining the resultant in the x and ydirections: • F1 = 2 N in the positive ydirection • F2 = 1,5 N in the positive xdirection • F3 = 1,3 N in the negative ydirection • F4 = 1 N in the negative xdirection SOLUTION x Step 1: First determine R First draw the Cartesian plane with the vectors in the xdirection: y Step 2: Secondly determine R Next we draw the Cartesian plane with the vectors in the ydirection: y y 2 2 F1 F3 1 1 x F4 R 0 0 F2 1 2 x y Step 3: Draw the resultant vectors, R x headtotail and R y y R 0 0 x y x R y R R 0 0 1 16 2 Step 4: Comparison of results To double check, we can replot all the vectors again as we did in the previous worked example to see that the outcome is the same: 2 1 1 2 2 x 1 F2 F1 x R F4 y R R 0 0 1 1.2. Resultant of perpendicular vectors F3 2 x Exercise 1 – 3: FACT 1. Sketch the resultant of the following force vectors using the tailtohead method: • F1 = 2,1 N in the positive ydirection • F2 = 1,5 N in the negative xdirection 2. Sketch the resultant of the following force vectors using the tailtohead method: • F1 = 12 N in the positive ydirection • F2 = 10 N in the positive xdirection • F3 = 5 N in the negative ydirection • F4 = 5 N in the negative xdirection When dealing with more than two vectors the procedure is repetitive. First find the resultant of any two of the vectors to be added. Then use the same method to add the resultant from the first two vectors with a third vector. This new resultant is then added to the fourth vector and so on, until there are no more vectors to be added. 3. Sketch the resultant of the following force vectors using the tailtohead method by first determining the resultant in the x and ydirections: • F1 = 2 N in the positive ydirection • F2 = 1,5 N in the negative ydirection • F3 = 1,3 N in the negative ydirection • F4 = 1 N in the negative xdirection 4. Sketch the resultant of the following force vectors using the tailtohead method by first determining the resultant in the x and ydirections: • F1 = 6 N in the positive ydirection • F2 = 3,5 N in the negative xdirection • F3 = 8,7 N in the negative ydirection • F4 = 3 N in the negative ydirection Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 23FD 2. 23FF 3. 23FG 4. 23FH www.everythingscience.co.za m.everythingscience.co.za Sketching tailtotail method In this method we draw the two vectors with their tails on the origin. Then we draw a line parallel to the first vector from the head of the second vector and vice versa. Where the parallel lines intersect is the head of the resultant vector that will also start at the origin. We will only deal with perpendicular vectors but this procedure works for any vectors. Let us apply this procedure to the same two vectors we used to illustrate the headtotail method: Chapter 1. Vectors in two dimensions 17 • F1 = 2 N in the positive ydirection • F2 = 1,5 N in the positive xdirection 1. We first draw a Cartesian plane with the first vector originating at the origin: y 2. Then we add the second vector but also originating from the origin so that the vectors are drawn tailtotail: y 2 2 F1 F1 1 1 0 0 1 2 x F2 0 0 3. Now we draw a line parallel to F1 from the head of F2 : 2 x 4. Next we draw a line parallel to F2 from the head of F1 : y y 2 2 F1 F1 1 1 F2 0 0 1 2 x 5. Where the two lines intersect is the head of the resultant vector which will originate at the origin so: y 2 F1 0 0 1 2 F2 0 0 1 2 x You might be asking what you would do if you had more than 2 vectors to add together. In this case all you need x by adding to do is first determine R all the vectors that are parallel to the x y by adding all the vecdirection and R tors that are parallel to the ydirection. Then you use the tailtotail method to x and R y. find the resultant of R R F2 1 18 1 x 1.2. Resultant of perpendicular vectors Exercise 1 – 4: 1. Sketch the resultant of the following force vectors using the tailtotail method: • F1 = 2,1 N in the positive ydirection • F2 = 1,5 N in the negative xdirection 2. Sketch the resultant of the following force vectors using the tailtotail method by first determining the resultant in the x and ydirections: • F1 = 2 N in the positive ydirection • F2 = 1,5 N in the negative ydirection • F3 = 1,3 N in the negative ydirection • F4 = 1 N in the negative xdirection 3. Sketch the resultant of the following force vectors using the tailtotail method by first determining the resultant in the x and ydirections: • F1 = 6 N in the positive ydirection • F2 = 3,5 N in the negative xdirection • F3 = 8,7 N in the negative ydirection • F4 = 3 N in the negative ydirection Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 23FJ 2. 23FK 3. 23FM www.everythingscience.co.za m.everythingscience.co.za Closed vector diagrams A closed vector diagram is a set of vectors drawn on the Cartesian using the tailtohead method and that has a resultant with a magnitude of zero. This means that if the first vector starts at the origin the last vector drawn must end at the origin. The vectors form a closed polygon, no matter how many of them are drawn. Here are a few examples of closed vector diagrams: 4 y F2 3 F3 F1 2 1 0 0 1 2 3 4 x In this case there were 3 force vectors. When drawn tailtohead with the first force starting at the origin the last force drawn ends at the origin. The resultant would have a magnitude of zero. The resultant is drawn from the tail of the first vector to the head of the final vector. Chapter 1. Vectors in two dimensions 19 In the diagram below there are 4 vectors that also form a closed vector diagram. y 4 F2 3 2 −4 −3 F1 1 −2 −1 −1 F3 1 2 3 4 x F4 −2 −3 −4 In this case with 4 vectors, the shape is a 4sided polygon. Any polygon made up of vectors drawn tailtohead will be a closed vector diagram because a polygon has no gaps. Using Pythagoras’ theorem to find magnitude If we wanted to know the resultant of the three blue vectors and the three red vectors in Figure 1.2 we can use the resultant vectors in the x and ydirections to determine this. The black arrow represents the resul x and R y . We can tant of the vectors R find the magnitude of this vector using the theorem of Pythagoras because the three vectors form a right angle triangle. If we had drawn the vectors to scale we would be able to measure the magnitude of the resultant as well. y 5 4 x R 3 2 y R R 1 0 1 2 3 4 x What we’ve actually sketched out already is our approach to finding the resultant of many vectors using components so remember this example when we get there a little later. 20 1.2. Resultant of perpendicular vectors Worked example 4: Finding the magnitude of the resultant QUESTION The force vectors in Figure 1.2 have the following magnitudes: 1 N, 1 N, 2 N for the blue ones and 2 N, 2 N and 1,5 N for the red ones. Determine the magnitude of the resultant. SOLUTION Step 1: Determine the resultant of the vectors parallel to the yaxis The resultant of the vectors parallel to the yaxis is found by adding the magnitudes (lengths) of three vectors because they all point in the same direction. The answer is y =1 N + 1 N + 2 N = 4 N in the positive ydirection. R Step 2: Determine the resultant of the vectors parallel to the xaxis The resultant of the vectors parallel to the xaxis is found by adding the magnitudes (lengths) of three vectors because they all point in the same direction. The answer is x =2 N + 2 N + 1,5 N = 5,5 N in the positive xdirection. R Step 3: Determine the magnitude of the resultant We have a right angled triangle. We also know the length of two of the sides. Using Pythagoras we can find the length of the third side. From what we know about resultant vectors this length will be the magnitude of the