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= V3 The total current passing through the circuit will be the sum of the individual currents passing through each resistor. IT = I1 + I2 + I3 If we return to the analogy of a river, a parallel circuit is the same as the river breaking into three streams, which later rejoin to one river again. The amount of water ﬂowing in the river is equal to the sum of the amounts of water ﬂowing in the individual streams. Ohm’s Law applies to resistors in parallel, just as it did to resistors in a series. The current ﬂowing through each resistor is equal to the total voltage drop divided by the resistance in that resistor. I1 = VT R1 and I2 = VT R2 and I3 = VT R3 Since IT = I1 + I2 + I3, + VT then IT = VT R3 R1 + VT = VT and VT R3 R1 RT + VT R2 + VT R2,. If we divide both sides of the ﬁnal equation by VT, we get the relationship between the total resistance of the circuit and the individual parallel resistances in the circuit. The total resistance is sometimes called the equivalent resistance. 1 RT = 1 R1 + 1 R2 + 1 R3 Consider the parallel circuit sketched below. The voltage drop for the entire circuit is 90. V. Therefore, the voltage drop in each of the resistors is also 90. V. 29 www.ck12.org The current through each resistor can be found using the voltage drop and the resistance of that resistor: I1 = VT R1 = 90: V 60: W = 1:5 A I2 = VT R2 = 90: V 30: W = 3:0 A I3 = VT R3 = 90: V 30: W = 3:0 A The total current through the circuit would be the sum of the three currents in the individual resistors. IT = I1 + I2 = I3 = 1:5 A + 3:0 A + 3:0 A = 7:5 A The equivalent resistance for this circuit is found using the equation above. 1 RT = 1 R1 + 1 R2 + 1 R3 = 1 60: W + 1 30: W + 1 30: W = 1 60: W + 2 60: W + 2 60: W = 5 60: W RT = 60: W 5 = 12 W The equivalent resistance for the
circuit could also be found by using the total voltage drop and the total current. RT = VT IT 7:5 A = 12 W = 90: W Summary • Parallel electrical circuits have multiple paths the current may take. • VT = V1 = V2 = V3. • IT = I1 + I2 + I3. + 1 • R3 = 1 R1 = 1 R2 1 RT. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=apHkG4T6QHM Follow up questions: 1. Why do the light bulbs glow less brightly when connected across a 120 V source in a series circuit than when connected across the same 120 V source in a parallel circuit? 2. Why do the other bulbs go dark when one bulb is removed in the series circuit but the other bulbs do not go dark when one bulb is removed in the parallel circuit? 30 www.ck12.org Review Concept 10. Parallel Circuits 1. Three 15.0W resistors are connected in parallel and placed across a 30.0 V potential difference. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through a single branch of the circuit? 2. A 12.0W and a 15.0W resistor are connected in parallel and placed across a 30.0 V potential. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through each branch of the circuit? 3. A 120.0W resistor, a 60.0W resistor, and a 40.0W resistor are connected in parallel and placed across a potential difference of 12.0 V. (a) What is the equivalent resistance of the parallel circuit? (b) What is the total current through the circuit? (c) What is the current through each branch of the circuit? • parallel circuit: A closed electrical circuit in which the current is divided into two or more paths and then returns via a common path to complete the circuit. • equivalent resistance: A single resistance that would cause the same power loss as the actual resistance values distributed throughout a circuit. References 1. Image copyright vilax, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. Light bulb: Image copyright Snez, 2013; composite created
by CK-12 Foundation - Samantha Bacic. http:// www.shutterstock.com. Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 31 CONCEPT 11 Resistors in Parallel Students will learn how to analyze and solve problems involving circuits with resistors in parallel. Students will learn how to analyze and solve problems involving circuits with resistors in parallel. www.ck12.org Key Equations Guidance 1 Rtotal = 1 R1 + 1 R2 + 1 R3 + : : : Resistors in Parallel: All resistors are connected together at both ends. There are many rivers (i.e. The main river branches off into many other rivers), so all resistors receive different amounts of current. But since they are all connected to the same point at both ends they all receive the same voltage. Example 1 A circuit is wired up with 2 resistors in parallel. Both resistors are directly connected to the power supply, so both have the same 20V across them. But they are on different ’rivers’ so they have different current ﬂowing through them. Lets go through the same questions and answers as with the circuit in series. Question: What is the total resistance of the circuit? = 1 Answer: The total resistance is Note: Total resistance for a circuit in parallel will always be smaller than smallest resistor in the circuit. 90W thus, Rtotal = 90W 90W = 10 10W = 1 90W + 9 90W + 1 = 1 R1 + 1 R2 10 = 9W 1 Rtotal Question: What is the total current coming out of the power supply? Answer: Use Ohm’s Law (V = IR) but solve for current (I = V =R). Itotal = Vtotal Rtotal = 20V 9W = 2:2A Question: How much power does the power supply dissipate? Answer: P = IV, so the total power equals the total voltage multiplied by the total current. Thus, Ptotal = ItotalVtotal = (2:2A)(20V ) = 44:4W. So the Power Supply outputs 44W (i.e. 44 Joules of energy per second). 32 www.ck12.org Concept 11. Resistors in Parallel Question: How much power is each resistor dissipating? Answer: Each resistor has different current across it, but the same voltage. So
, using Ohm’s law, convert the power formula into a form that does not depend on current. P = IV = V R Substituted I = V =R into the power R formula. P90W = V 2 10W = 40W 90W R90W Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should–Energy is always conserved. 90W = 4:4W ; P10W = V 2 R10W = (20V )2 V = V 2 = (20V )2 10W Question: How much current is ﬂowing through each resistor? Answer: Use Ohm’s law to calculate the current for each resistor. I90W = V90W R90W = 20V 90W = 0:22A I10W = V10W R10W = 20V 10W = 2:0A Notice that the 10W resistor has the most current going through it. It has the least resistance to electricity so this makes sense. Note: If you add up the currents of the individual ’rivers’ you get the total current of the of the circuit, as you should. Watch this Explanation MEDIA Click image to the left for more content. Simulation • http://simulations.ck12.org/Resistor/ 33 Time for Practice 1. Three 82 W resistors and one 12 W resistor are wired in parallel with a 9 V battery. a. Draw the schematic diagram. b. What is the total resistance of the circuit? 2. What does the ammeter read and which resistor is dissipating the most power? www.ck12.org 3. Given three resistors, 200 W; 300 W and 600 W and a 120 V power source connect them in a way to heat a container of water as rapidly as possible. a. Show the circuit diagram b. How many joules of heat are developed after 5 minutes? Answers to Selected Problems 1. b. 8:3 W 2. 0:8A and the 50 W on the left 3. part 2 43200J. 34 www.ck12.org Concept 12. Combined Series-Parallel Circuits CONCEPT 12 Combined Series-Parallel Circuits • Solve problems of combined circuits. Electrical circuits can become immensely complicated. This circuit is a polynomial plotter, which allows users to plot polynomials and evaluate functions at various x
values. Combined Series-Parallel Circuits Most circuits are not just a series or parallel circuit; most have resistors in parallel and in series. These circuits are called combination circuits. When solving problems with such circuits, use this series of steps. 1. For resistors connected in parallel, calculate the single equivalent resistance that can replace them. 2. For resistors in series, calculate the single equivalent resistance that can replace them. 3. By repeating steps 1 and 2, you can continually reduce the circuit until only a single equivalent resistor remains. Then you can determine the total circuit current. The voltage drops and currents though individual resistors can then be calculated. Example Problem: the total current through the circuit, and ﬁnd the current through each individual resistor. In the combination circuit sketched below, ﬁnd the equivalent resistance for the circuit, ﬁnd 35 www.ck12.org Solution: We start by simplifying the parallel resistors R2 and R3. = 1 180 W 1 R23 R23 = 99 W + 1 220 W = 1 99 W We then simplify R1 and R23 which are series resistors. RT = R1 + R23 = 110 W + 99 W = 209 W We can then ﬁnd the total current, IT = VT RT All the current must pass through R1, so I1 = 0:11 A. = 24 V 209 W = 0:11 A The voltage drop through R1 is (110 W)(0:11 A) = 12:6 volts. Therefore, the voltage drop through R2 and R3 is 11.4 volts. I2 = V2 220 W = 0:052 A R2 180 W = 0:063 A and I3 = V3 R3 = 11:4 V = 11:4 V Summary • Combined circuit problems should be solved in steps. Practice Video teaching the process of simplifying a circuit that contains both series and parallel parts. MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=In3NF8f-mzg Follow up questions: 1. In a circuit that contains both series and parallel parts, which parts of the circuit are simpliﬁed ﬁrst? 2. In the circuit drawn below, which resistors should be simpliﬁed ﬁrst? 36 www.ck12.org Concept 12. Combined Series-Parallel Circ
uits Review 1. Two 60.0W resistors are connected in parallel and this parallel arrangement is then connected in series with a 30.0W resistor. The combination is placed across a 120. V potential difference. (a) Draw a diagram of the circuit. (b) What is the equivalent resistance of the parallel portion of the circuit? (c) What is the equivalent resistance for the entire circuit? (d) What is the total current in the circuit? (e) What is the voltage drop across the 30.0W resistor? (f) What is the voltage drop across the parallel portion of the circuit? (g) What is the current through each resistor? 2. Three 15.0 Ohm resistors are connected in parallel and the combination is then connected in series with a 10.0 Ohm resistor. The entire combination is then placed across a 45.0 V potential difference. Find the equivalent resistance for the entire circuit. • combined or combination circuits: A route for the ﬂow of electricity that has elements of both series and parallel circuits. References 1. User:Linkaddict/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Huge_circuit.JPG. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 37 Physics Unit 14: Magnetism Patrick Marshall Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation,
www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents www.ck12.org Contents 1 Properties of Magnets 2 Magnetic Fields 1 6 iv www.ck12.org Concept 1. Properties of Magnets CONCEPT 1 Properties of Magnets • Describe magnetic ﬁelds around permanent magnets. • Understand ferromagnetism and magnetic domains. • Describe some properties of magnets. Some countries are using powerful electromagnets to develop high-speed trains, called maglev, or magnetic levitation, trains. These trains use the repulsive force of magnets to ﬂoat over a guide way, removing the friction of steel wheels and train tracks. Reducing this friction allows the trains to travel at much higher speeds. Properties of Magnets Any magnet, regardless of its shape, has two ends called poles where the magnetic effect is strongest. If a magnet is suspended by a ﬁne thread, it is found that one pole of the magnet will always point toward the north. This fact has been made use of in navigation since the eleventh century. The pole of the magnet that seeks the north pole is called the north pole of the magnet, while the opposite side is the south pole. It is a familiar fact that when two magnets are brought near one another, the magnets exert
a force on each other. The magnetic force can be either attractive or repulsive. If two north poles or two south poles are brought near each other, the force will be repulsive. If a north pole is brought near a south pole, the force will be attractive. 1 www.ck12.org The Earth’s geographic north pole (which is close to, but not exactly at the magnetic pole) attracts the north poles of magnets. We know, therefore, that this pole is actually the Earth’s magnetic south pole. This can be seen in the image above; the geographic north and south poles are labeled with barber shop poles, and the Earth’s magnetic poles are indicated with the double-headed arrow. Only iron and few other materials such as cobalt, nickel, and gadolinium show strong magnetic effects. These materials are said to be ferromagnetic. Other materials show some slight magnetic effect but it is extremely small and can be detected only with delicate instruments. Ferromagnetic Domains Microscopic examination reveals that a magnet is actually made up of tiny regions known as magnetic domains, which are about one millimeter in length and width. Each domain acts like a tiny magnet with a north and south pole. 2 www.ck12.org Concept 1. Properties of Magnets When the ferrous material is not magnetized, the domains are randomly organized so that the north and south poles do not line up and often cancel each other. When the ferrous material is placed in a magnetic ﬁeld, the domains line up with the magnetic ﬁeld so that the north poles are all pointed in the same direction and the south poles are all pointed in the opposite direction. In this way, the ferrous material has become a magnet. In many cases, the domains will remain aligned only while the ferrous material is in a strong magnetic ﬁeld; when the material is removed from the ﬁeld, the domains return to their previous random organization and the ferrous material loses any magnetic properties. Magnets that have magnetic properties while in the ﬁeld of another magnet but lose the magnetic properties when removed from the ﬁeld are called temporary magnets. Under certain circumstances, however, the new alignment can be made permanent and the ferrous substance becomes a permanent magnet. That is, the ferrous object remains a magnet even when removed from the other magnetic ﬁeld. The formation of temporary magnets allows
a magnet to attract a non-magnetized piece of iron. You have most likely seen a magnet pick up a paper clip. The presence of the magnet aligns the domains in the iron paper clip and it becomes a temporary magnet. Whichever pole of the magnet is brought near the paper clip will induce magnetic properties in the paper clip that remain as long as the magnet is near. Permanent magnets lose their magnetic properties when the domains are dislodged from their organized positions and returned to a random jumble. This can occur if the magnet is hammered on or if it is heated strongly. Magnetic Fields When we were dealing with electrical effects, it was very useful to speak of an electric ﬁeld that surrounded an electric charge. In the same way, we can imagine a magnetic ﬁeld surrounding a magnetic pole. The force that one magnet exerts on another can be described as the interaction between one magnet and the magnetic ﬁeld of the other magnet. Magnetic ﬁeld lines go from the north magnetic pole to the south magnetic pole. We deﬁne the magnetic ﬁeld at any point as a vector (represented by the letter B ) whose direction is from north to south magnetic poles. 3 Summary • Any magnet has two ends called poles where the magnetic effect is strongest. • The magnetic pole found at the north geographical pole of the earth is a south magnetic pole. • The force that one magnet exerts on another can be described as the interaction between one magnet and the magnetic ﬁeld of the other magnet. • Magnetic ﬁeld lines go from the north magnetic pole to the south magnetic pole. www.ck12.org Practice MEDIA Click image to the left for more content. http://www.darktube.org/watch/crealev-levitating-ﬂoating-flying-hovering-bouncing This video demonstrates magnetic levitation. 1. In the video, one object rests on top of the magnetic ﬁeld of another. Compare the friction between these two objects to the friction between a saucer and a table the saucer rests on. Review 1. The earth’s magnetic ﬁeld (a) has a north magnetic pole at exactly the same spot as the geographical north pole. (b) is what causes compasses to work. (c) is what causes electromagnets to work. (d) all of these are true. (e
) none of these are true. 2. A material that can be permanently magnetized is generally said to be (a) magnetic. (b) electromagnetic. (c) ferromagnetic. (d) none of these are true. 3. The force between like magnetic poles will be (a) repulsive. (b) attractive. (c) could be repulsive or attractive. 4. Why is a magnet able to attract a non-magnetic piece of iron? 5. If you had two iron rods and noticed that they attract each other, how could you determine if both were magnets or only one was a magnet? • magnet: A body that can attract certain substances, such as iron or steel, as a result of a magnetic ﬁeld. 4 www.ck12.org Concept 1. Properties of Magnets • magnetic pole: Either of two regions of a magnet, designated north and south, where the magnetic ﬁeld is strongest. Electromagnetic interactions cause the north poles of magnets to be attracted to the south poles of other magnets, and conversely. The north pole of a magnet is the pole out of which magnetic lines of force point, while the south pole is the pole into which they point. • ferromagnetic: A body or substance having a high susceptibility to magnetization, the strength of which depends on that of the applied magnetizing ﬁeld, and which may persist after removal of the applied ﬁeld. This is the kind of magnetism displayed by iron, and is associated with parallel magnetic alignment of adjacent domains. • magnetic ﬁeld: A ﬁeld of force surrounding a permanent magnet or a moving charged particle. • magnetic domain: An atom or group of atoms within a material that have some kind of “net” magnetic ﬁeld. • temporary magnet: A piece of iron that is a magnet while in the presence of another magnetic ﬁeld but loses its magnetic characteristics when the other ﬁeld is removed. • permanent magnet: A piece of magnetic material that retains its magnetism after it is removed from a magnetic ﬁeld. References 1. User:JakeLM/Wikipedia. http://commons.wikimedia.org/wiki/File:Maglev_june2005.jpg. CC-BY 2.5 2. Globe: Image copyright pdesign, 2013; Poles: Image copyright lineartestpilot, 2013;
Composite created by http://www.shutterstock.com. Used under licenses from Shutter- CK-12 Foundation - Samantha Bacic. stock.com 5 CONCEPT 2 www.ck12.org Magnetic Fields Students will learn the idea of magnetic ﬁeld lines, how they behave in the situation of permanent magnets and current carrying wires and also how to calculate the magnetic ﬁeld at an arbitrary distance from the wire. Students will learn the idea of magnetic ﬁeld lines, how they behave in the situation of permanent magnets and current carrying wires and also how to calculate the magnetic ﬁeld at an arbitrary distance from the wire. Key Equations µ0I 2pr Where µ0 = 4p 107 Tm/A Bwire = Magnetic ﬁeld at a distance r from a current-carrying wire Permeability of Vacuum (approximately same for air also) Guidance Permanent magnets (like refrigerator magnets) consist of atoms, such as iron, for which the magnetic moments (roughly electron spin) of the electrons are “lined up” all across the atom. This means that their magnetic ﬁelds add up, rather than canceling each other out. The net effect is noticeable because so many atoms have lined up. The magnetic ﬁeld of such a magnet always points from the north pole to the south. The magnetic ﬁeld of a bar magnet, for example, is illustrated below: If we were to cut the magnet above in half, it would still have north and south poles; the resulting magnetic ﬁeld would be qualitatively the same as the one above (but weaker). Charged particles in motion also generate magnetic ﬁelds. The most frequently used example is a current carrying wire, since current is literally moving charged particles. The magnitude of a ﬁeld generated by a wire depends on distance to the wire and strength of the current (I) (see ’Key Equations’ section) : Meanwhile, its direction can be found using the so called ﬁrst right hand rule : point your thumb in the direction of the current. Then, curl your ﬁngers around the wire. The direction your ﬁngers will point in the same direction as the ﬁeld. Be sure to use your right hand! 6 www.ck12.org Concept 2. Magnetic Fields Sometimes, it is
necessary to represent such three dimensional ﬁelds on a two dimensional sheet of paper. The following example illustrates how this is done. In the example above, a current is running along a wire towards the top of your page. The magnetic ﬁeld is circling the current carrying wire in loops which are perpendicular to the page. Where these loops intersect this piece of paper, we use the symbol J to represent where the magnetic ﬁeld is coming out of the page and the symbol N to. This convention can be used for all vector quantities: represent where the magnetic ﬁeld is going into the page ﬁelds, forces, velocities, etc. Example 1 You are standing right next to a current carrying wire and decide to throw your magnetic ﬁeld sensor some distance perpendicular to the wire. When you go to retrieve your sensor, it shows the magnetic ﬁeld where it landed to be 4 105 T. If you know the wire was carrying 300A, how far did you throw the sensor? Solution To solve this problem, we will just use the equation given above and solve for the radius. 7 www.ck12.org B = r = r = µoI 2pr µoI 2pB 4p 107 Tm/A 300 A 2p 4 105 T r = 1:5 m MEDIA Click image to the left for more content. Watch this Explanation Simulation Magnet and Compass (PhET Simulation) Time for Practice 1. Sketch the magnetic ﬁeld lines for the horseshoe magnet shown here. Then, show the direction in which the two compasses (shown as circles) should point considering their positions. In other words, draw an arrow in the compass that represents North in relation to the compass magnet. 8 www.ck12.org Concept 2. Magnetic Fields 2. Find the magnetic ﬁeld a distance of 20 cm from a wire that is carrying 3 A of electrical current. 3. In order to measure the current from big power lines the worker simply clamps a device around the wire. This provides safety when dealing with such high currents. The worker simply measures the magnetic ﬁeld and deduces the current using the laws of physics. Let’s say a worker uses such a clamp and the device registers a magnetic ﬁeld of 0.02 T. The clamp is 0.05 m from the wire. What is the electrical
current in the wire? Answers to Selected Problems 1. Both pointing away from north 2. 3 106T 3. 5000 A 9 Physics Unit 15: Electromagnetism Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James H Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: April 6, 2014 iii Contents
www.ck12.org 1 4 8 13 18 22 26 Contents 1 Electromagnetic Induction 2 Electromagnets 3 Current and Magnetism 4 Electric Motors 5 Electromotive Force 6 Electric Generators 7 Transformers iv www.ck12.org Concept 1. Electromagnetic Induction CONCEPT 1 Electromagnetic Induction • Deﬁne electromagnetic induction. • Explain how electromagnetic induction occurs. • Describe the current produced by electromagnetic induction. • Identify ways that electromagnetic induction is used. The girl on the left in this photo is riding a stationary bike. She’s getting exercise, but that’s not the real reason she’s riding the bike. She’s using her muscle power to generate electricity through a process called electromagnetic induction. What Is Electromagnetic Induction? Electromagnetic induction is the process of generating electric current with a magnetic ﬁeld. It occurs whenever a magnetic ﬁeld and an electric conductor, such as a coil of wire, move relative to one another. As long as the conductor is part of a closed circuit, current will ﬂow through it whenever it crosses lines of force in the magnetic ﬁeld. One way this can happen is illustrated in the Figure 1.1. The sketch shows a magnet moving through a wire coil. You can watch an animated version of the illustration at this URL: http://jsticca.wordpress.com/2009/09/01 /the-magnet-car/ Q: What is another way that a coil of wire and magnet can move relative to one another and generate an electric current? A: The coil of wire could be moved back and forth over the magnet. The Current Produced by a Magnet The device with the pointer in the circuit above is an ammeter. It measures the current that ﬂows through the wire. The faster the magnet or coil moves, the greater the amount of current that is produced. If more turns were added to the coil or a stronger magnet were used, this would produce more current as well. 1 www.ck12.org FIGURE 1.1 The Figure 1.2 shows the direction of the current that is generated by a moving magnet. If the magnet is moved back and forth repeatedly, the current keeps changing direction. In other words, alternating current (AC) is produced. Alternating current is electric current that keeps reversing direction. FIGURE 1.2 How Electromagnetic Induction Is Used Two
important devices depend on electromagnetic induction: electric generators and electric transformers. Both devices play critical roles in producing and regulating the electric current we depend on in our daily lives. Electric generators use electromagnetic induction to change kinetic energy to electrical energy. They produce electricity in power plants. Electric transformers use electromagnetic induction to change the voltage of electric current. Some transformers increase voltage and other decrease voltage. Q: How do you think the girl on the exercise bike in the opening photo is using electromagnetic induction? A: As she pedals the bike, the kinetic energy of the turning pedals is used to move a conductor through a magnetic ﬁeld. This generates electric current by electromagnetic induction. 2 www.ck12.org Summary Concept 1. Electromagnetic Induction • Electromagnetic induction is the process of generating electric current with a magnetic ﬁeld. It occurs whenever a magnetic ﬁeld and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic ﬁeld. • The current produced by electromagnetic induction is greater when the magnet or coil moves faster, the coil has more turns, or the magnet is stronger. If the magnet or coil is moved back and forth repeatedly, alternating current is produced. • Electric generators and electric transformers use electromagnetic induction to generate electricity or change the voltage of electric current. Vocabulary • electromagnetic induction : Process of generating electric current with a changing magnetic ﬁeld. Practice Simulate electromagnetic induction at the following URL. Then answer the questions below. http://micro.magnet.fsu.edu/electromag/java/faraday2/ 1. How is electric current created in the simulation? What type of current is it? 2. How is electric current measured in the simulation? 3. What happens when you stop moving the magnet? Review 1. What is electromagnetic induction? When does it occur? 2. How could you increase the amount of current produced by electromagnetic induction? 3. Explain how a moving magnet and a coil of wire can be used to produce alternating current. 4. List two devices that use electromagnetic induction. References 1. Christopher Auyeung.. CC-BY-NC-SA 3.0 2. Christopher Auyeung.. CC-BY-NC-SA 3.0 3 CONCEPT 2 www.ck12.org Electromagnets • Deﬁne and describe a solenoid. • Deﬁne and describe an electromagnet
. • Determine the direction of the magnetic ﬁeld inside a solenoid given the direction of current ﬂow in the coil wire. • Understand why an electromagnet has a stronger magnetic ﬁeld than a solenoid. One of the most famous electric car companies is Tesla, named after Nikola Tesla. These electric cars, and all others, require an electromagnet to run the engine. Electromagnets A long coil of wire consisting of many loops of wire and making a complete circuit is called a solenoid. The magnetic ﬁeld within a solenoid can be quite large since it is the sum of the ﬁelds due to the current in each individual loop. 4 www.ck12.org Concept 2. Electromagnets The magnetic ﬁeld around the wire is determined by a hand rule. Since this description doesn’t mention electron ﬂow, we must assume that the current indicated by I is conventional current (positive). Therefore, we would use a right hand rule. We grasp a section of wire with our right hand pointing the thumb in the direction of the current ﬂow and our ﬁngers will curl around the wire in the direction of the magnetic ﬁeld. Therefore, the ﬁeld points down the cavity in these loops from right to left as shown in the sketch. If a piece of iron is placed inside the coil of wire, the magnetic ﬁeld is greatly increased because the domains of the iron are aligned by the magnetic ﬁeld of the current. The resulting magnetic ﬁeld is hundreds of time stronger than the ﬁeld from the current alone. This arrangement is called an electromagnet. The picture below shows an electromagnet with an iron bar inside a coil. Our knowledge of electromagnets developed from a series of observations. In 1820, Hans Oersted discovered that a current-carrying wire produced a magnetic ﬁeld. Later in the same year, André-Marie Ampere discovered that a coil of wire acted like a permanent magnet and François Arago found that an iron bar could be magnetized by putting it inside coil of current-carrying wire. Finally, William Sturgeon found that leaving the iron bar inside the coil greatly increased the magnetic ﬁeld. Two major advantages of electromagnets are that they are extremely
strong magnetic ﬁelds, and that the magnetic ﬁeld can be turned on and off. When the current ﬂows through the coil, it is a powerful magnet, but when the current is turned off, the magnetic ﬁeld essentially disappears. Electromagnets ﬁnd use in many practical applications. Electromagnets are used to lift large masses of magnetic materials such as scrap iron, rolls of steel, and auto parts. 5 www.ck12.org The overhead portion of this machine (painted yellow) is a lifting electromagnet. It is lowered to the deck where steel pipe is stored and it picks up a length of pipe and moves it to another machine where it is set upright and lowered into an oil well drill hole. Electromagnets are essential to the design of the electric generator and electric motor and are also employed in doorbells, circuit breakers, television receivers, loudspeakers, electric dead bolts, car starters, clothes washers, atomic particle accelerators, and electromagnetic brakes and clutches. Electromagnets are commonly used as switches in electrical machines. A recent use for industrial electromagnets is to create magnetic levitation systems for bullet trains. Summary • A solenoid is a long coil of wire consisting of many loops of wire that makes a complete circuit. • An electormagnet is a piece of iron inside a solenoid. • While the magnetic ﬁeld of a solenoid may be quite large, an electromagnet has a signiﬁcantly larger magnetic ﬁeld. • Electromagnets’ magnetic ﬁelds can be easily turned off by just halting the current. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=S6oop6RXg9w Follow up questions. 1. What components are needed to make a homemade electromagnet? 2. What objects were attracted by the electromagnet in the video? Review 1. Magnetism is always present when electric charges ___________. 2. What happens to the strength of an electromagnet if the number of loops of wire is increased? 3. What happens to the strength of an electromagnet if the current in the wire is increased? 4. Which direction does the magnetic ﬁeld point in the solenoid sketched here? 6 www.ck12
.org Concept 2. Electromagnets • solenoid: A current-carrying coil of wire that acts like a magnet when a current passes through it. • electromagnet: A temporary magnet consisting of an iron or steel core wound with a coil of wire, through which a current is passed. • magnetic levitation: The suspension of an object above a second object by means of magnetic repulsion. References 1. Image copyright Dongliu, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. Coil: Image copyright Giant Stock, 2013; modiﬁed by CK-12 Foundation - Samantha Bacic. http://www. shutterstock.com. Used under license from Shutterstock.com 3. Image copyright Zigzag Mountain Art, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 4. Image copyright Ingvar Tjostheim, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 5... CC BY-NC-SA 6. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 7 CONCEPT 3 Current and Magnetism Students will learn to analyze and solve problems involving current carrying wires in magnetic ﬁelds. Students will learn to analyze and solve problems involving current carrying wires in magnetic ﬁelds. www.ck12.org Key Equations Fwire = LIB sin(q) Force on a Current Carrying Wire In this equation, L refers to the length of the wire, I to the electric current, B the magnitude of the magnetic ﬁeld and q is the angle between the direction of the current and the direction of the magnetic ﬁeld. µ0I 2pr Where µ0 = 4p 107 Tm/A Bwire = Magnetic ﬁeld at a distance r from a current-carrying wire Permeability of Vacuum (approximately same for air also) Force on a Wire Since a wire is nothing but a collection of moving charges, the force it will experience in a magnetic ﬁeld will simply be the vector sum of the forces on the individual charges. If the wire is straight — that is, all the charges are moving in the same direction — these forces will all point in the same direction, and so will their sum. Then, the direction of the force can be found using the
second right hand rule, while its magnitude will depend on the length of the wire (denoted L ), the strength of the current, the strength of the ﬁeld, and the angle between their directions: Two current-carrying wires next to each other each generate magnetic ﬁelds and therefore exert forces on each other: 8 Concept 3. Current and Magnetism MEDIA Click image to the left for more content. www.ck12.org Example 1 Example 2 A wire loop and an inﬁnitely long current carrying cable are placed a distance r apart. The inﬁnitely long wire is carrying a current I1 to the left and the loop is carrying a current I2 CCW. The dimensions of the wire loop are shown in the diagram illustrating the situation below. What is the magnitude and direction of the net force on the loop (the mass of the wires are negligible)? Solution In this problem, it is best to start by determining the direction of the force on each segment of the loop. Based on the ﬁrst right hand rule, the magnetic ﬁeld from the inﬁnite cable points into the page where the loop is. This means that the force on the top segment of the loop will be down toward the bottom of the page, the force on the left segment will be to right, the force on the bottom segment will be toward the top of the page, and the force on the right segment will be to the left. The forces on the left and right segments will balance out because both segments are the same distance from the cable. The forces from the top and bottom section will not balance out because the wires are different distances from the cable. The force on the bottom segment will be stronger than the one on the top segment because the magnetic ﬁeld is stronger closer to the cable, so the net force on the loop will be up, toward the top of the page. Now we will begin to calculate the force’s magnitude by ﬁrst determining the strength of the magnetic ﬁeld at the bottom and top segments. All we really have to do is plug in the distances to each segment into the equation we already know for the magnetic ﬁeld due to a current carrying wire. 9 www.ck12.org B = Bbottom = Btop = µoI 2pr µoI1 2pR µoI1 2p2R