resultant vector. The resultant is: Rx2 + Ry2 = R2 (Pythagoras’ theorem) (5,5)2 + (4)2 = R2 R = 6,8 Step 4: Quote the final answer Magnitude of the resultant: 6,8 N Note: we did not determine the resultant vector in the worked example above because we only determined the magnitude. A vector needs a magnitude and a direction. We did not determine the direction of the resultant vector. Chapter 1. Vectors in two dimensions 21 Graphical methods ESBK9 Graphical techniques In grade 10 you learnt how to add vectors in one dimension graphically. We can expand these ideas to include vectors in twodimensions. The following worked example shows this. Worked example 5: Finding the magnitude of the resultant in two dimensions graphically QUESTION y = 4 N in the positive ydirection and R x = 3 N in the positive xGiven two vectors, R direction, use the tailtohead method to find the resultant of these vectors graphically. SOLUTION Step 1: Choose a scale and draw axes The vectors we have do not have very big magnitudes so we can choose simple scale, we can use 1 N : 1 cm as our scale for the drawing. Then we draw axes that the vector diagram should fit in. The largest vector has length 4 N and both vectors are in the positive direction so we can draw axes from the origin to 5 and expect the vectors to fit. x Step 2: Draw R x is 3 N so the arThe magnitude of R row we need to draw must be 3 cm long. The arrow must point in the positive xdirection. y 5 4 3 y 5 2 4 1 3 x R 0 0 2 1 0 0 22 1 2 3 4 5 x 1.2. Resultant of perpendicular vectors 1 2 3 4 5 x y Step 3: Draw R y is 4 so the arrow The length of R we need to draw must be 4 cm long. The arrow must point in the positive ydirection. The important fact to note is that we are implementing the headtotail method so the vector must start at x. the end (head) of R Step 4: Draw the resultant vector, R The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a x to the head vector from the tail of R of Ry . y y 5 5 4 4 3 3 y R R 2 y R 2 1 1 0 0 1 2 3 4 5 x x R θ 0 x R 0 1 2 3 4 5 x Step 5: Measure the resultant, R We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to the is 5 cm long therefore the magnitude actual result. In the last diagram the resultant, R of the vector is 5 N. The direction of the resultant, θ, we need to measure from the diagram using a protractor. The angle that the vector makes with the xaxis is 53◦ . Step 6: Quote the final answer is 5 N at 53◦ from the positive xdirection. R In the case where you have to find the resultant of more than two vectors first apply the tailtohead method to all the vectors parallel to the one axis and then all the vectors y from all the vectors parallel to the other axis. For example, you would first calculate R x from all the vectors parallel to the xaxis. After that parallel to the yaxis and then R you apply the same procedure as in the previous worked example to the get the final resultant. Chapter 1. Vectors in two dimensions 23 Worked example 6: Finding the magnitude of the resultant in two dimensions graphically QUESTION Given the following three force vectors, determine the resultant force: • F1 = 3,4 N in the positive xdirection • F2 = 4 N in the positive xdirection • F3 = 3 N in the negative ydirection SOLUTION x Step 1: Determine R First we determine the resultant of all the vectors that are parallel to the xaxis. There are two vectors F1 and F2 that we need to add. We do this using the tailtohead method for colinear vectors. y 1 F1 0 0 1 F2 2 3 4 5 6 7 8 x x , that would give us the same outcome is: The single vector, R y 1 x R 0 0 1 2 3 4 5 6 7 8 x y Step 2: Determine R y = F3 . There is only one vector in the ydirection, F3 , therefore R Step 3: Choose a scale and draw axes The vectors we have do not have very big magnitudes so we can choose simple scale, we can use 1 N : 1 cm as our scale for the drawing. Then we draw axes that the diagram should fit on. The longest vector has length 7,4 N. We need our axes to extend just further than the vectors aligned with each axis. Our axes need to start at the origin and go beyond 7,4 N in the positive xdirection and further than 3 N in the negative ydirection. Our scale choice of 1 N : 1 cm means that our axes actually need to extend 7,4 cm in the positive xdirection and further than 3 cm in the negative ydirection 24 1.2. Resultant of perpendicular vectors y 1 0 1 2 3 4 5 6 7 8 x −1 −2 −3 −4 Step 4: Draw Rx x is 7,4 N so the arrow we need to draw must be 7,4 cm long. The The magnitude of R arrow must point in the positive xdirection. y 1 x R 0 1 2 3 4 5 6 7 8 x −1 −2 −3 −4 Step 5: Draw Ry y is 3 N so the arrow we need to draw must be 3 cm long. The The magnitude of R arrow must point in the negative ydirection. The important fact to note is that we are x. implementing the headtotail method so the vector must start at the end (head) of R Chapter 1. Vectors in two dimensions 25 y 1 x R 0 1 2 3 4 5 6 7 8 x −1 −2 y R −3 −4 Step 6: Draw the resultant vector, R The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a vector from the tail of x to the head of R y. R y 1 x R 0 1 θ 2 3 4 5 6 7 8 x −1 −2 −3 y R −4 Step 7: Measure the resultant, R We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to is 8,0 cm long therefore the the actual result. In the last diagram the resultant, R magnitude of the vector is 8,0 N. The direction of the resultant we need to measure from the diagram using a protractor. The angle that the vector makes with the xaxis is 22◦ . Step 8: Quote the final answer is 8,0 N at −22◦ from the positive xdirection. R 26 1.2. Resultant of perpendicular vectors Worked example 7: Finding the resultant in two dimensions graphically QUESTION Given the following three force vectors, determine the resultant force: • F1 = 2,3 N in the positive xdirection • F2 = 4 N in the positive ydirection • F3 = 3,3 N in the negative ydirection • F4 = 2,1 N in the negative ydirection SOLUTION x Step 1: Determine R x = F1 . There is only one vector in the xdirection, F1 , therefore R y Step 2: Determine R Then we determine the resultant of all the vectors that are parallel to the yaxis. There are three vectors F2 , F3 and F4 that we need to add. We do this using the tailtohead method for colinear vectors. 4 y , that would give The single vector, R us the same effect is: 4 y 3 2 y 1 3 F3 2 F2 y R −1 x 1 −2 F4 −1 x −2 Chapter 1. Vectors in two dimensions 27 Step 3: Choose a scale and draw axes We choose a scale 1 N : 1 cm for the drawing. Then we draw axes that the diagram should fit in. We need our axes to extend just further than the vectors aligned with each axis. Our axes need to start at the origin and go beyond 2,3 N in the positive xdirection and further than 1,4 N in the negative ydirection. Our scale choice of 1 N = 1 cm means that our axes actually need to extend 2,3 cm in the positive xdirection and further than 1,4 cm in the negative ydirection x Step 4: Draw R x is 2,3 N so the arThe magnitude of R row we need to draw must be 2,3 cm long. The arrow must point in the positive xdirection. y 1 x R 1 2 3 x −1 −2 y 1 1 2 3 x −1 −2 y Step 5: Draw R y is 1,4 N so the arThe magnitude of R row we need to draw must be 1,4 cm long. The arrow must point in the negative ydirection. The important fact to note is that we are implementing the headtotail method so the vector must x. start at the end (head) of R Step 6: Draw the resultant vector, R The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a x to the head vector from the tail of R y. of R y y 1 1 x R x R 1 −1 2 3 1 x −1 y R R 2 3 x y R −2 −2 Step 7: Measure the resultant, R We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to the is 2,7 cm long therefore magnitude of the vector. In the last diagram the resultant, R the magnitude of the vector is 2,7 N. 28 1.2. Resultant of perpendicular vectors The direction of the resultant we need to measure from the diagram using a protractor. The angle that the vector makes with the xaxis is 31 degrees. Step 8: Quote the final answer is 2,7 N at −31◦ from the positive xdirection. R Worked example 8: Finding the resultant in two dimensions graphically QUESTION A number of tugboats are trying to manoeuvre a submarine in the harbour but they are not working as a team. Each tugboat is exerting a different force on the submarine. Given the following force vectors, determine the resultant force: • F1 = 3,4 kN in the positive xdirection • F2 = 4000 N in the positive ydirection • F3 = 300 N in the negative ydirection • F4 = 7 kN in the negative ydirection SOLUTION Step 1: Convert to consistent S.I. units To use the graphical method of finding the resultant we need to work in the same units. Strictly speaking in this problem all the vectors are in newtons but they have different factors which will affect the choice of scale. These need to taken into account and the simplest approach is to convert them all to a consistent unit and factor. We could use kN or N, the choice does not matter. We will choose kN. Remember that k represents a factor of ×103 . Chapter 1. Vectors in two dimensions 29 F1 and F4 do not require any adjustment because they are both in kN. To convert N to kN we use: kN = ×103 N 1 = kN ×103 N = ×10−3 kN To convert the magnitude of F2 to kN: F2 = 4000 N F2 = 4000 × 10−3 kN F2 = 4 kN Therefore F2 = 4 kN in the positive ydirection. To convert the magnitude of F3 to kN: F3 = 300 N F3 = 300 × 10−3 kN F3 = 0,3 kN Therefore F3 = 0,3 kN in the negative ydirection. So: • F1 = 3,4 kN in the positive xdirection • F2 = 4 kN in the positive ydirection • F3 = 0,3 kN in the negative ydirection • F4 = 7 kN in the negative ydirection Step 2: Choose a scale and draw axes The vectors we have do have very big magnitudes so we need to choose a scale that will allow us to draw them in a reasonable space, we can use 1 kN : 1 cm as our scale for the drawings. x Step 3: Determine R x = F1 . There is only one vector in the xdirection, F1 , therefore R y Step 4: Determine R Then we determine the resultant of all the vectors that are parallel to the yaxis. There are three vectors F2 , F3 and F4 that we need to add. We do this using the tailtohead method for colinear vectors. 30 1.2. Resultant of perpendicular vectors y , that would give The single vector, R us the same outcome is: y y 4 F3 3 4 3 F2 2 2 1 1 x F4 −1 −2 y R −1 −3 −2 −4 −3 x −4 Step 5: Draw axes Then we draw axes that the diagram should fit on. We need our axes to extend just further than the vectors aligned with each axis. Our axes need to start at the origin and go beyond 3,4 kN in the positive xdirection and further than 3,3 kN in the negative ydirection. Our scale choice of 1 kN : 1 cm means that our axes actually need to extend 3,4 cm in the positive xdirection and further than 3,3 cm in the negative ydirection y x Step 6: Draw R The length of Rx is 3,4 kN so the arrow we need to draw must be 3,4 cm long. The arrow must point in the positive xdirection. y 1 x R 1 2 3 4 x −1 −2 1 −3 1 2 3 4 x −4 −1 −2 −3 −4 Chapter 1. Vectors in two dimensions 31 y Step 7: Draw R The length of Ry is 3,3 kN so the arrow we need to draw must be 3,3 cm long. The arrow must point in the negative ydirection. The important fact to note is that we are implementing the headtotail method so the vector must start x. at the end (head) of R Step 8: Draw the resultant vector, R The resultant vector is the vector from the tail of the first vector we drew directly to the head of the last vector we drew. This means we need to draw a x to the head vector from the tail of R of Ry . y y 1 1 x R x R 1 2 3 4 x 1 θ 2 −2 4 x −1 −1 y R 3 −2 R y R −3 −3 −4 −4 Step 9: Measure the resultant, R We are solving the problem graphically so we now need to measure the magnitude of the vector and use the scale we chose to convert our answer from the diagram to is 4,7 cm long therefore the the actual result. In the last diagram the resultant, R magnitude of the vector is 4,7 kN. The direction of the resultant we need to measure from the diagram using a protractor. The angle that the vector makes with the xaxis is 44◦ . Step 10: Quote the final answer is 4,7 kN at −44◦ from the positive xdirection. R Algebraic methods ESBKB Algebraic addition and subtraction of vectors In grade 10 you learnt about addition and subtraction of vectors in one dimension. The following worked example provides a refresher of the concepts. 32 1.2. Resultant of perpendicular vectors Worked example 9: Adding vectors algebraically QUESTION A force of 5 N to the right is applied to a crate. A second force of 2 N to the left is also applied to the crate. Calculate algebraically the resultant of the forces applied to the crate. SOLUTION Step 1: Draw a sketch A simple sketch will help us understand the problem. 5N 2N Step 2: Decide which method to use to calculate the resultant Remember that force is a vector. Since the forces act along a straight line (i.e. the xdirection), we can use the algebraic technique of vector addition. Step 3: Choose a positive direction Choose the positive direction to be to the right. This means that the negative direction is to the left. Rewriting the problem using the choice of a positive direction gives us a force of 5 N in the positive xdirection and force of 2 N in the negative xdirection being applied to the crate. Step 4: Now define our vectors algebraically F1 = 5 N F2 = −2 N Step 5: Add the vectors Thus, the resultant force is: F1 + F2 = (5) + (−2) =3N Step 6: Quote the resultant Remember that in this case a positive force means to the right: 3 N to the right. We can now expand on this work to include vectors in two dimensions. Worked example 10: Algebraic solution in two dimensions QUESTION Chapter 1. Vectors in two dimensions 33 A force of 40 N in the positive xdirection acts simultaneously (at the same time) to a force of 30 N in the positive ydirection. Calculate the magnitude of the resultant force. SOLUTION Step 1: Draw a rough sketch As before, the rough sketch looks as follows: lt su t an re 30 N α 40 N Step 2: Determine the length of the resultant Note that the triangle formed by the two force vectors and the resultant vector is a rightangle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let R represent the length of the resultant vector. Then: Fx2 + Fy2 = R2 Pythagoras’ theorem (40)2 + (30)2 = R2 R = 50 N Step 3: Quote the resultant The magnitude of the resultant force is then 50 N. Direction For two dimensional vectors we have only covered finding the magnitude of vectors algebraically. We also need to know the direction. For vectors in one dimension this was simple. We chose a positive direction and then the resultant was either in the positive or in the negative direction. In grade 10 you learnt about the different ways to specify direction. We will now look at using trigonometry to determine the direction of the resultant vector. We can use simple trigonometric identities to calculate the direction. We can calculate the direction of the resultant in the previous worked example. 