Now we will calculate the net force on the loop using the equation given above. We’ll consider up the positive direction. SF = I2L( SF = Fbottom Ftop SF = I2LBbottom I2LBtop SF = I2L(Bbottom Btop) µoI1 2pR µoI1I2L 2pR µoI1I2L 4pR µoI1 2p2R 1 2 SF = SF = (1 ) ) start by summing the forces on the loop substitute in the values for each of the force terms factor the equation substitute in the values for the magnetic ﬁeld factor the equation again simplify to get the answer Watch this Explanation MEDIA Click image to the left for more content. Time for Practice 1. A vertical wire, with a current of 6:0 A going towards the ground, is immersed in a magnetic ﬁeld of 5:0 T pointing to the right. What is the value and direction of the force on the wire? The length of the wire is 2:0 m. 10 www.ck12.org Concept 3. Current and Magnetism 2. 3. A futuristic magneto-car uses the interaction between current ﬂowing across the magneto car and magnetic ﬁelds to propel itself forward. The device consists of two ﬁxed metal tracks and a freely moving metal car (see illustration above). A magnetic ﬁeld is pointing downward with respect to the car, and has the strength of 5:00 T. The car is 4:70 m wide and has 800 A of current ﬂowing through it. The arrows indicate the direction of the current ﬂow. a. Find the direction and magnitude of the force on the car. b. If the car has a mass of 2050 kg, what is its velocity after 10 s, assuming it starts at rest? c. If you want double the force for the same magnetic ﬁeld, how should the current change? 4. A horizontal wire carries a current of 48 A towards the east. A second wire with mass 0:05 kg runs parallel to the ﬁrst, but lies 15 cm below it. This second wire is held in suspension by the magnetic ﬁeld of the ﬁrst wire above it. If each wire has a length of half a meter, what is the magnitude and direction of
the current in the lower wire? 5. Show that the formula for the force between two current carrying wires is F = µoLi1i2 2pd, where d is the distance between the two wires, i1 is the current of ﬁrst wire and L is the segment of length of the second wire carrying a current i2. (Hint: ﬁnd magnetic ﬁeld emanating from ﬁrst wire and then use the formula for a wire immersed in that magnetic ﬁeld in order to ﬁnd the force on the second wire.) 6. Two long thin wires are on the same plane but perpendicular to each other. The wire on the y axis carries a current of 6:0 A in the y direction. The wire on the x axis carries a current of 2:0 A in the +x direction. Point, P has the co-ordinates of (2:0; 2; 0) in meters. A charged particle moves in a direction of 45o away from the origin at point, P, with a velocity of 1:0 107 m=s: a. Find the magnitude and direction of the magnetic ﬁeld at point, P. b. If there is a magnetic force of 1:0 106 N on the particle determine its charge. c. Determine the magnitude of an electric ﬁeld that will cancel the magnetic force on the particle. 7. A long straight wire is on the x axis and has a current of 12 A in the x direction. A point P, is located 2:0 m above the wire on the y axis. a. What is the magnitude and direction of the magnetic ﬁeld at P. b. If an electron moves through P in the x direction at a speed of 8:0 107 m=s what is the magnitude and direction of the force on the electron? Answers to Selected Problems 1. Down the page; 60 N 2. a. To the right, 1:88 104 N b. 91:7 m=s c. It should be doubled 3. East 1:5 104 A 4.. 5. a. 8 107 T b. 1:3 106 C 11 6. a. 1:2 106 T; +z b. 1:5 1017 N; y www.ck12.org 12 www.ck12.org Concept 4. Electric Motors CONCEPT 4 Electric Motors • Explain the design and operation
of an electric motor. As gas prices continue to rise, electric cars and hybrids are becoming increasingly popular. These cars are certainly a part of our future. On the left in the image above is an all-electric vehicle, and on the right is a hybrid vehicle that uses gas part time and electricity part time. Electric Motors In an earlier concept, we described and calculated the force that a magnetic ﬁeld exerts on a current carrying wire. Since you are familiar with Newton’s third law of motion, you know that if the magnetic ﬁeld exerts a force on the current carrying wire, then the current carrying wire also exerts a force of equal magnitude and opposite direction on the magnetic ﬁeld. In the sketch above, a circuit is connected to a battery, with one part of the circuit placed inside a magnetic ﬁeld. When current runs through the circuit, a force will be exerted on the wire by the magnetic ﬁeld, causing the wire to 13 www.ck12.org move. If we choose to consider electron current in this case, the electrons ﬂow from the back of the sketch to the front while the magnetic ﬁeld is directed upward. Using the left hand rule for this, we ﬁnd that the force on the wire is to the right of the page. Had we chosen to consider the current to be conventional current, then the current would be ﬂowing from the front of the sketch to the back and we would use the right hand rule. The force on the wire would, once again, be toward the right. This movement is harnessed in electrical motors. Electrical motors change electrical energy into mechanical energy. The motor consists of an electrical circuit with part of the wires inside a magnetic ﬁeld. This can be seen below. Positive charges move through the circuit in the direction of the light purple arrows. When the charges move up through the part of the coil that is right next to the north pole, the right hand rule tells us that the wire suffers the force, F, pushing the wire in the direction of the blue arrow, toward the back of the sketch. On the other side of the coil, where the charges are moving down through the ﬁeld, the right hand rule shows the force would push this side of the coil toward the front. These two forces are working together, rotating the coil in the direction of the circular red arrow.
Where the rotating coil (in grey) meets the wires attached to the power source (black), we ﬁnd a split ring commutator. The coil turns, but the commutator and power source do not. As the coil turns, it moves off of the blue box connector and as it continues to turn, it connects to the other blue box connector. As the coil turns, it reverses its connections to the external circuit. Therefore, inside the coil, the current is always ﬂowing in the same direction because the left side of the coil is always attached to the left side of the external circuit. This allows the coil, or armature, to continue to spin the same direction all the time. In electrical motors, these coils often consist of not just one, but many wires, as can be seen here: 14 www.ck12.org Concept 4. Electric Motors Summary • When current runs through the circuit, a force will be exerted on the wire. • An electrical motor changes electrical energy into mechanical energy. • A split ring commutator keeps the current in the coil ﬂowing in the same direction even though the coil changes sides every half turn. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=elFUJNodXps In the video, a simple electrical motor is constructed. Follow up questions. 1. Who ﬁrst built an electric motor? 2. What size battery was used in the video motor? 3. The rover Curiosity is on the surface of what body? Review 1. Which way will the wire be pushed when current passes through the wire? a. up b. down 15 c. left d. right e. None of these. www.ck12.org 2. Which way will the coil spin when current passes through the wire? a. clockwise b. counterclockwise • electric motor: An electricmotor converts electricity into mechanical motion. • commutator: A split - ring commutator (sometimes just called a commutator) is a simple and clever device for reversing the current direction through an armature every half turn. • armature: A revolving structure in an electric motor or generator, wound with the coils that carry the current. References 1. Myrtle Beach TheDigitel and User:Mariordo/Wikimedia Commons. http://commons.wikimedia.org/wiki/F ile:Nissan_Le
af_%26_Chevy_Volt_charging_trimmed.jpg. CC-BY 2.0 16 www.ck12.org Concept 4. Electric Motors 2... CC BY-NC-SA 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 4. CK-12 Foundation - Samantha Bacic.. CC BY-NC-SA 3.0 5. Image copyright Sim Kay Seng, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 6. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 7. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 17 CONCEPT 5 www.ck12.org Electromotive Force • Deﬁne EMF. • Calculate EMF for a wire moving in a magnetic ﬁeld. Electrical generators convert mechanical energy into electrical energy. Every electrical generator needs some method for spinning the coil inside the magnetic ﬁeld. Hydroelectric generators use water pressure to spin the coil while windmills, of course, use the wind to spin the coil. The image here is a combination of steam turbine and generator. The steam can be produced by burning coal or diesel fuel or by a nuclear reaction and the steam then turns the coil and generates electricity. Electromotive Force When an individual charge ﬂies through a magnetic ﬁeld, a force is exerted on the charge and the path of the charge bends. In the case shown in the sketch below, the charge is positive and the right hand rule shows us the force will be upward, perpendicular to both the ﬁeld and the path of the charge. 18 www.ck12.org Concept 5. Electromotive Force If a wire that is part of a complete circuit is moved through a magnetic ﬁeld, the force on the individual electrons in the wire occurs in exactly the same manner. Since the electrons in the wire are negatively charged, the force would be in the opposite direction but otherwise the situation is the same. When the wire is pulled downward through the magnetic ﬁeld, the force on the electrons cause them to move within the wire. Since the charges are negative, the left hand rule shows that the electrons would move as diagrammed in the sketch. (Point ﬁngers
in the direction of the magnetic ﬁeld, point thumb in the direction of wire movement, and palm shows direction of electron ﬂow.) No current will ﬂow, of course, unless the section of wire is part of a complete circuit. This process allows us to convert mechanical energy (the motion of the wire) into electrical energy (the current). This is the opposite of what happens in an electric motor where electrical energy is converted into mechanical energy. In order to maintain a constant current ﬂow, it is necessary to have a potential difference or voltage in the circuit. The voltage or potential difference is also frequently referred to electromotive force. The term electromotive force, 19 like many historical terms, is a misnomer. Electromotive force is NOT a force, it is a potential difference or potential energy per unit charge and is measured in volts. The potential difference in the case of moving a wire through a magnetic ﬁeld is produced by the work done on the charges by whatever is pushing the wire through the ﬁeld. The EMF (or voltage) depends on the magnetic ﬁeld strength, B, the length of the wire in the magnetic ﬁeld, l, and the velocity of the wire in the ﬁeld. www.ck12.org EMF = Blv This calculation is based on the wire moving perpendicularly through the ﬁeld. If the wire moves an angle to the ﬁeld, then only the component of the wire perpendicular to the ﬁeld will generate EMF. Example Problem: A 0.20 m piece of wire that is part of a complete circuit moves perpendicularly through a magnetic ﬁeld whose magnetic induction is 0.0800 T. If the speed of the wire is 7.0 m/s, what EMF is induced in the wire? Solution: EMF = Blv = (0:0800 N=A m)(0:20 m)(7:0 m=s) = 0:11 N m=C = 0:11 J=C = 0:11 V Summary • If a wire that is part of a complete circuit is moved through a magnetic ﬁeld, the magnetic ﬁeld exerts a force on the individual electrons in the wire, which causes a current to ﬂow. • The potential difference in the case of moving a wire through
a magnetic ﬁeld is produced by the work done on the charges by whatever is pushing the wire through the ﬁeld. • The EMF (or voltage) depends on the magnetic ﬁeld strength, B, the length of the wire in the magnetic ﬁeld, l, and the velocity of the wire in the ﬁeld, EMF = Blv. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=0OHmMVBLXTI Follow up questions. 1. We have been discussing the process of generating electricity by moving a wire through a magnetic ﬁeld. What happens if the wire is held steady and the magnetic ﬁeld moves instead? 2. When a loop of wire is turned circularly in a magnetic ﬁeld, what type of current is produced? Review 1. Which of the following units are equivalent to those of EMF produced in a generator? (a) T m=s (b) V m2=s 20 www.ck12.org (c) J=s (d) A W (e) T m Concept 5. Electromotive Force 2. A straight wire 0.500 m long is moved straight up through a 0.400 T magnetic ﬁeld pointed in the horizontal direction. The speed of the wire is 20.0 m/s. (a) What EMF is induced in the wire? (b) If the wire is part of a circuit with a total resistance of 6:00 W, what is the current in the circuit? 3. A straight wire, 25.0 m long, is mounted on an airplane ﬂying at 125 m/s. The wire moves perpendicularly through earth’s magnetic ﬁeld (B = 5:00 105 T ). What is the EMF induced in the wire? 4. A straight wire, 30.0 m long, moves at 2.00 m/s perpendicularly through a 1.00 T magnetic ﬁeld. (a) What is the induced EMF? (b) If the total resistance of the circuit is 15:0 W, what is the current in the circuit? • electromotive force: The potential energy per unit charge that produces a ﬂow of electricity in a circuit, expressed in volts. • induced current: The electric current
generated in a loop of conducting material by movement of the loop across a magnetic ﬁeld. References 1. Courtesy of the NRC. http://commons.wikimedia.org/wiki/File:Modern_Steam_Turbine_Generator.jpg. Public Domain 2. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 21 CONCEPT 6 Electric Generators www.ck12.org • Explain how an electric generator works. • Explain the difference between an electric generator and an electric motor. • Explain the difference between peak and effective voltage and current from a generator. These large machines are electric generators. This particular row of generators is installed in a hydroelectric power station. The insides of these generators are coils of wire spinning in a magnetic ﬁeld. The relative motion between the wire and the magnetic ﬁeld is what generates electric current. In all generators, some mechanical energy is used to spin the coil of wire in the generator. In the case of hydroelectric power, the coil of wire is spun by water falling from higher PE to lower PE. Windmills and steam turbines are used in other types of power generators to spin the coil. Electric Generators Electric generators convert mechanical energy to electric energy. The generator consists of some number of wire loops wrapped around an iron core and placed in a strong magnetic ﬁeld. The loops of wire and the iron core are called the armature. The armature is mounted so that it can rotate freely inside the magnetic ﬁeld. Mechanical energy is used to spin the armature in the ﬁeld so that the wire loops cut across the ﬁeld and produce electric current. The EMF of this current is calculated by EMF = Blv. 22 www.ck12.org Concept 6. Electric Generators Consider the coil and magnetic ﬁeld sketched above. When the right hand side of the coil moves up through the ﬁeld, the left hand rule indicates that the electron ﬂow will be from the front to the back in that side of the coil. The current generated will have the greatest EMF as the wire is cutting perpendicularly across the ﬁeld. When the wire reaches the top of its arc, it is moving parallel to the �
��eld and therefore, not cutting across the ﬁeld at all. The EMF at this point will be zero. As that same wire then cuts down through the ﬁeld as it continues to spin, the left hand rule indicates that the electron ﬂow will be from the back to the front in that side of the coil. In this second half of the arc, the direction of the electron ﬂow has reversed. The magnitude of the EMF will reach maximum again as the wire cuts perpendicularly down through the ﬁeld and the EMF will become zero again as the wire passes through the bottom of the arc. The current produced as the armature goes around will resemble a sine wave where the EMF reaches a maximum in one direction, then goes to zero, then goes to a maximum in the other direction. This type of current is called alternating current. By having more and more loops of wire on the armature, the crests and troughs overlap and ﬁll in until a constant current is produced. A direct current is one that always ﬂows in the same direction rather than alternating back and forth. Batteries produce direct currents. A generator can also produce direct current by using a split ring commutator that changes external connections every half turn of the armature so that even though the current in the coil changes direction, every time the current in the coil changes direction, the external connection switches so that the external current always goes in the same direction. Generators and motors are almost identical in construction but convert energy in opposite directions. Generators convert mechanical energy to electrical energy and motors convert electrical energy to mechanical. Because of the alternating direction in alternating current, the average value is less than the power supplied by a direct current. In fact, the average power of an AC current is one-half its maximum power and one-half the power of an equivalent DC current. The effective current of an AC generator is 0.707 times its maximum current. The same is true for the effective voltage of an AC generator. 23 Ieff = 0:707 Imax Veff = 0:707 Vmax Example Problem: An AC generator develops a maximum voltage of 34.0 V and delivers a maximum current of 0.170 A. www.ck12.org a. What is the effective voltage of the generator? b. What is the effective current delivered by the generator? c. What is the resistance in the circuit? Solution
: a. Veff = 0:707 Vmax = (0:707)(34:0 V ) = 24:0 V b. Ieff = 0:707 Imax = (0:707)(0:17 A) = 0:120 A c. R = V 0:120 A = 200: W I = 24:0 V Summary • Electric generators convert mechanical energy to electric energy. • The generator consists of some number of wire loops wrapped around an iron core and placed in a strong magnetic ﬁeld. • The loops of wire and the iron core are called the armature. • The armature is mounted so that it can rotate freely inside the magnetic ﬁeld. • Mechanical energy is used to spin the armature in the ﬁeld so that the wire loops cut across the ﬁeld and produce electric current. • The current produced as the armature goes around will resemble a sine wave where the EMF reaches a maximum in one direction, then goes to zero, then goes to a maximum in the other direction. This type of current is called alternating current. • A generator can also produce direct current by using a split ring commutator that changes external connections every half turn of the armature so that even though the current in the coil changes direction, every time the current in the coil changes direction, the external connection switches so that the external current always goes in the same direction. • The effective current of an AC generator is 0.707 times its maximum current. • The effective voltage of an AC generator is 0.707 times its maximum voltage. Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=RFOMpOM1WHQ Follow up questions. 1. Which of the two generators in the video (an AC generator and a DC generator) involves a magnetic ﬁeld? 2. Which of the two generators in the video involves a wire-wrapped armature? 3. What is the difference between the DC generator and the AC generator? 24 Concept 6. Electric Generators www.ck12.org Review 1. What three things are necessary to produce EMF mechanically? (a) magnet, force lines, and magnetic ﬁeld (b) EMF, conductor, and magnetic ﬁeld (c) conducting wire, magnetic ﬁeld, and relative motion (d) conducting wire, electrical ﬁ
eld, and relative motion (e) none of these will produce EMF mechanically 2. Increasing which of the following will increase the output of a generator? (a) EMF (b) strength of the magnetic ﬁeld (c) resistance of the conductor (d) load on the meter (e) none of these 3. The current in the rotating coil of all generators is (a) AC (b) DC (c) pulsating AC (d) pulsating DC 4. A generator in a power plant develops a maximum voltage of 170. V. (a) What is the effective voltage? (b) A 60.0 W light bulb is placed across the generator. A maximum current of 0.70 A ﬂows through the bulb. What effective current ﬂows through the bulb? (c) What is the resistance of the light bulb when it is working? 5. The effective voltage of a particular AC household outlet is 117 V. (a) What is the maximum voltage across a lamp connected to the outlet? (b) The effective current through the lamp is 5.50 A. What is the maximum current in the lamp? • direct current: An electric current ﬂowing in one direction only. • alternating current: An electric current that reverses direction in a circuit at regular intervals. • electric generator: An electric generator is a device that converts mechanical energy to electrical energy. • armature: The rotating part of a generator, consisting essentially of copper wire wound around an iron core. References 1. Image copyright James L. Davidson, 2013. http://www.shutterstock.com. Public Domain 2. Galvanometer: Image copyright scropy, 2013; composite created by CK-12 Foundation. http://www.shutters tock.com. Used under license from Shutterstock.com 3. CK-12 Foundation - Samantha Bacic.. CC-BY-NC-SA 3.0 25 CONCEPT 7 • Describe the function of a transformer. • Explain the relationship between turns and voltage ratio. • Solve mathematical problems involving transformers. www.ck12.org Transformers Power loss in long transmission lines is related to the magnitude of the current. Speciﬁcally, the power loss can be decreased by decreasing the magnitude of the current. The amount of power passed through transmission lines can be calculated by multiplying voltage by current. The same power can be transmitted using a very high voltage and a very low current as with a
low voltage and high current. Since power companies do not wish to waste power as it is transmitted to homes and businesses, they deliberately ’step up’ the voltage and reduce the current before transmitting the power over extended distances. That type of power transmits well without great loss of energy but it cannot be used in household appliances. It becomes necessary to convert it back (’step down’) to low voltage and high current for household use. That is the job of electrical transformers – those big gray barrels you see on power poles. Transformers When we move a wire through a magnetic ﬁeld, a force is exerted on the charges in the wire and a current is induced. Essentially the same thing happens if we hold the wire steady and move the magnetic ﬁeld by moving the magnet. Yet a third way of causing relative motion between the charges in a wire and a magnetic ﬁeld is to expand or contract the ﬁeld through the wire. When a current begins to ﬂow in a wire, a circular magnetic ﬁeld forms around the wire. Within the ﬁrst fractional second when the current begins to ﬂow, the magnetic ﬁeld expands outward from the wire. If a second wire is placed nearby, the expanding ﬁeld will pass through the second wire and induce a brief current in the wire. 26 www.ck12.org Concept 7. Transformers Consider the sketch above. When the knife switch is closed, current begins to ﬂow in the ﬁrst circuit and therefore, a magnetic ﬁeld expands outward around the wire. When the magnetic ﬁeld expands outward from the wire on the right side, it will pass through the wire in the second circuit. This relative motion between wire and ﬁeld induces a current in the second circuit. The magnetic ﬁeld expands outward for only a very short period of time and therefore, only a short jolt of current is induced in the second circuit. You can leave the knife switch closed and the current will continue to ﬂow in the ﬁrst circuit but no current is induced in the second circuit because the ﬁeld is constant and therefore there is no relative motion between the ﬁeld and the wire in the second circuit. When the knife switch is opened, the current in the ﬁr
st circuit ceases to ﬂow and the magnetic ﬁeld collapses back through the wire to zero. As the magnetic ﬁeld collapses, it passes through the wire and once again we have relative motion between the wire in the second circuit and the magnetic ﬁeld. Therefore, we once again have a short jolt of current induced in the second circuit. This second jolt of induced current will be ﬂowing in the opposite direction of the ﬁrst induced current. We can produce an alternating current in the second circuit simply by closing and opening the knife switch continuously in the ﬁrst circuit. Obviously, a transformer would have little use in the case of DC current because current is only induced in the second circuit when the ﬁrst circuit is started or stopped. With AC current, however, since the current changes direction 60 times per second, the magnetic ﬁeld would constantly be expanding and contracting through the second wire. A transformer is a device used to increase or decrease alternating current voltages. They do this with essentially no loss of energy. A transformer has two coils, electrically insulated from each other as shown in the sketch. One coil is called the primary coil and the other is called the secondary coil. When the primary coil is connected to a source of AC voltage, the changing current creates a varying magnetic ﬁeld. The varying magnetic ﬁeld induces a varying EMF in the secondary coil. The EMF induced in the secondary coil is called the secondary voltage and is proportional to the primary voltage. The secondary voltage also depends on the ratio of turns on the secondary coil to turns on the primary coil. 27 www.ck12.org secondary voltage primary voltage VS VP = = number of turns on secondary number of turns on primary NS NP If the secondary voltage is larger than the primary voltage, the transformer is called a step-up transformer. If the voltage out of the transformer is smaller than the voltage in, then the transformer is called a step-down transformer. In an ideal transformer, the electric power put into the primary equals the electric power delivered by the secondary. The current that ﬂows in the primary depends on how much current is required by the secondary circuit. VPIP = VSIS IS IP = VP VS = NP NS Example Problem: A particular step-up transformer has 200 turns on the primary coil and 3000 turns on the secondary coil. a. If the voltage on the primary
coil is 90.0 V, what is the voltage on the secondary coil? b. If the current in the secondary circuit is 2.00 A, what is the current in the primary coil? c. What is the power in the primary circuit? d. What is the power in the secondary circuit? Solution: = NP NS = NP NS VS = (90:0 V )(3000) a. VP VS IS IP = 30:0 A b. IP c. PP = VPIP = (90:0 V )(30:0 A) = 2700 W d. PS = VSIS = (1350 V )(2:00 A) = 2700 W = 200 3000 = 200 3000 90:0 V VS 2:00 A IP (200) = 1350 V 28 www.ck12.org Summary Concept 7. Transformers • When a current begins to ﬂow in a wire, a circular magnetic ﬁeld forms around the wire. • Within the ﬁrst fractional second when the current begins to ﬂow, the magnetic ﬁeld expands outward from the wire. • If a second wire is placed nearby, the expanding ﬁeld will pass through the second wire and induce a brief current in the wire. • A transformer is a device used to increase or decrease alternating current voltages. • A transformer has two coils, electrically insulated from each other. One coil is called the primary coil and the other is called the secondary coil. • The varying magnetic ﬁeld induces a varying EMF in the secondary coil. • The EMF induced in the secondary coil is called the secondary voltage and is proportional to the primary voltage. The secondary voltage also depends on the ratio of turns on the secondary coil to turns on the primary coil. secondary voltage primary voltage = number of turns on secondary number of turns on primary Practice MEDIA Click image to the left for more content. http://www.youtube.com/watch?v=-v8MYAFl7Mw Follow up questions. 1. What type of transformer is used at the power station where the electric power is generated? 2. What type of transformer is used at power sub-stations? 3. What type of transformer is used inside cell phone chargers? Review 1. A step-down transformer has 7500 turns on its primary and 125 turns on its secondary. The voltage across the primary is 7200 V. (a) What is
the voltage across the secondary? (b) The current in the secondary is 36 A. What current ﬂows in the primary? 2. The secondary of a step-down transformer has 500 turns. The primary has 15,000 turns. (a) The EMF of the primary is 3600 V. What is the EMF of the secondary? (b) The current in the primary is 3.0 A. What is the current in the secondary? 3. An ideal step-up transformer’s primary coil has 500 turns and its secondary coil has 15,000 turns. The primary EMF is 120 V. (a) Calculate the EMF of the secondary. (b) If the secondary current is 3.0 A, what is the primary current? (c) What power is drawn by the primary? 29 www.ck12.org • transformer: An electric device consisting essentially of two or more windings wound on the same core, which by electromagnetic induction transforms electric energy from one circuit to another circuit such that the frequency of the energy remains unchanged while the voltage and current usually change. • primary coil: A coil to which the input voltage is applied in a transformer. • secondary coil: The coil in a transformer where the current is induced. • step-up transformer: A step-up transformer is one that increases voltage. • step-down transformer: A step-down transformer is one that decreases voltage. References 1. Image copyright Sylvie Bouchard, 2013. http://www.shutterstock.com. Used under license from Shutter- stock.com 2. CK-12 Foundation - Samantha Bacic, using image copyright Sergii Korolko, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 3. CK-12 Foundation - Christopher Auyeung.. CC-BY-NC-SA 3.0 30rors are also used in supermarkets, underground parking areas etc. to cover a wider field of view to be observed. Fig. 1.35: Parabolic dish aerial. (vii) Shaving mirror A concave mirror is used as a shaving or a make-up mirror. The mirror is placed at a distance less than its focal length, from the face. A virtual upright, magnified image of the face is seen in the mirror (Fig. 1.36). B face F Fig. 1.36: Shaving mirror. C (b) Convex mirrors A convex mirror
is used as a rear view driving mirror. Any object in front of the mirror, i.e, lying behind the car, forms an upright diminished virtual image between F and P (Fig. 1.37). For example, another car behind is seen upright and small in size in the mirror. The driver of a car has a wider view of all the objects lying behind. 24 image C F M I P B O object (car) v u Fig. 1.37: Rear view driving mirror. A convex mirror covers a wider field of view so that the light rays from a wide angle can be observed. A plane mirror of the same size as the convex mirror helps us to see the objects lying in a limited area (smaller). 1.8 Project work Construction of a solar concetrator Suggested materials: Mild steel wires of about 3 mm in thickness, aluminium foil, superglue Assembly 1. Cut the mild steel wire so as to make three circles of diameter 1.0 m each. 2. Make the three circles using the wire. 3. Weld the wires to form a spherical framework. 4. Cover the framework using the aluminium foil. 5. Cut the framework to produce two equal ½ sphere. 25 6. Mount one of the ½ sphere on a stand such that it can be rotated in all directions. 7. Using a thicker mild steel wire, make a framework that will be able to support a small sufuria and connect it to the framework above. Make sure that the sauce pan is at ½ centre of the sphere i.e. at focal point. 8. Direct the ½ sphere and the sauce pan towards the sun. 9. Fill the sauce pan with water at 15 minutes interval. 10. Measure the temperature at 15 minutes interval. Such an arrangement is called a solar concentrator. 26 Topic summary • The reflecting surface of a plane mirror is a straight plane, concave mirror curves in, convex mirror curves outwards and parabolic mirror is a section of a paraboloid. • The centre of the reflecting surface is the pole, P, of a curved mirror. • The centre of curvature, C, of a curved mirror is the centre of the sphere of which the mirror forms a part. • The principal axis of a curved mirror is a line passing through the pole, P, and the centre of curvature, C. • The radius of curvature, r, of a curved mirror is the radius of the sphere of which the mirror forms a part. •
The principal focus, F, is a fixed point on the principal axis where a set of incident rays parallel and close to the principal axis converge in a concave mirror and appear to diverge on in a convex mirror. • The principal focus is real for a concave mirror and virtual for a convex mirror. • The focal length, f, of a curved mirror is the distance from its pole to the principal focus. • Focal length of a concave mirror is half the radius of curvature of the mirror. • A parabolic concave reflector has a single point focus and the caustic curve is not formed. • Curved mirrors obey the laws of reflection just like plane mirrors. • The focal length, f, of a curved mirror is half the radius of curvature, r. • Magnification (m) is defined as the ratio of the height of the image to the height of the object. Magnification = height of image (IM) height of object (OB) = image distance (v) object distance (u) m = 1 f IM OB 1 = u = v u + 1 v • The nature, size and the position of the image formed in a concave mirror depends on the position of the object in front of the mirror. See the Table 1.1 below. 27 Table 1.1 Object Image Nature of image Size of image compared to object F At infinity (far away) Beyond C At C Between C and F Beyond C At F Between F and C At C At infinity (far away) Real and inverted Diminished Real and inverted Diminished Real and inverted Same size Real and inverted Magnified Real and inverted Magnified Between F and P Behind the mirror Virtual and upright Magnified • In a convex mirror, for all the positions of the object at a measurable distance from the mirror, an upright, diminished, virtual image is always formed between F and P. • Cinema projectors, search lights, car head lights, astronomical telescopes, solar concentrators, dish aerials are a few applications of curved reflecting surfaces. 28 Topic Test 1 1. The focal length of a curved mirror is the distance between the principal focus and the centre of curvature. the pole of the mirror and the centre of curvature. the pole of the mirror and the principal focus. the object and the image. A. B. C. D. 2. In a curved mirror, how is the focal length (f) related to the radius of curvature (r) of the mirror?