34 1.2. Resultant of perpendicular vectors Worked example 11: Direction of the resultant QUESTION A force of 40 N in the positive xdirection acts simultaneously (at the same time) to a force of 30 N in the positive ydirection. Calculate the magnitude of the resultant force. SOLUTION Step 1: Magnitude We determined the magnitude of the resultant vector in the previous worked example to be 50 N. The sketch of the situation is: nt a ult res 30 N α 40 N Step 2: Determine the direction of the resultant To determine the direction of the resultant force, we calculate the angle α between the resultant force vector and the positive xaxis, by using simple trigonometry: opposite side adjacent side 30 tan α = 40 α = tan−1 (0,75) tan α = α = 36,87◦ Step 3: Quote the resultant The resultant force is then 50 N at 36,9◦ to the positive xaxis. Exercise 1 – 5: Algebraic addition of vectors 1. A force of 17 N in the positive xdirection acts simultaneously (at the same time) to a force of 23 N in the positive ydirection. Calculate the resultant force. 2. A force of 23,7 N in the negative xdirection acts simultaneously to a force of 9 N in the positive ydirection. Calculate the resultant force. 3. Four forces act simultaneously at a point, find the resultant if the forces are: Chapter 1. Vectors in two dimensions 35 • F1 = 2,3 N in the positive xdirection • F2 = 4 N in the positive ydirection • F3 = 3,3 N in the negative ydirection • F4 = 2,1 N in the negative ydirection 4. The following forces act simultaneously on a pole, if the pole suddenly snaps in which direction will it be pushed: • F1 = 2,3 N in the negative xdirection • F2 = 11,7 N in the negative ydirection • F3 = 6,9 N in the negative ydirection • F4 = 1,9 N in the negative ydirection Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 23FN 2. 23FP 3. 23FQ 4. 23FR www.everythingscience.co.za 1.3 m.everythingscience.co.za Components of vectors ESBKC In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors which when added give that original vector. These vectors which sum to the original are called components of the original vector. The process of breaking a vector into its components is called resolving into components. In practise it is most useful to resolve a vector into components which are at right angles to one another, usually horizontal and vertical. Think about all the problems we’ve solved so far. If we have vectors parallel to the x and yaxes problems are straightforward to solve. → Any vector can be resolved into a horizontal and a vertical component. If R is a vector, → → → then the horizontal component of R is Rx and the vertical component is Ry . y 3 R 2 y R θ 1 x R 0 0 1 2 x 3 When resolving into components that are parallel to the x and yaxes we are always dealing with a rightangled triangle. This means that we can use trigonometric identities to determine the magnitudes of the components (we know the directions because they are aligned with the axes). From the triangle in the diagram above we know that 36 1.3. Components of vectors cos(θ) = Rx R sin θ = and Rx = cos(θ) R Rx = R cos(θ) Ry R Ry = sin(θ) R Ry = R sin(θ) Rx = R cos(θ) Ry = R sin(θ) Note that the angle is measured counterclockwise from the positive xaxis. Worked example 12: Resolving a vector into components QUESTION A force of 250 N acts at an angle of 30◦ to the positive xaxis. Resolve this force into components parallel to the x and yaxes. SOLUTION Step 1: Draw a rough sketch of the original vector 0N 25 30◦ Step 2: Determine the vector components Next we resolve the force into components parallel to the axes. Since these directions are perpendicular to one another, the components form a rightangled triangle with the original force as its hypotenuse. y 0N 100 25 Fy 30◦ 0 0 Fx 100 200 300 x Notice how the two components acting together give the original vector as their resultant. Chapter 1. Vectors in two dimensions 37 Step 3: Determine the magnitudes of the component vectors Now we can use trigonometry to calculate the magnitudes of the components of the original displacement: Fy = 250 sin(30◦ ) Fx = 250 cos(30◦ ) and = 125 N = 216,5 N Remember Fx and Fy are the magnitudes of the components. Fx is in the positive xdirection and Fy is in the positive ydirection. Worked example 13: Resolving a vector into components QUESTION A force of 12,5 N acts at an angle of 230◦ to the positive xaxis. Resolve this force into components parallel to the x and yaxes. SOLUTION Step 1: Draw a rough sketch of the original vector y 15 10 5 −15 −10 −5 −5 F −10 5 10 15 x −15 Step 2: Determine the vector components Next we resolve the force into components parallel to the axes. Since these directions are perpendicular to one another, the components form a rightangled triangle with the original force as its hypotenuse. 38 1.3. Components of vectors Now we can use trigonometry to calculate the magnitudes of the components of the original force: Fy = 12,5 sin(230◦ ) Fx = 12,5 cos(230◦ ) and = −9,58 N = −8,03 N Notice that by using the full angle we actually get the correct signs for the components if we use the standard Cartesian coordinates. Fx is in the negative xdirection and Fy is in the negative ydirection. Exercise 1 – 6: 1. Resolve each of the following vectors into components: • F1 =5 N at 45◦ to the positive xaxis. • F2 =15 N at 63◦ to the positive xaxis. • F3 =11,3 N at 127◦ to the positive xaxis. • F4 =125 N at 245◦ to the positive xaxis. 2. Resolve each of the following vectors into components: • F1 =11 × 104 N at 33◦ to the positive xaxis. • F2 =15 GN at 28◦ to the positive xaxis. • F3 =11,3 kN at 193◦ to the positive xaxis. • F4 =125 × 105 N at 317◦ to the positive xaxis. Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 23FS 2. 23FT www.everythingscience.co.za m.everythingscience.co.za Vector addition using components ESBKD Components can also be used to find the resultant of vectors. This technique can be applied to both graphical and algebraic methods of finding the resultant. The method is straightforward: 1. make a rough sketch of the problem; 2. find the horizontal and vertical components of each vector; x; 3. find the sum of all horizontal components, R Chapter 1. Vectors in two dimensions 39 y; 4. find the sum of all the vertical components, R 5. then use them to find the resultant, R. → Consider the two vectors, F1 and F2 , in Figure 1.3, together with their resultant, R. y 5 F 2 4 R 3 2 1 1 F 0 0 1 2 3 4 5 7 x 6 Figure 1.3: An example of two vectors being added to give a resultant. Each vector in Figure 1.3 can be broken down into one component in the xdirection (horizontal) and one in the ydirection (vertical). These components are two vectors which when added give you the original vector as the resultant. This is shown in Figure 1.4 below: F1x F2x x R y y R R 3 F2x 1 F1y F1y 2 1 F 0 0 1 2 F1x 3 4 5 6 Figure 1.4: Adding vectors using components. We can see that: F1 = F1x + F1y F2 = F2x + F2y y =R x + R R 40 1.3. Components of vectors 7 x F2y 2 4 F F2y 5 x = F1x + F2x But, R y = F1y + F2y and R In summary, addition of the xcomponents of the two original vectors gives the xcomponent of the resultant. The same applies to the ycomponents. So if we just added all the components together we would get the same answer! This is another important property of vectors. Worked example 14: Adding vectors using components QUESTION If in Figure 1.4, F1 =5,385 N at an angle of 21,8◦ to the horizontal and F2 =5 N at an angle of 53,13◦ to the horizontal, find the resultant force, R. SOLUTION Step 1: Decide how to tackle the problem The first thing we must realise is that the order that we add the vectors does not matter. Therefore, we can work through the vectors to be added in any order. We also draw up the following table to help us work through the problem: Vector F1 F2 Resultant xcomponent ycomponent Total Step 2: Resolve F1 into components We find the components of F1 by using known trigonometric ratios. First we find the magnitude of the vertical component, F1y : Secondly we find the magnitude of the horizontal component, F1x : F1y F1 F1y sin(21,8◦ ) = 5,385 F1y = (sin(21,8◦ )) (5,385) F1x F1 F1x cos(21,8◦ ) = 5,385 F1x = (cos(21,8◦ )) (5,385) cos(θ) = sin(θ) = = 5,00 N = 2,00 N N 2.00 N 85 5,3 θ 5.00 N Chapter 1. Vectors in two dimensions 41 The components give the sides of the right angle triangle, for which the original vector, F1 , is the hypotenuse. Vector F1 F2 xcomponent 5,00 N Resultant 5,385 N ycomponent 2,00 N Resultant Step 3: Resolve F2 into components We find the components of F2 by using known trigonometric ratios. First we find the magnitude of the vertical component, F2y : Secondly we find the magnitude of the horizontal component, F2x : F2x F2 F 2x cos(53,13◦ ) = 5 F2x = (cos(53,13◦ )) (5) cos(θ) = F2y F2 F 2y sin(53,13◦ ) = 5 F2y = (sin(53,13◦ )) (5) sin(θ) = = 3,00 N = 4,00 N 5,0 0N 4,31 N θ 3,23 N Vector F1 F2 Resultant xcomponent 5,00 N 3,00 N ycomponent 2,00 N 4,00 N Total 5,385 N 5N Step 4: Determine the components of the resultant vector Now we have all the components. If we add all the horizontal components then → we will have the xcomponent of the resultant vector, Rx . Similarly, we add all the → vertical components then we will have the ycomponent of the resultant vector, Ry . Rx = F1x + F2x = 5,00 N + 3,00 N = 2,00 N + 4,00 N = 8,00 N = 6,00 N → Therefore, Rx is 8 N to the right. 42 Ry = F1y + F2y 1.3. Components of vectors → Therefore, Ry is 6 N up. Vector F1 F2 Resultant xcomponent 5,00 N 3,00 N 8,00 N ycomponent 2,00 N 4,00 N 6,00 N Total 5,385 N 5N Step 5: Determine the magnitude and direction of the resultant vector Now that we have the components of the resultant, we can use the Theorem of Pythagoras to determine the magnitude of the resultant, R. R2 = (Ry )2 + (Rx )2 = (6,00)2 + (8,00)2 = 100,00 R = 10,00 N 6N 8N 10 N α The magnitude of the resultant, R is 10,00 N. So all we have to do is calculate its direction. We can specify the direction as the angle the vectors makes with a known direction. To do this you only need to visualise the vector as starting at the origin of a coordinate system. We have drawn this explicitly below and the angle we will calculate is labelled α. Using our known trigonometric ratios we can calculate the value of α: tan α = 6,00 8,00 α = tan−1 6,00 8,00 α = 36,9◦ Step 6: Quote the final answer → ◦ R is 10 m at an angle of 36, 9 to the positive xaxis. Chapter 1. Vectors in two dimensions 43 Worked example 15: Resultant using components QUESTION Determine, by resolving into components, the resultant of the following four forces acting at a point: • F1 =3,5 N at 45◦ to the positive xaxis. • F2 =2,7 N at 63◦ to the positive xaxis. • F3 =1,3 N at 127◦ to the positive xaxis. • F4 =2,5 N at 245◦ to the positive xaxis. SOLUTION Step 1: Sketch the problem Draw all of the vectors on the Cartesian plane. This does not have to be precisely accurate because we are solving algebraically but vectors need to be drawn in the correct quadrant and with the correct relative positioning other. y 3 2 F2 F1 1 F3 −3 −2 −1 −1 F4 −2 1 2 3 x −3 We are going to record the various components in a table to help us manage keep track of the calculation. For each vector we need to determine the components in the x and ydirections. Vector F1 F2 F3 F4 R 44 xcomponent ycomponent 1.3. Components of vectors Total 3,5 N 2,7 N 1,3 N 2,5 N Step 2: Determine components of F1 y 3 2 F1 1 −3 −2 1 −1 −1 2 3 x −2 −3 Firstly we find the magnitude of the vertical component, F1y : Secondly we find the magnitude of the horizontal component, F1x : F1y F1 F 1y sin(45◦ ) = 3,5 F1y = (sin(45◦ )) (3,5) F1x F1 F 1x cos(45◦ ) = 3,5 F1x = (cos(45◦ )) (3,5) = 2,47 N = 2,47 N cos(θ) = sin(θ) = Step 3: Determine components of F2 y 3 2 F2 1 −3 −2 −1 −1 1 2 3 x −2 −3 Chapter 1. Vectors in two dimensions 45 Secondly we find the magnitude of the horizontal component, F2x : Firstly we find the magnitude of the vertical component, F2y : F2y F2 F2y sin(63◦ ) = 2,7 F2y = (sin(63◦ )) (2,7) sin(θ) = F2x F2 F 2x cos(63◦ ) = 2,7 F2x = (cos(63◦ )) (2,7) cos θ = = 2,41 N = 1,23 N Step 4: Determine components of F3 y 3 2 1 F3 −3 −2 −1 −1 1 2 3 x −2 −3 Firstly we find the magnitude of the vertical component, F3y : F3y F3 F3y sin(127◦ ) = 1,3 F3y = (sin 127◦ ) (1,3) sin(θ) = = 1,04 N Step 5: Determine components of F4 46 1.3. Components of vectors Secondly we find the magnitude of the horizontal component, F3x : F3x F3 F 3x cos(127◦ ) = 1,3 F3x = (cos 127◦ ) (1,3) cos(θ) = = −0,78 N y 3 2 1 −3 −2 −1 −1 F4 −2 1 3 x 2 −3 Secondly we find the magnitude of the horizontal component, F4x : Firstly we find the magnitude of the vertical component, F3y : F4y F4 F 4y sin(245◦ ) = 2,5 F4y = (sin(245◦ )) (2,5) sin(θ) = F4x F4 F 4x cos(245◦ ) = 2,5 F4x = (cos(245◦ )) (2,5) cos(θ) = = −2,27 N = −1,06 N Step 6: Determine components of resultant Sum the various component columns to determine the components of the resultant. Remember that if the component was negative don’t leave out the negative sign in the summation. Vector F1 F2 F3 F4 R xcomponent 2,47 N 1,23 N −0,78 N −1,06 N 1,86 N ycomponent 2,47 N 2,41 N 1,04 N −2,27 N 3,65 N Total 3,5 N 2,7 N 1,3 N 2,5 N Now that we have the components of the resultant, we can use the Theorem of Pythagoras to determine the magnitude of the resultant, R. R2 = (Ry )2 + (Rx )2 = (1,86)2 + (3,65)2 = 16,78 R = 4,10 N Chapter 1. Vectors in two dimensions 47 We can also determine the angle with the positive xaxis. tan(α) = 1,86 3,65 α = tan−1 ( 3,65 ) 1,86 α = 27,00◦ Step 7: Quote the final answer The resultant has a magnitude of 4,10 N at and angle of 27,00◦ to the positive xdirection. Informal experiment:Force board Aim: Determine the resultant of three nonlinear forces using a force board. Apparatus and materials: You will need: • blank paper • force board • 4 spring balances • assortment of weights 48 • gut or string • four pulleys1.3. Components of vectors Method: Before beginning the detailed method think about the strategy. By connecting a cord to the ring, running it over a pulley and hanging weights off it you can get a force exerted on the ring. The more weights or heavier the weight you hang the greater the force. The force is in the direction of the cord. If you run a number of cords over pulleys you are exerting more forces, all in different directions, on the ring. So we have a system where we can change the magnitude and direction of the forces acting on the ring. By putting a spring balance between the cord and the ring, we can measure the force. Putting a piece of paper under the ring allows us to draw the directions and the readings on the spring balances allows us to measure the magnitude. We are going to use this information to measure the forces acting on the rings and then we can determine the resultants graphically. 1. Set up the forceboard and place a piece of paper under the ring. 2. Set up four different forces by connecting a spring balance to the ring on the one side and some cord on the other side. Run the cord over a pulley and attach some weights to it. Work in a group to do this effectively. 3. Draw a line along each cord being careful not to move any of them. 4. Make a note of the reading on each spring balance. 5. Now remove the paper. 6. Working on the paper, draw each line back towards the where the centre of the ring had been. The lines should all intersect at a point. Make this point the centre of your Cartesian coordinate system. 7. Now choose an appropriate scale to relate the length of arrows to the readings on the spring balances. Using the appropriate spring balance and correct line on the paper, draw an arrow to represent each of the forces. Results: For two different choices of 3 of the force vectors we will determine the resultant. To determine the resultant we need to add the vectors together. The easiest way to do this is to replicate the vectors using a ruler and protractor and draw them tailtohead. Conclusions and questions: Note the direction and the magnitudes of the resultants of the various combinations. 1. How does the calculated resultant compare to the vector that wasn’t used to calculate the resultant in each case? 2. What general relationship should exist between the resultant and the fourth vector and why do you think this is the case? 3. Would this be the same if we had more or less forces in the problem? Justify your answer. Chapter 1. Vectors in two dimensions 49 See simulation: 23FV at www.everythingscience.co.za See video: 23FW at www.everythingscience.co.za 1.4 Chapter summary ESBKF See presentation: 23FX at www.everythingscience.co.za • A vector has a magnitude and direction. • Vectors can be used to represent many physical quantities that have a magnitude and direction, like forces. • Vectors may be represented as arrows where the length of the arrow indicates the magnitude and the arrowhead indicates the direction of the vector. • Vectors in two dimensions can be drawn on the Cartesian plane. • Vectors can be added graphically using the headtotail method or the tailtotail method. • A closed vector diagram is a set of vectors drawn on the Cartesian using the tailtohead method and that has a resultant with a magnitude of zero. • Vectors can be added algebraically using Pythagoras’ theorem or using components. • The direction of a vector can be found using simple trigonometric calculations. • The components of a vector are a series of vectors that, when combined, give the original vector as their resultant. • Components are usually created that align with the Cartesian coordinate axes. For a vector F that makes an angle of θ with the positive xaxis the xcomponent y = R sin(θ). x = R cos(θ) and the ycomponent is R is R Exercise 1 – 7: 1. Draw the following forces as vectors on the Cartesian plane originating at the origin: • F1 = 3,7 N in the positive xdirection • F2 = 4,9 N in the positive ydirection 2. Draw the following forces as vectors on the Cartesian plane: • F1 = 4,3 N in the positive xdirection • F2 = 1,7 N in the negative xdirection • F3 = 8,3 N in the positive ydirection 3. Find the resultant in the xdirection, Rx , and ydirection, Ry for the following forces: 50 1.4. Chapter summary • F1 = 1,5 N in the positive xdirection • F2 = 1,5 N in the positive xdirection • F3 = 2 N in the negative xdirection 4. Find the resultant in the xdirection, Rx , and ydirection, Ry for the following forces: • F1 = 4,8 N in the positive xdirection • F2 = 3,2 N in the negative xdirection • F3 = 1,9 N in the positive ydirection • F4 = 2,1 N in the negative ydirection 5. Find the resultant in the xdirection, Rx , and ydirection, Ry for the following forces: • F1 = 2,7 N in the positive xdirection • F2 = 1,4 N in the positive xdirection • F3 = 2,7 N in the negative xdirection • F4 = 1,7 N in the negative ydirection 6. Sketch the resultant of the following force vectors using the tailtohead method: • F1 = 4,8 N in the positive ydirection • F2 = 3,3 N in the negative xdirection 7. Sketch the resultant of the following force vectors using the tailtohead method: • F1 = 0,7 N in the positive ydirection • F2 = 6 N in the positive xdirection • F3 = 3,8 N in the negative ydirection • F4 = 11,9 N in the negative xdirection 8. Sketch the resultant of the following force vectors using the tailtohead method by first determining the resultant in the x and ydirections: • F1 = 5,2 N in the positive ydirection • F2 = 7,5 N in the negative ydirection • F3 = 4,8 N in the positive ydirection • F4 = 6,3 N in the negative xdirection 9. Sketch the resultant of the following force vectors using the tailtohead method by first determining the resultant in the x and ydirections: • F1 = 6,7 N in the positive ydirection • F2 = 4,2 N in the negative xdirection • F3 = 9,9 N in the negative ydirection • F4 = 3,4 N in the negative ydirection 10. Sketch the resultant of the following force vectors using the tailtotail method: Chapter 1. Vectors in two dimensions 51 • F1 = 6,1 N in the positive ydirection • F2 = 4,5 N in the negative xdirection 11. Sketch the resultant of the following force vectors using the tailtotail method by first determining the resultant in the x and ydirections: • F1 = 2,3 N in the positive ydirection • F2 = 11,8 N in the negative ydirection • F3 = 7,9 N in the negative ydirection • F4 = 3,2 N in the negative xdirection 12. Four forces act simultaneously at a point, find the resultant if the forces are: • F1 = 2,3 N in the positive xdirection • F2 = 4,9 N in the positive ydirection • F3 = 4,3 N in the negative ydirection • F4 = 3,1 N in the negative ydirection 13. Resolve each of the following vectors into components: a) F1 =105 N at 23,5◦ to the positive xaxis. b) F2 =27 N at 58,9◦ to the positive xaxis. c) F3 =11,3 N at 323◦ to the positive xaxis. d) F4 =149 N at 245◦ to the positive xaxis. e) F5 =15 N at 375◦ to the positive xaxis. f) F6 =14,9 N at 75,6◦ to the positive xaxis. g) F7 =11,3 N at 123,4◦ to the positive xaxis. h) F8 =169 N at 144◦ to the positive xaxis. 14. A point is acted on by two forces and the resultant is zero. The forces a) have equal magnitudes and directions. b) have equal magnitudes but opposite directions. c) act perpendicular to each other. d) act in the same direction. 15. A point in equilibrium is acted on by three forces. Force F1 has components 15 N due south and 13 N due west. What are the components of force F2 ? a) 13 N due north and 20 N due west b) 13 N due north and 13 N due west c) 15 N due north and 7 N due west d) 15 N due north and 13 N due east N F2 W 20 N F1 S 52 1.4. Chapter summary E 16. Two vectors act on the same point. What should the angle between them be so that a maximum resultant is obtained? a) 0◦ b) 90◦ c) 180◦ d) cannot tell 17. Two forces, 4 N and 11 N, act on a point. Which one of the following cannot be the magnitude of a resultant? a) 4 N b) 7 N c) 11 N d) 15 N 18. An object of weight W is supported by two cables attached to the ceiling and wall as shown. The tensions in the two cables are T1 and T2 respectively. Tension T1 = 1200 N. Determine the tension T2 by accurate construction and measurement or by calculation. 45◦ T1 70◦ T2 W 19. An object X is supported by two strings, A and B, attached to the ceiling as shown in the sketch. Each of these strings can withstand a maximum force of 700 N. The weight of X is increased gradually. 45◦ 30◦ B A X a) Draw a rough sketch of the triangle of forces, and use it to explain which string will break first. b) Determine the maximum weight of X which can be supported. Think you got it? Get this answer and more practice on our Intelligent Practice Service 1. 23FY 7. 23G6 13a. 23GD 13g. 23GM 18. 23GT 2. 23FZ 8. 23G7 13b. 23GF 13h. 23GN 19. 23GV 3. 23G2 9. 23G8 13c. 23GG 14. 23GP www.everythingscience.co.za 4. 23G3 10. 23G9 13d. 23GH 15. 23GQ 5. 23G4 11. 23GB 13e. 23GJ 16. 23GR 6. 23G5 12. 23GC 13f. 23GK 17. 