A. f = r 2 B. f = 2r C. f = r D. None of the above. 3. In a concave mirror, when the object is placed between the principal focus and the pole, the image formed is A. real and diminished C. virtual and upright B. real and magnified D. virtual and inverted 4. In a concave mirror, when the object is placed at 12 cm from the pole a real image is formed at 36 cm from the pole of the mirror. The magnification produced by the mirror is A. 0.25 B. 4.00 C. 0.33 D. 3.00 5. The effect of formation of a caustic curve in a concave mirror can be minimized by A. cutting off the marginal rays. B. cutting off the axial rays. C. cutting off all the rays parallel to the principal axis. D. cutting off the rays passing through the centre of curvature of the mirror. 6. In a convex mirror, the image formed is always A. C. real and upright. real and inverted. B. virtual and upright. D. virtual and inverted. 7. In a concave mirror, when a real object is placed at 2f (Fig. 1.38), the image formed will be A. at infinity. B. between p and f. C. between P and 2f. D. at 2f. 8. Draw a sketch of a concave mirror of radius of curvature 20 cm. Mark on the diagram the pole, the centre of curvature, the principal axis, the principal focus and the focal length. 29 PF2FobjectFig. 1.38 9. Describe a parabolic mirror. What is the advantage of a parabolic mirror over a spherical concave mirror. 10. Define the following terms: pole, principal axis, principal focus, focal length, focal plane in relation to a concave mirror. 11. Copy and complete the following diagrams (Fig. 1.39 (a) to (d)) to show the path of the reflected rays. Fig. 1.39 12. A concave mirror has a focal length of 6 cm and an object 2 cm tall is placed 10 cm from the pole of the mirror. By means of an accurate ray diagram, find the position and size of the image formed. Is the image real or virtual? Explain your answer. 13. Describe an experiment to determine the magnification of an image in a concave mirror.
14. Fig. 1.40 shows the graph of real image distance v against the object distance u for a concave mirror. Fig. 1.40 Explain why the coordinate P is for a magnified image. 15. (a) Define magnification. 30 PFC(a)PFC(b)PFC(c)PFC (d)0uPv (b) An object of height 3 cm is placed at 20 cm in front of a concave mirror. The real image formed is 10 cm from the pole of the mirror. Calculate the height of the image formed. 16. A concave mirror has a focal length of 6 cm and a real object 3 cm tall is placed 15 cm from the pole. Calculate the distance of the image from the pole if the size of the image is 2 cm. 17. Explain, with the aid of a neat ray diagram how a concave mirror can be used as a ‘make-up’ mirror. (The object may be represented by an arrow-head). 18. Motorists use a convex mirror, rather than a plane mirror, as a rear-view mirror. State one advantage and one disadvantage of using a convex mirror. 19. Name the type of mirror used in the following: (a) car head lights. (b) solar concentrators. (c) underground car parking area. 20. For the following Figure 1.41 (a) and (b), copy and complete the ray diagram to locate the position of the point image I for the point object placed at O. (a) Fig. 1.41 (b) 21. The graph (Fig. 1.42) shows how the image distance v and the object distance u vary for a concave mirror. (a) When the object is placed at a distance of 60 cm from the mirror Fig. 1.42 31 PCOCOv(cm)0u(cm)10203040506070802030405060708010 (i) what is the distance of the image from the mirror? (ii) Is the image magnified, diminished or the same size as the object? (iii) what is the magnification produced by the mirror? (b) The object is now moved until it is 20 cm from the mirror. State and explain what happens to (i) the image distance. (ii) the magnification produced by the mirror. 22. A man uses a showing mirror with a focal length of 72 cm
to view the images of his face. If his face is 18 cm from the mirror, determine the image distance and the magnification of his face. 23. The real image produced by a concave mirror is observed to be six times longer than the object when the object is 34.2 cm in front of a mirror. Determine the radius of curvature of this mirror. 24. An object and a concave mirror are used to produce a sharp image of the object on the screen. The table below shows the magnification and image distance of the object. Magnification m Image distance v (cm) 0.25 20 1.5 40 2.5 56 3.5 72 Table 1.2 (a) Plot a graph of m against v (b) Use the graph to find: i. The image distance when M=1.0 ii. The object distance iii. The focal length of the mirror 32 TOPIC 2 Refraction of light in thin lenses Unit Outline • Definition of a lens • Types of lenses • Terms used in thin lenses • Image formation by converging and diverging lenses • The lens formula • Sign convetion • Magnification • Power of lenses • Defects of lenses • Simple telescope Introduction In secondary 1, we learnt about refraction of light through a medium. In this topic, we are going to specifically focus refraction in thin lenses. How the thin lenses are applied in optical instruments such as microscope, glasses, cameras and others. 2.1 Definition of a lens Activity 2.1 What is a lens? (Work in pairs or in groups) Materials • Water • Round bottomed flask • Plain paper • Retort stand • Sun • Magnifying lens Steps 1. What is a lens? 2. Fill a volumetric flask with clear water. 3. Cork the flask. 4. Tilt the flask such that the neck of the cork is horizontal. 5. Place a source of light (sun, bulbs) above the flask. 33 6. Place a white paper under the flask preferably on the ground. 7. Move the flask to or away the white paper. 8. What happens on the plain paper? 9. Discuss the results in your groups. 10. Replace the round bottomed flask with a magnifying lens. What happens to the paper? Explain. Sun Clamp Round bottomed flask Stand Cork Water Piece of paper Fig. 2.1 A lens is a transparent medium bounded by two spherical surface or a planed curved surface. 2.2 Types of
lenses To identify and describe types and shapes of lenses Activity 2.2 (Work in groups) Materials • Charts showing converging and diverging beam through lens. • Convex lenses Steps • Plane lenses • Spherical lenses 1. Place some lenses available in your school on a labeled white plane paper. Trace their outlines. What is the shape of the lenses? 2. Feel the lenses in Fig. 2.2 below with your fingers. What do you feel? What are their shapes? Why do you think they are made in such shapes? 34 3. Name the above lenses. Fig. 2.2 4. Identify the lenses in Fig. 2.3 below. What are their shapes? Why do you think they are made in such shapes? Name them. (a) (b) (c) Fig. 2.3: Types of concave lenses There are two main groups of lenses. A type that is thick in the middle and thin at the edges, causing rays of light to converge. This is called converging or convex lenses. The other type is thin in middle and thick at the edges causing the rays of light to diverge. This lens is called diverging or concave lens. Concave lenses are of different shapes as shown in Fig. 2.4. A bi-convex or double convex lens has both its surfaces ‘curving out’. (Fig. 2.4). (a) (b) (c) Bi-convex or double convex Plano - convex Concavo - convex Concave lenses are also of different shapes (See Fig. 2.5) Fig. 2.4: Types of convex lenses (a) (b) Bi-concave plano-concave Fig. 2.5: Types of lenses 35 A bi-concave or double concave lens has both its surfaces ‘curving in’. Other concave lenses are plano-concave and convexo-concave or diverging meniscus (Fig. 2.5). When a beam of light is incident on the lenses, rays tend to converge at a point when they pass through a convex lens and diverge through a concave lens. (Fig. 2.6) (a) Incident rays Refracted rays Principal axis P F C 2F (b) C 2F Incident rays Refracted rays Lens F P Principal axis f f Fig. 2.
6: Refraction of light through lenses 2.3 Terms used in thin lenses Activity 2.3 To find out the meaning of the terms used in thin lenses (Work in pairs or in groups) Materials • Magnifying lens • Double convex lenses • Sunshine Steps • Piece of paper 1. This activity is done outside the classroom during daylight. 2. Place a dry tissue paper (or dry leaves) on a flat open ground. 3. Place the lens on the tissue paper. 4. Slowly lift the lens upwards away from the paper until a spot of light is formed on the paper. Fig. 2.7: Burning a paper using a lens 36 5. Hold the lens at this position for some time. 6. Observe what happens to the paper. Explain why the paper burns. 7. Repeat the activity with a concave lens. What do you observe? What happens to the beam of light? The paper starts to burn. This shows that a convex lens brings to a focus point light energy from the sun and since light is in form of energy, a lot of it is concentrated at a point. This point where the rays are brought together after passing through the convex lens is called principal focus. This point is real. When the activity is repeated with a concave lens, nothing happens. That means the principal focus of a concave lens is virtual. The following are other common terms used in thin lenses: (a) The centre of curvature (C) The centre of curvature of the surface of a lens is the centre of the sphere of which the lens forms a part (Fig. 2.8 (a) and (b)). For each spherical lens there are two centres of curvature (C1, C2) due to the two curved surface. (b) The radius of curvature (r) The radius of curvature of the surface of a lens is the radius of the sphere of which the surface forms a part (Fig. 2.8 (a) and (b)). Each surface has its own radius of curvature (r1 or r2). (c) Principal axis The principle axis of a lens is a line passing through the two centres of curvature (c1 and c2) as shown in Fig. 2.8. r1 c1 c2 r2 r1 c1 c2 r2 (a) Convex lens (b) Concave lens Fig. 2.8: Principal axis (d) The principal focus A prism always dev
iates the light passing through it towards its base. A convex lens may be regarded as being made up of large portions of triangular prisms as 37 shown below. The emergent beam, therefore, becomes convergent in a convex lens (Fig. 2.9 (a)). The reverse is the effect in a concave lens (Fig. 2.9(b)). (a) (b) Fig. 2.9: Action of lenses compared with prisms (i) Principal focus of a convex lens Consider a set of incident rays parallel and close to the principal axis of a convex lens (Fig. 2.10). These rays, after refraction through the lens, pass through point F on the principal axis. Since all the rays converge at this point, it is called principal focus. Since this point can be projected on a screen, it is said to be a real principal focus. Incident rays Refracted rays Principal axis P F C 2F f Fig. 2.10: Principal focus on a convex lens (ii) Principal focus of a concave lens For a set of incident rays parallel and close to the principal axis of a concave lens, the refracted rays appear to diverge from a fixed point on the principal axis. This point is called the principal focus F, of a concave lens (Fig. 2.11). This principal focus is virtual since it cannot be projected on a screen. Incident rays Refracted rays Lens C 2F F P Principal axis f Fig. 2.11: Principal focus on a concave lens 38 (e) The focal plane When a set of parallel rays are incident on a convex lens at an angle to the principal axis, as shown in Fig. 2.12, the refracted rays converge to a point, on a line passing through F and perpendicular to the principal axis. The plane passing through F is the focal plane. Lens axis Focal plane 90º F Fig. 2.12: Focal plane of a convex lens 3 1 F F (f) The optical centre (P) F 2 P The optical centre of a lens is a point which lies exactly in the middle of the lens (PA = PB) as shown in Fig. 2.13(a) and 2.13(b). Light rays going through this point go straight through without any deviation or displacement. (a) (b) A P B P Fig. 2.13: Optical centre of a convex and concave lenses (g) The
focal length of a lens (f) This is the distance from the optical centre to the principal focus of the lens (see Fig. 2.10(a) and 2.11(b). Biconvex and biconcave lenses have a focal length on each side of the lens. The concept of centres of curvature of the surfaces is required only in drawing the principal axis. Otherwise these points are referred to as 2F, as they are situated at a distance twice the focal length from the centre of the lens (PC = 2PF). 39 Exercise 2.1 1. Distinguish between converging and diverging lenses. 2. Define the following: (a) Principal axis (b) Optical centre 3. Differentiate between the principal focus of the concave and convex lens. 4. How many principal foci does a biconcave lens have? 2.4 Image formation by converging lenses Activity 2.4 (Work in groups) To find out and describe the image formed by converging lenses Materials • Convex lenses • Tree, Screen (white wall can act as screen) Steps 1. Place a convex lens between a screen and a far away object e.g. a candle. (See fig 2.14). 2. Adjust the distance between the lens and the screen until the image of the candle is observed on the screen. 2F F v F 2F u Fig. 2.14: Image formation by a far object 3. What are the characteristics of the image formed? 4. In groups, discuss the formation of the image using ray diagrams. Ray diagrams are used to illustrate how and where the image is formed. The following are the important incident rays and their corresponding refracted rays used in the construction of ray diagrams. 40 Ray 1: A ray of light parallel and close to the principal axis, passes through the principal focus F (Fig. 2.15). P F F P Ray 2: A ray of light through the principal focus F emerges parallel to the principal axis after refraction (Fig. 2.16). Fig. 2.15: Ray 1 F P P F Fig. 2.16: Ray 2 Ray 3: A ray through the optical centre, P is undeviated after refraction through the lens (Fig. 2.17). P P Fig. 2.17: Ray 3 2.5 Locating images by simple ray diagrams and describing their characteristics To locate the image of an object, we need a minimum of two incident