23GS m.everythingscience.co.za Chapter 1. Vectors in two dimensions 53 CHAPTER Newton’s laws 2.1 Introduction 56 2.2 Force 56 2.3 Newton’s laws 74 2.4 Forces between masses 109 2.5 Chapter summary 119 2 2 Newton’s laws 2.1 Introduction ESBKG In this chapter we will learn how a net force is needed to modify the motion of an object. We will recall what a force is and learn about how force and motion are related. We are also introduced to Newton’s three laws and we will learn more about the force of gravity. Key Mathematics Concepts • Ratio and proportion — Physical Sciences, Grade 10, Science skills • Equations — Mathematics, Grade 10, Equations and inequalities • Units and unit conversions — Physical Sciences, Grade 10, Science skills 2.2 Force ESBKH What is a force? ESBKJ A force is anything that can cause a change to objects. Forces can do things like: • change the shape of an object, • accelerate or stop an object, and • change the direction of a moving object. A force can be classified as either a contact force or a noncontact force. Figure 2.1: Contact forces 56 2.1. Introduction A contact force must touch or be in contact with an object to cause a change. Examples of contact forces are: • the force that is used to push or pull things, like on a door to open or close it • the force that a sculptor uses to turn clay into a pot • the force of the wind to turn a windmill Figure 2.2: Contact forces A noncontact force does not have to touch an object to cause a change. Examples of noncontact forces are the forces due to: • gravity, like the Earth pulling the Moon towards itself; • electricity, like a proton and an electron attracting each other; and • magnetism, like a magnet pulling a paper clip towards itself. The unit of force in the international system of units (S.I. units) is the newton (symbol N). This unit is named after Sir Isaac Newton who first defined force. Force is a vector quantity and so it has a magnitude and a direction. We → − use the symbol F for force. Figure 2.3: Noncontact forces This chapter will often refer to the resultant force acting on an object. The resultant force is simply the vector sum of all the forces acting on the object. It is very important to remember that all the forces must be acting on the same object. The resultant force is the force that has the same effect as all the other forces added together. Chapter 2. Newton’s laws 57 Different types of forces in physics A lot of the physics topics you will study revolve around the impact or effect of forces. Although there are many different forces, will learn some fundamental principles for approaching the problems and applications in this book no matter which force applies Physics is the study of the natural world and you probably know a lot more physics than you think. You see things happening everyday that are governed by the laws of physics but you probably aren’t thinking about physics at the time. If you throw a stone up in the air it eventually falls to the ground. A lot of physics can be learnt by analysing an everyday situation. We are going to learn about some forces in the next few sections but before we start lets describe an everyday situation in which they all play a role so you can visualise what is happening. You need a table and a three books that all have different masses. Take any book and put it on the table. Nothing happens, the book just rests on the table if the table is flat. If you slowly lift one side of the table so that the top of the table is tilted the book doesn’t move immediately. As you lift the table more and more the book suddenly starts to slide off the table. You can repeat this with all three books and see how much you have to tilt the table before the books start to slide. This real world situation illustrates a lot of the physics we want to learn about in this chapter. The normal force 58 2.2. Force ESBKK When an object is placed on a surface, for example think of the case of putting a book on a table, there are a number of forces acting. Firstly, if the table were not there the book would fall to the floor. The force that causes this is gravity. The table stops the book falling to the floor. The only way this can happen is for the table to exert a force on the book. The force that the table exerts on the book must balance out the force of gravity. This tells us a few things immediately! Gravity is a force pulling the book down, it is a vector. The force that the table exerts must balance this out and it can only do this if it has the same magnitude and acts in the opposite direction. This occurs often, gravity pulls a person towards the earth but when you are standing on the ground something must be balancing it, if you put a heavy box on the ground the gravitational force is balanced. If you put a brick on water it will sink because nothing balances the gravitational force. We give the force that a surface (any surface) exerts to balance the forces on an object in contact with that surface the normal force. The normal force is a force that acts on the object as a result of the interaction with the surface and is perpendicular to the surface. This last part might be seem unexpected (counterintuitive) because if we tilt the table slightly the direction of the gravitational force hasn’t changed but the direction of the normal force has a little (the normal is not always directly opposite gravity). Don’t panic, this will all make sense before the end of this chapter. Remember: the normal force is always perpendicular (at a right angle) to the surface. DEFINITION: Normal force , is the force exerted by a surface on an object in contact with it. The normal force, N N N Friction forces Why does a box sliding on a surface eventually come to a stop? The answer is friction. Friction arises where two surfaces are in contact and moving relative to each other. For an everyday example, press your hands together and move one backwards and forwards, we have two surfaces in contact with one moving relative to the other. Your Chapter 2. Newton’s laws 59 hands get warm, you will have experienced this before and probably rub your hands together in winter to warm them up. The heat is generated through friction. Friction arises because the surfaces interact with each other. Think about sandpaper with lots of bumps on the surface. If you rub sandpaper the bumps will slot into any groove . When the surface of one object slides over the surface of another, each body exerts a frictional force on the other. For example if a book slides across a table, the table exerts a frictional force onto the book and the book exerts a frictional force onto the table. Frictional forces act parallel to surfaces. DEFINITION: Frictional force Frictional force is the force that opposes the motion of an object in contact with a surface and it acts parallel to the surface the object is in contact with. rough table box applied force The magnitude of the frictional force depends on the surface and the magnitude of the normal force. Different surfaces will give rise to different frictional forces, even if the normal force is the same. Frictional forces are proportional to the magnitude of the normal force. Ffriction ∝ N For every surface we can determine a constant factor, the coefficient of friction, that allows us to calculate what the frictional force would be if we know the magnitude of the normal force. We know that static friction and kinetic friction have different magnitudes so we have different coefficients for the two types of friction: • µs is the coefficient of static friction • µk is the coefficient of kinetic friction A force is not always large enough to make an object move, for example a small applied force might not be able to move a heavy crate. The frictional force opposing the motion of the crate is equal to the applied force but acting in the opposite direction. This frictional force is called static friction. When we increase the applied force (push harder), the frictional force will also increase until it reaches a maximum value. When the applied force is larger than the maximum force of static friction the object will move. The static frictional force can vary from zero (when no other forces are present and the object is stationary) to a maximum that depends on the surfaces. 