rays from the object. From the three standard rays discussed above, any two incident rays and their corresponding refracted rays can be drawn to locate the image. If the refracted rays converge, a real image is obtained. If the refracted rays diverge, then a virtual image is obtained. 41 2.5.1 Convex lens Activity 2.5 (Work in groups) Materials To design and describe the characteristics of images formed by convex lenses when the object is at infinity • White screen • Lens • An object at infinity (landscape or a tree) • Metre rule Instructions 1. In this activity, you will design and carry out an investigation to describe the characteristics of images formed by convex lenses when the object is at infinity. 2. Modify the set-up, we used in Activity 1.6 with materials provided to you. Sketch the new set-up. 3. Write a brief procedure for your investigation. Conduct the investigations. Draw the image formed. 4. Describe its characteristics. 5. Suggest some possible sources of errors in your investigation and explain how they can be minimised. Write a report and present it in a class discussion. Note: The distance from the centre of the lens to the screen is nearly equal to the focal length, f, of the lens. (a) Object far away from the lens (at infinity) Since the object is at infinity, all the rays from the object, incident on the lens are almost parallel. The refracted rays converge at a point on the focal plane, as shown in Fig. 2.18. Image characteristics A diminished, real, inverted image is formed at F. F P F I M Fig. 2.18: Object OB at infinity 42 (b) Object OB just beyond C (2F) Activity 2.6 (Work in groups) Materials • Screen Steps To describe images formed by convex lens when the object is beyond 2F and at 2F • Lens • Candle 1. Mark the positions of the principal focus F and 2F on both the sides of the lens with a piece of chalk. 2. Place a lit candle on the table along the principal axis of the lens, slightly away from 2F. 3. Place a white screen, on the other side of the lens, perpendicular to the principal axis of the lens and adjust its position to and fro to the screen and observe what happens. What are the characteristics of the image formed? Real image Lens Object (candle) Screen 2F F v P
F u 2F Fig. 2.19: Object beyond 2F 4. Repeat step 3 by placing the candle at 2F and observe what happens. What are the characteristics of the images formed? Fig 2.20 shows the ray diagram to locate the images when the object is beyond C. B 0 2F F P F 2F I M Fig. 2.20: Object OB just beyond 2F 43 Image characteristics A diminished, real, inverted image is formed between F and 2F. (c) Object OB at 2F The ray diagram when the object (candle) was placed at 2F is as shown in Fig. 2.22 below. B 0 2F F P F 2F I M Fig. 2.21: Object OB at 2F Image characteristics A real, inverted image of the same size as the object is formed at 2F. (d) Object OB between 2F and F To design and describe the images formed by convex lens when the object is between F and 2F and at F Activity 2.7 (Work in groups) Materials • Candle • Lens • Screen Instructions 1. Modify the set-up as used in Activity 2.6 by placing the candle between 2F and F. 2. Draw the set-up. 3. Write a brief procedure for your investigation. Carry out the investigation and describe the characteristics of the images formed. 4. Suggest some possible sources of errors in your investigation and explain how they can be minimised. 5. Write a report and present it to a class. 44 The simple ray diagram when the object is between F and 2F is as shown in Fig. 2.22. B 0 2F F P F 2F I M Fig. 2.22: Object OB between 2F and F Image characteristics A real, inverted and magnified image is formed beyond 2F. (e) Object OB at F When the object was at F, the refracted rays are nearly parallel and converge at infinity as shown in Fig. 2.23 below. B O F P F parallel rays Fig. 2.23: Object OB at F Image characteristics A real, inverted, magnified image is formed far away from the lens i.e. at infinity. (cannot be described) (f) Object OB between F and P Activity 2.8 (Work in groups) Materials To describe the image formed by convex lens when the object is between F and P • Candle • Lens • Screen Steps 1. Repeat Activity 2.7 keeping
the candle close to the lens, between F and P. Can you get an image on the screen? Describe its characteristics. 45 2. Is the image real or virtual? 3. Where is the image formed? 4. Explain your observations. Lens Eye F Object F Upright, virtual and enlarged image Fig. 2.24: Object between F and P The image formed is virtual and cannot be projected on the screen. An enlarged, upright image can be seen through the lens on the same side with the object (Fig. 2.24) above. A simple ray diagram to locate the image when the object is placed between F and P is as shown in Fig. 2.25 below. M B I F O P F Fig. 2.25: Object OB between F and the lens. Image characteristics A magnified, upright and virtual image is formed on the same side as the object. 2.5.2 Concave lens When the object is at infinity, an upright, diminished and virtual image is formed at principal focus F. For all other positions of the object OB, an upright, diminished, virtual image is always formed between F and P (Fig. 2.26). 46 B M I P O F Fig. 2.26: Image formation by a concave lens. Example 2.1 A convex lens has a focal length of 2 cm and a real object 6 cm tall is placed 18 cm from the centre of the lens. By means of an accurate scale diagram, find the position, size and nature of the image formed. Solution Using rays 1 and 3 of the image construction, two incident rays are drawn from B and the corresponding refracted rays through the lens. The refracted rays converge at M where the image of B is formed. Scale chosen for object and image values: 1 cm = 6 cm. BBBBBB OOOOOO FFFFFF 2F2F2F2F PPPP FFFF 2F2FF2F2FF2F2FF III uuuu vvvv MMMMMM Fig. 2.27: Graphical construction of images formed by convex lens The image of O is magnified, inverted and formed at I. IM is the real image formed at 6 cm from the lens. The height of the image is 2 cm. Since the scale is 1 cm represents 6 cm, the image is 36 cm from the lens and the height of the image is 12 cm (Fig. 2.27) above. Example 2.2
A concave lens has a focal length of 2 cm and real object 1.0 cm tall is placed at 3 cm from the centre of the lens. By means of an accurate scale diagram, find the position, size and the nature of the image formed. 47 Solution Scale chosen: 1 cm to represent 1 cm Similar to Example 2.1, draw minimum two incident rays from B and the corresponding refracted rays. Since the refracted rays diverge, a virtual image is formed. The image is 1.2 cm from the lens and the height of the image is 0.4 cm (Fig. 2.28). It is diminished, upright and virtual. Fig. 2.28 Exercise 2.2 1. Name two features of the image formed by a convex lens when: (a) The object is between F and optical centre (b) (c) The object is at infinity. The object is at F. 2. Sketch a ray diagram to show image formation for an object placed between 2 F and F of a converging lens. State four characteristics of the image. 3. (a) If a convex lens picks up rays from a very distant object, where is the image formed? (b) If the object is moved towards the lens, what happens to the position and size of the image? 4. An object 2 cm high is placed 2 cm away from a convex lens of focal length 6 cm. By using an accurate drawing on graph paper, find the position, height and type of the image. 2.6 The lens formula Activity 2.9 Lens formula (Work in groups) 1. Using reference materials or internet research about the relationship between focal length, f, object distance, u, and image distance, v. 2. What is the lens formular? Derive it. 48 BuvOFFPIM 3. How is it important to learning of convex and concave mirrors or curved reflecting surfaces? The lens formula is a formula relating the focal length, image and object distance. Consider a convex lens of focal length, f, which forms a real image IM of an object OB as shown in Fig. 2.29. B 2F O D F P u I 2F M F v f Fig. 2.29: Lens formula Triangles OBP and IMP are similar (3 angles are equal) ∴ OB IM = OP IP ……………………………………… (1) Draw a line DP perpendicular to the principal axis where DP = BO. Triangles PDF and IMF
are similar (3 angles are equal) ∴ DP IM = PF IF …………………………………… (2) Since DP = OB, from equations (1) and (2), = OP IP u v = PF IF f v – f Cross multiplying, uv – uf = vf Dividing both sides by uvf uv uvf 1 – = ⇒ – = u uf uvf vf uvf 1 f 1 v. Hence 1 f 1 = + u 1 v. This is the Lens formula, where u stands for the distance of the object from the optical centre. v stands for the distance of the image from the optical centre. f stands for the focal length of the lens. 49 2.7 Sign Convention (Real is positive) We can adopt a method or a convention to describe the upward motion and downward motion. For example let the distances up be negative and down positive or vice versa. ∴ 3 m up = -3 m 3 m down = +3 m There are several sign conventions used when the distances of the object and the image are measured from the lens. In this book, we shall adopt the real is positive in which: 1. All the distances are measured from the optical centre. 2. The distances of the real objects and the real images measured from the optical centre are taken as positive, while those of virtual objects and virtual images are taken as negative. From this convention, the focal length of a convex lens is positive and that of a concave lens is negative. See Fig. 2.30 (a) and (b). P F F +f (a) P – f (b) Fig. 2.30: Real and virtual focal lengths of lenses Example 2.3 An object is placed 24 cm from the centre of a convex lens of focal length 20 cm. Calculate the distance of the image from the lens. Solution From 20 6 – 5 120 – = 1 u 1 24 = 1 120 The image distance (v) = 120 cm 50 Example 2.4 An object is placed 2 cm from the centre of a concave lens of focal length 20 cm. Calculate the distance of the image from the lens. Solution From lens formula 20 1 – 10 20 – = = –9 20 v = –20 9 = –2.2 cm, v = 2.2 cm v is negative because the image is virtual. 2.8 Magnification formula of the lens The term magnification refers to how many times an image
is bigger than the object. Linear magnification (m) is defined as the ratio of the height of the image to the height of the object. To derive magnification formula Activity 2.10 (Work in groups) Material • Graph papers Steps 1. Draw three vertical lines on a graph paper. 18 cm 6 cm 2 cm A B C Fig. 2.31 2. How many times is line B bigger than A. 3. How many times is line B bigger than C. 4. What are the units of these comparisons? 51 Earlier in this unit we have done activities where we saw that the size of images formed by lenses are either bigger or smaller than the object. The increase or decrease in size of an object is called magnification. That is Linear magnification (m) = height of the image height of the object = = image distance (v) object distance (u) = h1 h0 v u Note: Since magnification is a ratio, it does not have units Sometimes it becomes difficult to measure the height of the image or the object accurately. In such cases, magnification can be calculated in terms of distances u and v. For example, consider a convex lens where a magnified image is formed (Fig. 2.32). B 2F O F P F 2F I R u v M Fig. 2.32: Magnification Since triangles OBP and IMP are similar (3 angles are equal), the ratios of corresponding sides are equal i.e, IM IP IM OB = OB OP = IP OP = v u ∴ Hence magnification, m = IM OB = v u Magnification (m) = image distance (v) object distance (u) or m = v u 52 v Therefore m = = u also equal to the ratio of image to object distances v – u h1 h0, therefore, the ratio of image to object sizes hi –– ho is measured from the optical centre. Example 2.5 An object of height 2 cm is placed 20 cm infront of a convex lens. A real image is formed 80 cm from the lens. Calculate the height of the image. Solution hi h0 m = = ∴ hi = hi 2 = 80 20 = 8 cm ⇒ v u 8 1 80 × 2 20 10 1 Example 2.6 An object placed 30 cm from a convex lens produces an image of magnification 1. What is the focal length of the lens? Solution Magnification, m = OB IM = OP IP = 1. (Fig. 2.33)
Since m = 1; then v = u This occurs when object is at 2f. Hence 2f = 30 ∴ f = 15 cm B O P 30 cm I M u v Fig. 2.33: Image formed by convex lens 53 Example 2.7 An object of height 1.2 cm is placed 2 cm from a convex lens and real image is formed at 36 cm from the lens. Calculate (a) the focal length of the lens (b) magnification produced by the lens (c) the size of the image. Solution (a) From lens formula 36 = 1 f 1 f 18 + 1 36 = 19 36 = 1 f 36 19 = 1 f ⇒ f = 1.89 cm Focal length of the lens = 9 cm (b) m = v u = 36 2 = 3 (c) m = hi h0 ∴ hi = 3 × 1.2 = 3.6 Size of the image = 3.6 cm Example 2.8 An object of height 2 cm is placed 8 cm from a convex lens and a virtual image is formed on the same side as the object at 24 cm from the lens. Calculate (a) the focal length of the lens (b) the height of the image formed. Solution (a) From lens formula24 = 3 – 1 24 = 1 f 1 f ∴ focal length, f = 12 cm (v = –24 cm because the image is virtual) ⇒ 1 12 = 1 f 54 (b) Magnification m = v u = h1 h0 ⇒ –24 8 = h1 2 (negative sign indicate image is virtual) ∴ hi = 24 × 2 8 = 6 cm Example 2.9 A convex lens produces a real image of an object and the image is 3 times the size of the object. The distance between the object and the image is 80 cm. Calculate the focal length of the lens. Solution Magnification m = v u = 3 ∴ v = 3u ………… (1) u + v = 80 ………… (2) Solving equations (1) and (2) u + 3u = 80 ⇒ 4u = 80 ∴ u = 20 cm Hence v = 3u = 60 cm From lens formula 1 f 1 = + u 3 + 1 60 = ∴ focal length, f = 15 cm 2.9 Power of a lens 1 v = + = = 1 20 4 60 1 60 1 15 To explain what
is the power of a lens Activity 2.11 (Work in groups) Material • A lens Steps 1. Discuss with your classmates what the power of the lens is. 2. Is it possible to increase the power of a lens? Discuss. 3. Share your findings with other classmates. 55 The ability to collect rays of light and focus them at a point in the case of a converging, or to diverge them so that they appear to come from a point in the case of diverging lens is called the power of a lens. It is calculated from its focal length using the formula Power = 1 f The unit for power is the dioptre represented by the symbol D. The f must be in S.I units of length. Example 2.10 A lens has a focal length of 25 cm. Find the power of the lens. Solution f = 25 cm = 0.25 m. The focal length of convex lens = +ve (It forms real image) 1 ∴ Power = NB: For a concave lens f = -ve (because a concave lens forms a virtual image) +0.25 = +4 m–1 ∴ Power = 1 -0.25 = -4 m–1 Exercise 2.3 1. Define the terms: principal axis, optical centre and focal length of a convex lens. 2. With the help of a diagram, show the action of a convex lens as a converging lens. 3. The focal length of a diverging lens is 15 cm. With the help of a diagram explain the meaning of this statement. 4. Fig. 2.34 below shows a convex lens of focal length 15 cm and two rays of light parallel to the principal axis. Copy and complete the diagram to show the path of these rays as they pass through the lens. Label the position of the principal focus as F. P axis Fig. 2.34 5. Draw ray diagrams showing how a convex lens could be used to produce (a) a real and diminished image (b) a virtual and magnified image of a real object. 56 6. Fig. 2.35 is drawn to scale. One incident ray from the object is parallel to the principal axis and other ray passes through the principal focus of a convex lens. Copy and complete the diagram to show the path of the ray through the lens. Hence determine (i) position of the image (ii) the magnification produced by the lens. Convex lens Object Fig. 2.
35 7. Copy the table below and put a tick () in three of the boxes to describe the image formed by a diverging lens. Table 2.1 Magnified Diminished Upright Inverted Virtual Real 8. Draw a diagram to show how a convex lens produce a virtual image. 9. Fig. 2.36 shows two rays of light approaching a thin diverging lens. Copy and complete the diagram and show the path of the rays as they pass through and emerge out of the lens. Label the position of the principal focus F. Fig. 2.36 10. Fig. 2.37 is drawn to scale. An object OB placed in front of a convex lens of focal length 5.0 cm. Copy and complete the diagram and (a) show the position of the image (b) find the size of the image 57 Convex lens B O F F Fig. 2.37 11. A convex lens is used to form an upright, magnified image of an object placed 6 cm from the lens. Calculate the focal length of the lens, if the magnification produced is 4. 12. An object 3 cm high is placed 20 cm from a lens of focal length –25 cm. Find the position, size and the nature of the image formed. 13. At what distance must an object be placed from a convex lens of focal length 20 cm so as to get real image 4 times the size of the object? 14. An object 3 cm high is placed 30 cm from a convex lens of focal length 20 cm (a) Find the position, size and the nature of the image formed (b) If the same object is now moved by 20 cm towards the lens, calculate the magnification produced by the lens. 15. A convex lens forms a focused image on a screen when the distance between an illuminated object and the screen is 1 m. The image is 0.25 times the size of the object. Calculate (a) the object distance from the lens (b) the focal length of the lens used. 16. An object 3 cm high is placed 150 cm from a screen. Calculate the focal length of the lens that has to be placed between the object and the screen, so as to produce a real image 6 cm high on the screen. 17. An object 6 cm high is placed 30 cm from a diverging lens of focal length 15 cm. With the help of a scale diagram determine (a) the position of the image. (
b) the magnification produced by the lens. 18. A real object placed 8 cm in front of a converging lens produces an image at a distance of 2 cm from the lens and on the same side as the object. Calculate the focal length of the lens. 19. A diverging lens of focal length 24 cm forms an image at 18 cm from the lens. Calculate the distance of the object from the lens. 20. In an experiment to determine the focal length of a converging lens, a student obtains the results shown in Table 2.3. 58 Table 2.2 u (cm) v (cm) 21.0 50.0 24.0 40.0 33.0 22.5 36.0 25.0 45.0 22.0 60.0 20.0 (i) Plot a graph of u (x-axis) against v (y-axis) (ii) Using the graph, determine the focal length of the lens. 2.10 Applications of thin lenses 1. The human eye The human eye consists of a nearly spherical ball of about 2.5 cm diameter except for a slight bulge at the front. Fig. 2.38 shows the cross-section of the human eye with the optic nerve leading to the brain. Retina Optic nerve Suspensory ligaments Iris Lens Pupil Cornea ciliary muscles Fig. 2.38: A cross-section of a human eye. The front portion of the eye is known as the cornea and is slightly bulged outwards and is transparent in nature. Behind the cornea, there is a diaphragm called the iris, with a hole in the middle known as the pupil. Behind the iris is a crystalline lens. This is a biconvex converging lens made of a large number of jelly-like layers which are flexible and transparent in nature. The lens is suspended inside the eye by the help of suspensory ligaments which fasten it to the ciliary muscles. These muscles control the shape of the lens. The lens forms a real, diminished and inverted image on the retina. Light falling on the retina produces a sensation in the cells which then send the electrical signals to the brain by the nerve known as the optic nerve. The amount of light reaching the retina is regulated by the size of the pupil. When a bright object is viewed, the iris reduces the size of the pupil so as to admit less light, whereas in dark light the iris contracts
so as to admit as much light as possible. An image formed by the eye lens leaves an impression on the retina for about 0.1 second. This persistence of the vision enables us to “see” cinema or television pictures which appear to change smoothly from one image to the next without 59 any interruption. In a cinema theatre or television screen about 20 pictures are projected per second. During the time interval between the pictures, the eye “remembers” the previous picture. Image formation in the eye When one looks at far objects, such as a tree, the eye lens becomes thinner and the focal length of the lens increases. The ciliary muscle is relaxed and the lens has the longest focal length. It is able to focus rays from distant objects onto the retina (Fig. 2.39 (a)). To view the objects close to the eye, the lens becomes thicker and the focal length of the lens decreases. The contraction of the ciliary muscle reduces tension in the lens and the lens becomes more curved with short focal length and is more powerful. The lens now focuses images of near objects onto the retina (Fig. 2.39 (b)). For both far view and close view the image formed is real, diminished and inverted. This process by which the lens of the eye changes its focal length and produces focused images of both distant and near objects on the retina is called accommodation. (a) Far view – thin lens (b) Near view – thick lens Fig. 2.39: Accommodation of the eye In the eye, the distance between the eye lens and the retina remains the same, whereas the lens automatically changes its focal length according to the distance of the object. This effect is brought about by changing the shape of the ciliary muscle attached to the lens. Defects of vision A normal human eye can accommodate the range of distances from far off objects to objects close to the eye. There is, however, a limit to the power of accommodation of the eye. As a person grows older, the power of accommodation gradually decreases. Also, despite the ability of the eye to adjust its focal length by changing the lens shape, some eyes cannot produce clear images over the normal range of vision. This type of defect may arise due to the eyeball being slightly too long or too short compared to the normal spherical ball or due to the curvature of the cornea being defective. 60 The most common defects of vision are short-sightedness (myopia) and longsightedness (hypermetrop
ia) (see Fig. 2.40, (a) (b) ). (a) Short-sightedness (b) Long-sightedness Fig. 2.40: Short-sightedness and long-sightedness (a) Short-sightedness or Myopia A person suffering from short-sightedness can only see nearby objects. The image of a distant object is formed in front of the retina as shown in Fig. 2.33 (a). This defect arises due to the eyeball being too long or more refraction takes place at the cornea and hence the focal length of the eye lens becomes short. In order to correct this defect, a concave lens of appropriate focal length should be used. This lens diverges the rays from a distant objects so that they appear to come from a virtual image formed at a point closer to the lens. The eye can focus on this virtual image, as shown in Fig. 2.41 (b). Far off objects at infinity convex lens Virtual image (a) Myopic eye (b) Corrective measure Fig. 2.41: Short-sightedness and corrective measure. (b) Long-sightedness or Hypermetropia A person suffering from long-sightedness can see distant objects clearly, but cannot see distinctly objects lying closer than a certain distance. The image of a nearby object is formed behind the retina as shown in Fig. 2.42 (a). The defect arises due to the eyeball being too short or due to the curvature of the cornea being defective and the focal length of the eye lens becoming longer. In order to correct this defect, a convex lens of appropriate focal length should be used. This lens converges the ray from a near object so that they appear to come from a virtual image formed at a point far off from the lens. The eye focuses on this virtual image as shown in Fig. 2.42 (b). 61 Near object Virtual image (a) Hypermetropic eye (b) Corrective measure Fig. 2.42: Long-sightedness and the corrective measure. 2. The lens camera A camera is a device used to take photographs. A human eye, though in principle, is similar to a camera, is far superior than the finest camera ever made by man. Fig. 2.43 (a) shows some parts of a lens camera. Fig. 2.43 (b) shows a commercial camera. Film Lens Diaphragm Shutter (a) (b) Fig
. 2.43: A lens camera A camera consists of a converging lens and a light sensitive film or plate enclosed in a light-tight box, blackened from inside. The lens focuses light from an object to form a real, diminished and inverted image on the film. Focusing of objects is done, by adjusting the distance between the lens and the films. The amount of light entering the camera through the lens is regulated with the help of a diaphragm, with an adjustable opening in the middle. Light is admitted by the shutter, which opens for different required intervals of time and then closes automatically. During this interval of time, the film is exposed to light from the object. The film contains light sensitive chemicals that change on exposure to light. The film is developed to get what is called a negative. From the negative a photograph (positive) may be printed. Comparison of a lens camera with the human eye Similarities 1. Both use converging lenses 2. Both produce real, inverted, diminished images. 62 3. Both can control the amount of light entering the device. 4. Both are black inside. Differences Camera Eye 1. Focal length of the lens is constant. 1. Focal length of the lens changes with the thickness of the lens. 2. Distance between the lens and the 2. Distance between the lens and the retina film can be altered. is a constant. 3. Focuses objects between a few centimetres from the lens to infinity. 3. Focuses objects between 25cm from the lens to infinity. 4. Form permanent images at the film. 4. Form temporary images at the retina. Example 2.11 A lens camera is used to take photograph of a distant building. A well focused image is formed on the film. The lens of the camera is 6 cm from the film. (a) What is the focal length of the lens? Give reasons for your answer. (b) If the camera is then used to take a photograph of a person 2.0 m away from the lens, without moving the camera, in which direction should the lens be moved in order to produce the best possible image? Solution (a) Focal length of the lens = 6 cm. Since the object is a distant building,the light rays incident on the lens are almost parallel and are brought to focus at the principal focus of the lens. (b) As the object distance is only 2.0 m i.e. object distance, u, has decreased as compared to the
distance in part (a). Hence the image distance, v, must be increased. To achieve this, the lens has to be moved forward towards the person. 3. Simple microscope A magnifying glass also known as a simple microscope is an instrument used to view the details of very small objects. It consists of a single converging lens of short focal length. When an object is placed within the focal length of such a lens, a magnified image which is virtual and upright is formed on the same side of the object. This image can be viewed by placing the eye close to the lens. The distance of the object from the lens is adjusted till an enlarged image is formed at a distance D, which is about 25 cm from the unaided eye. The distance D is referred to as 63 the least distance of distinct vision. Fig. 2.44 below illustrates the operations of a simple microscope. M I B O F D Fig. 2.44: A simple microscope. F Eye The action of a simple microscope The object OB, when viewed by an unaided eye, cannot be brought closer to the eye, than the distance D (Fig. 2.45 (a)). Otherwise the image as seen by the eye will not be clearly visible. When the same object is viewed through the magnifying glass, it moves nearer to the eye so that a magnified image is formed at the same distance D as before (Fig. 2.45 (b)). Therefore a simple microscope enables us to bring an object very close to the eye making it appear magnified and yet clearly visible. B O M I D v (a) B O (b) Eye F u Eye Fig. 2.45: Working of a simple microscope 64 Magnifying power of a simple microscope Linear magnification m = IM OB = v u Linear magnification of a lens is also called magnifying power of the instrument. In a simple microscope, the image distance v is negative, as the image is virtual. 1 f Hence, from the lens formula = – 1 v 1 u Multiplying throughout by v and simplifying From the above expression the shorter the focal length of the lens, the greater is the magnifying power of the instrument. Hence a simple microscope uses a converging lens of short focal length. Example 2.12 Calculate the magnification produced by a lens of focal length 5.0 cm used in a simple microscope, the least distance of distinct vision being 25 cm. Solution In this example, the image distance v = D = 25 cm. Magn
ification, m = 1 + v f = 1 + ( 25 5 ) = 1 + 5 = 6 Hence the magnification produced by the lens = 6. 4. Compound microscope In a simple microscope, the magnifying power cannot be increased beyond a certain limit, by decreasing the focal length of the lens. This is due to the mechanical difficulties of using a lens of very short focal length. A compound microscope uses two separate converging lenses, placed coaxially within two sliding tubes, to obtain a higher magnifying power. The lens O, nearer the object is called the objective lens and the lens E closer to the eye, is called the eyepiece lens. Though both these lenses are of short focal lengths, the eyepiece has a comparatively larger focal length than the objective lens. The final image formed is magnified, virtual and inverted as shown in Fig. 2.46. 65 Objective (O) Eye Eyepiece (E) 2F B O F I F 2F B' M Fig. 2.46: A compound microscope Action of a compound microscope The object OB is placed between F and 2F of the objective lens. A real, inverted, magnified image O′ B′ is formed beyond 2F of the objective lens. The position of the eyepiece lens is adjusted so that this image O′B′ falls within its focal length. The eyepiece then acts as a magnifying glass and produces a final magnified, virtual and inverted image IM at a distance of distinct vision D from the eye, placed very close to the eyepiece. If m1 is the magnification produced by the objective lens and m2 is the magnification produced by the eyepiece lens, then the magnification produced by the system of lenses m is given by, m = m1 × m2 If the first image O′B′ formed by the objective lens is exactly at the principal focus Fe of the eyepiece lens, then the final image IM will be formed at infinity. The image will be inverted and well enlarged. At this position, the compound microscope is said to be in normal adjustment. A good compound microscope produces a very high magnification. High magnification microscopes are usually used in research work in science (see Fig. 2.47). 66 Fig. 2.47: High magnification microscope for research work Example 2.13 In a compound microscope, the focal length of the objective lens is 2.0 cm and that of the eyepiece is 2.2 cm and they are placed at a