60 2.2. Force fsmax tic 3 2 St a Frictional Force (N ) 4 Kinetic 1 0 0 1 2 3 4 Applied Force (N ) For static friction the force can vary up to some maximum value after which friction has been overcome and the object starts to move. So we define a maximum value for the static friction: fsmax = µs N . When the applied force is greater than the maximum, static frictional force, the object moves but still experiences friction. This is called kinetic friction. For kinetic friction the value remains the same regardless of the magnitude of the applied force. The magnitude of the kinetic friction is: fk = µk N . Remember that static friction is present when the object is not moving and kinetic friction while the object is moving. For example when you drive at constant velocity in a car on a tar road you have to keep the accelerator pushed in slightly to overcome the friction between the tar road and the wheels of the car. However, while moving at a constant velocity the wheels of the car are rolling, so this is not a case of two surfaces “rubbing” against each other and we are in fact looking at static friction. If you should break hard, causing the car to skid to a halt, we would be dealing with two surfaces rubbing against each other and hence kinetic friction. The higher the value for the coefficient of friction, the more ’sticky’ the surface is and the lower the value, the more ’slippery’ the surface is. Friction is very useful. If there was no friction and you tried to prop a ladder up against a wall, it would simply slide to the ground. Rock climbers use friction to maintain their grip on cliffs. The brakes of cars would be useless if it wasn’t for friction! Early humans made use of friction to create fire. Friction can create a lot of heat and Chapter 2. Newton’s laws 61 the early humans used this fact when they rubbed two sticks together to start a fire. When you rub your hands together fast and pressing hard you will feel that they get warm. This is heat created by the friction. You can use this to start a fire. To start a fire you need two pieces of wood, one long straight, round piece approximately the same thickness as your finger and about 40 cm long as well as a thicker flat piece of wood. The flat thick piece of wood needs a hole that the long straight one can fit into. Then you put the flat piece on the ground, the long straight one in the hole and rub it between your hands applying downwards pressure to increase the normal force and the amount of friction. Where the two pieces of wood rub against each other the friction results in many tiny pieces of wood being rubbed off and getting hot. See video: 23GW www.everythingscience.co.za at After a while the hole will start to smoke. At this point the smoking wood bits, called the ember, need to be gently tipped out of the hole into a small bed of dry grass. You cover the ember completely and blow gently. The grass should start to burn. Then you use the burning grass to light some dry twigs, and keep working your way up to bigger pieces of wood. To make it even easier, a bow from wood with a string can be used to cause the wood to turn. By twisting the string from the bow around the long piece of wood it can be driven without requiring a person use their hands. Worked example 1: Static friction QUESTION A box resting on a surface experiences a normal force of magnitude 30 N and the coefficient of static friction tween the surface and the box, µs , is 0,34. What is the maximum static frictional force? SOLUTION Step 1: Maximum static friction We know that the relationship between the maximum static friction, fsmax , the coefficient of static friction, µs and the normal, N , to be: fsmax = µs N We have been given that µs = 0,34 and N = 30 N. This is all of the information required to do the calculation. Step 2: Calculate the result 62 2.2. Force fsmax = µs N = (0,34)(30) = 10,2 The maximum magnitude of static friction is 10,2 N. Worked example 2: Static friction QUESTION The forwards of your school’s rugby team are trying to push their scrum machine. The normal force exerted on the scrum machine is 10 000 N. The machine isn’t moving at all. If the coefficient of static friction is 0,78 what is the minimum force they need to exert to get the scrum machine to start moving? SOLUTION Step 1: Minimum or maximum The question asks what the minimum force required to get the scrum machine moving will be. We don’t know a relationship for this but we do know how to calculate the maximum force of static friction. The forwards need to exert a force greater than this so the minimum amount they can exert is in fact equal to the maximum force of static friction. Step 2: Maximum static friction We know that the relationship between the maximum static friction, fsmax , the coefficient of static friction, µs and the normal, N , to be: fsmax = µs N We have been given that µs = 0,78 and N = 10 000 N. This is all of the information required to do the calculation. Step 3: Calculate the result fsmax = µs N = (0,78)(10 000) = 7800 N The maximum magnitude of static friction is 7800 N. Chapter 2. Newton’s laws 63 Worked example 3: Kinetic friction QUESTION The normal force exerted on a pram is 100 N. The pram’s brakes are locked so that the wheels cannot turn. The owner tries to push the pram but it doesn’t move. The owner pushes harder and harder until it suddenly starts to move when the applied force is three quarters of the normal force. After that the owner is able to keep it moving with a force that is half of the force at which it started moving. What is the magnitude of the applied force at which it starts moving and what are the coefficients of static and kinetic friction? SOLUTION Step 1: Maximum static friction The owner of the pram increases the force he is applying until suddenly the pram starts to move. This will be equal to the maximum static friction which we know is given by: fsmax = µs N We are given that the magnitude of the applied force is three quarters of the normal force magnitude, so: 3 N 4 3 = (100) 4 = 75 N fsmax = Step 2: Coefficient of static friction We now know both the maximum magnitude of static friction and the magnitude of the normal force so we can find the coefficient of static friction: fsmax = µs N 75 = µs (100) µs = 0,75 Step 3: Coefficient of kinetic friction The magnitude of the force required to keep the pram moving is half of the magnitude of the force required to get it to start moving so we can determine it from: 1 max f 2 s 1 = (75) 2 = 37,5 N fk = We know the relationship between the magnitude of the kinetic friction, magnitude of the normal force and coefficient of kinetic friction. We can use it to solve for the 64 2.2. Force coefficient of kinetic friction: f k = µk N 37,5 = µk (100) µk = 0,375 Worked example 4: Coefficient of static friction QUESTION A block of wood experiences a normal force of 32 N from a rough, flat surface. There is a rope tied to the block. The rope is pulled parallel to the surface and the tension (force) in the rope can be increased to 8 N before the block starts to slide. Determine the coefficient of static friction. SOLUTION Step 1: Analyse the question and determine what is asked The normal force is given (32 N) and we know that the block does not move until the applied force is 8 N. We are asked to find the coefficient for static friction µs . Step 2: Find the coefficient of static friction F f = µs N 8 = µs (32) µs = 0,25 Note that the coefficient of frict