distance of 8.0 cm. A real object of size 1.0 mm is placed 3.0 cm from the objective lens. (a) Use the lens formula in turn for each lens to find the position of the final image formed. (b) Calculate (i) the magnification produced by the arrangement of these lenses and (ii) the size of the final image viewed by the eye? Solution (a) For the objective lens Solving this equation gives v = 6 cm As shown in Fig. 2.48, the real image I1M1 is formed at 6 cm from the objective lens. I1M1 acts as an object for the eyepiece (u = 2 cm). 1 For the eyepiece lens f 1 2.2 1 u 1 2 Solving this equation gives v = –22 cm 1 v 1 v = + = – 67 The negative sign shows that the image formed by the eyepiece is virtual and is formed on the same side as the object I1M1. The final image I2M2 is at a distance of 22 cm from the eyepiece (see Fig. 2.48). O 3 cm 6 cm 2 cm E Eye I2 M2 B O 22 cm I1 M1 8 cm Fig. 2.48: Arrangement of lenses in a compound microscope (b) (i) The magnification produced by the system of lenses m = m1 × m2 for the eyepiece. for the objective lens and m2 = where m1 = v u v u m = 6 3 × 22 2 = 2 × 11 = 22 (ii) The size of the final image = size of the object × m = 1 × 22 = 22 mm Topic summary • A lens is a transparent medium bound between two surfaces of definite geometrical shape. • Thin lenses may either be converging or diverging. • A convex lens is thicker at its centre than its edges and converges the light incident on it. • A concave lens is thicker at its edges than at the centre and diverges the light incident on it. • The following are some of the important terms used in spherical lenses: principal axis, optical centre, principal focus, focal length. • The focal length of a convex lens is positive, while that of a concave lens is negative. • The characteristics of the image formed by a converging lens depends on the position of the object (see Table 2.3). 68 Position of object At infinity (far
away) Beyond 2F At 2F Between 2F and F At F Between F and P Table 2.3 Position of image F Nature of image formed real and inverted diminished Size of image formed compared to object Between F and 2F At 2F Beyond 2F At infinity (far away) Same side as object real and inverted diminished real and inverted Same size real and inverted Magnified real and inverted Magnified Virtual and upright Magnified • A diverging lens always forms a virtual, upright, diminished image between F and P (except when the object is at infinity). • Magnification (m) is defined as the ratio of the height of the image to the height of the object magnification = height of image height of object = • Lens formula is given by 1 u + 1 v = image distance object distance 1 f where u is the object distance, v the image distance and f the focal length of the lens. • Short-signtedness and long-sightedness are two most common defects of a human eye • A lens camera, simple microscope, compound microscope are some examples of optical instruments. 69 Topic Test 2 1. Describe an experiment to illustrate that white light is composite in nature. 2. Fig. 2.49, drawn to scale, shows two rays starting from the top of an object OB incident on a converging lens of focal length 2 cm. B O Fig. 2.49: Equilateral glass prism (a) Copy and complete the diagram to determine where the image is formed. (b) Add one more incident ray from B through the principal focus and draw the corresponding refracted ray through the lens. (c) Calculate the magnification produced by the lens. 3. Fig. 2.50 shows an object placed at right angles to the principal axis of a thin converging lens. Fig. 2.50: Equilateral glass prism (a) Calculate the position of the image formed. (b) Give an application of this arrangement of a lens. (c) Describe the nature of the image formed. 4. Describe with the aid of a ray diagram, how an image is formed in a (i) simple microscope (ii) lens camera. 5. A converging lens is used to form an upright image, magnified 5 times of an object placed 6 cm from the lens. Determine the focal length of the lens 70 FO6 cm8 cmLensBF 6. Fig. 2.51 shows two converging lenses L1 and L2 placed 8 cm from each other
. The focal length of the lens L1 is 2 cm and that of L2 is 2.8 cm. An object 1.0 cm high is placed 3 cm from lens L1. Fig. 2.51: Equilateral glass prism (a) Construct a ray diagram to scale, on a graph paper to show the position of the final image as seen by the eye of a person. (b) Determine the magnification obtained by this arrangement. 71 ObjectEye3 cm8 cmL1L2 72 UNIT 2 Forces and Turning Effects Topics in the unit Topic 3: Moment of a Force Topic 4: Centre of Gravity and Equilibrium Learning outcomes Knowledge and Understanding • The effects of forces and centre of gravity. Skills • Design tests to locate the centre of gravity of regular objects by method of balancing and locate the centre of gravity of irregular shaped objects by means of a plumb-line. • Observe carefully. • Predict what might happen. • Use appropriate measures. • Draw a simple diagram to show moment of a force Interpret results accurately. • Calculate problems related to moments of forces. • Report findings appropriately. Attitudes • Appreciate the applications of moment of forces. Key inquiry questions • Why the pivot is important in taking moment of force? • What do you understand by couple forces? • Why cars are designed to have a wide base? • Why an object cannot be in equilibrium if it is in motion? • Why an overloaded vehicle is prone to overturn? 73 TOPIC 3 Moment of a Force Unit Outline • Moment of force • Principle of moments • Couple • Determination of centre of mass of a regular object • Applications of moment of a force Introduction One of the effects of a force we learnt about in Secondary 1 is that it produces a turning effect on body. But how can we quantify the turning effect? What are some of the applications of this effect in our daily lives? In this topic, we will seek answers to these questions. 3.1 Moment of a force A number of simple machines like levers, pliers, spanners and so on do work when a force produces a turning effect on some of their parts. It is important to know where the force should be applied for the machine to be more efficient in doing work. The following activity will help us to investigate the turning effect of a force. To investigate the turning effect of a force Activity 3.1 (Work in groups) Materials • A ruler. Steps 1. Balance a ruler on a finger. At what point did the
ruler balance? Why do you think it balances at that point? 2. Press the ruler at one end. Observe what happens to the ruler. 3. Repeat the experiment by pressing the ruler on the other end. Why do you think the ruler behaves in such a manner? 74 Moment of a Force Fig. 3.1: Balancing a ruler on a finger In both cases, the ruler turns about the finger. When the force F1 is applied at one end, the ruler turns in anticlockwise direction about the finger. When the force F2 is applied at the other end, the ruler turns in the clockwise direction. Both F1 and F2 produce a turning effect on the ruler about the finger. The turning effect of a force about a point is called the moment of the force about that point. This moment depends on the force applied and its distance from the point. The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. The moment of a force about a point is either clockwise or anticlockwise about the point. In Fig. 3.1, the anticlockwise moment about the finger is F1 × d1. The clockwise moment about the finger is F2 × d2. Moment of a force about a point = Force × Perpendicular distance from the point to the line of action of the force = F × d SI unit of moment of a force Moment of a force = Force (N) × Perpendicular distance (d). Therefore, the SI unit of moment is newton metre (Nm). Moment of a force is a vector quantity since it has both magnitude and direction. Example 3.1 A student applies a force of 10 N to the handle of a door, which is 0.8 m from the hinges of the door (Fig. 3.2). Calculate the moment of the force. 75 Solution Moment of a force about a point hinges = Force × perpendicular distance from the point to the force. = (10 × 0.8) Nm = 8 Nm in the clockwise direction. door 0.8 m F Fig. 3.2: Moment in opening the door Example 3.2 Calculate the moment of the force about the fulcrum when a pet dog of mass 10 kg is at a distance of 1.2 m from the fulcrum of the seesaw as shown in Fig. 3.3. mass = 10 kg ful
crum Solution F = Weight of dog = mg 1.2 m Fig. 3.3: Moment of force = 10 kg × 10 N/kg = 100 N Moment of the force about the fulcrum = Force × perpendicular distance from the fulcrum = 100 N × 1.2 m = 120 Nm in the clockwise direction. Exercise 3.1 1. Define ‘moment of a force’ and state its SI unit. 2. A force of 20 N is applied to open the gate of a fence as shown in Fig. 3.4. Calculate the moment of the force about the hinges if the force is applied at the edge of the gate. 76 hinges 1.2 m 0.3 m F 3. Give the scientific reasons for the following: Fig. 3.4: Moment in opening the gate (a) The handle of a door is fixed far from the hinges. (b) A pair of garden shears has small blades and long handles. (c) A lighter boy is able to produce same moment as that of a heavier girl on the seesaw as shown in Fig. 3.5 below. girl boy Fig. 3.5: Moment on a seesaw 4. A person applies a force of 500 N and produces a moment of force of 300 Nm about the wheels of a wheel cart (Fig. 3.6). Calculate the perpendicular distance, d, from the line of action of the force to the wheels. l o a d F = 500N weight d Fig. 3.6: Moments in a wheelcart 77 3.2 The principle of moments The principle of moment gives the relationship between two moments that are at the same turning point (fulcrum). Activity 3.2 (Work in groups) Materials To investigate the principle of moments • A metre rule • Three 100 g mass • string • support e.g. clamp Steps 1. Suspend a uniform metre rule from a firm support e.g. clamp, at the 50 cm mark, i.e. at its mid point G as shown in Fig. 3.7 (a) using a string. 2. Suspend a 100 g mass at a point A as shown in Fig. 3.7 (b). Why do you think the ruler balances as shown? (a) O (b) 50 cm 100 cm 0 P P A 100 cm 100 g mass F = 1 N Fig. 3.7: Principle of moments. 3. Now suspend a 200 g
mass at a point B near the 0 cm mark (Fig. 3.8 (a)). What has happened? The system turns in the anticlockwise direction. Now adjust the position of B till the system balances horizontally as shown in Fig. 3.8 (b). Explain the observations. (a) B O 200 g mass f = 2N P A 100 cm (b) 0 B P A F = 1N 2N F = 1N Fig. 3.8: Principle of moments. The metre rule turns in the clockwise direction. There is a moment of force in the clockwise direction due to the force acting vertically downwards at point A. Moment of the force in the clockwise direction = Force × perpendicular distance = 1.0 × PΑ 78 Moment due to the 2 N force about P is 2.0 × PB in the anticlockwise direction. Measure the distance PA and PB. Compare the values of 1.0 × PA and 2.0 × PB. What can you say about these values? We note that the two moments are equal in magnitude and opposite in direction. The clockwise moment of the 1 N force about point P is equal to the anticlockwise moment of the 2 N force about point P. Activity 3.3 (Work in groups) Materials To design and investigate the principle of moments with more than two forces • Four mass • A metre rule • string • Firm support e.g. clamp Instructions 1. In this activity, you will design and carry out an investigation to investigate the principle of moments with more than two forces. 2. By modifying the set-up, we used in Activity 3.2, with the materials provided, i.e using its masses, conduct an investigation. Sketch the new set-up. 3. Write a brief procedure and carefully execute the procedure to determine principle of moments with four masses. 4. By applying the relevant formula and relationships. Calculate the moments 0 about point P. 5. Compare your values with other groups. 6. What are some possible sources of errors? How can they be minimised in your investigation? Write a report and present it in a class discussion. 0 D F2 P 50 cm C F1 B A 100 cm F4 F3 Fig. 3.9: Balanced metre rule under action of forces. The sum of the clockwise moments = F3 × PA + F4 × PB. The sum of the anticlockwise moments = F1 × PC + F2 × PD. What can
you say about F1 × PC + F2 × PD and F3 × PA + F4 × PB? 79 A B 100 cm C F1 P 50 cm D F2 Fig. 3.9: Balanced metre rule under action of forces. F3 F4 From Activity 3.2 and 3.3 we can conclude that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point, when the metre rule is balanced. In Activity 3.3, we saw how a body can be balanced by a number of forces. When a body is balanced under the action of a number of forces, it is said to be in equilibrium. The results of Activities 3.2 and 3.3 are summarised in what is known as the principle of moments. It states that, when a body is in equilibrium under the action of forces, the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point. Activity 3.4 (Work in groups) Materials • A metre rule Steps To verify the principle of moments • Seven, 50 g masses 1. Consider a uniform metre rule suspended at its mid point P, which is the 50 cm mark. Suspend 3 masses; 200 g, 100 g and 50 g and adjust their positions A, B and C till the system is in equilibrium as shown in Fig. 3.10. 2. Calculate the distances PA, PB and PC in metres and enter the values in a table. 0 A F1 2N P B C 100 cm 50 cm F2 1N F3 0.5 N Fig. 3.9: Verifying the principle of moments. 3. Repeat the experiment by changing the positions of A or B or C or all the three so that the metre rule balances horizontally in each case. Record the results in Table 3.1. Table 3.1 F1 × PA (Nm) F2 × PB + F3 × PC (Nm) PA (m) PB (m) PC (m) 1 2 3 4 4. Complete the table and explain the results. 80 It is seen that the last two columns are equal for each set of results proving that the sum of the clockwise moments is equal to the sum of the anticlockwise moments about point P. F2 × PB + F3 × PC = F1 × PA Example 3.3 A uniform metre rule is pivoted at its centre P, and 3
masses are placed at A, B and C as shown in Fig. 3.10. Find the value for the weight W of the mass M placed at C so that the metre rule is balanced horizontally. 0 cm 50 cm A 2N 10 cm P B 1N 30 cm 40 cm Fig. 3.10: Principle of moment Mass M 100 cm C W Solution Taking moments about P, when the metre rule is in equilibrium. Sum of the clockwise moments = Sum of the anticlockwise moments W × 0.4 = (2.0 × 0.3) + (1.0 × 0.1) (2.0 × 0.3) + (1.0 × 0.1) 0.4 W = = 1.75 N Example 3.4 John, Joyce and Janet sat on a seesaw as shown in Fig. 3.11 below. Where is John, whose mass is 30 kg seated so that the seesaw is balanced horizontally if the masses of Joyce and Janet are 50 kg and 20 kg respectively? Joyce Janet P John 1m pivot 2 m d Fig. 3.11: See saw at balance Solution John’s weight = 600 N, Joyce’s weight = 500 N, Janet’s weight = 200 N Taking moments about the pivot, Sum of clockwise moments = Sum of the anticlockwise moments about the pivot about the pivot 81 600 × d = 500 × 2 + 200 × 1 600 d = 1 000 + 200 d = 1 200 600 = 2 m John should sit at a distance of 2 m from the pivot. Example 3.5 The uniform plank of wood in Fig. 3.12 is balanced at its center by the forces shown. Determine the value of W in kg. 2.6 N 24 cm 6 cm 2 N W Fig. 3.12 Solution Note that the 2.3 N produces an anticlockwise moment. Sum of clockwise moments = Sum of anticlockwise moments 0.24 × (2 + W) = 2.6 × 0.30 0.48 + 0.24W = 0.78 ⇒ W = 0.78 – 0.48 = 1.25 N 0.24 = 0.125 kg To determine the mass of an object using the principle of moments Activity 3.5 (Work in groups) Materials • A metre rule • A known mass Steps • An unknown mass • Support 1. Suspend a uniform metre rule at its mid point P. Suspend the object of mass
m, using a string, from a point A. Suspend a known mass M on the other side of the metre rule and adjust the position of the mass M till the metre rule is horizontal as shown in Fig. 3.13. 82 0 cm A object of mass, m P 50 cm B 100 cm mass M mg Mg Fig. 3.13: Finding unknown mass m. 2. Record the distances PA and PB. Repeat the experiment by changing the position of the object or the mass M. Enter the readings of M, PA and PB in a tabular form as shown in Table 3.2. Table 3.2 Mass M(g) PA(cm) PB(cm) M.PB m = (g) PA 1 2 3 3. Calculate the mean value for the mass of the object from the last column. Mean Taking moments about P, (mg) × PA = (Mg) × PB, m × PA = M × PB, (g cancels out) M × PB PA ∴ m = Exercise 3.2 1. State the principle of moments. 2. A boy of mass 20 kg sits on one side of a log of wood and 10 m away from the pivot. A girl of mass 30 kg sits on the opposite side of the log. How far is the girl from the pivot. 3. State two conditions for a system to be in a state of equilibrium. 4. Weights of 25 N, 28 N and 8 N were suspended on a uniform plank of wood pivoted at its center on a knife-edge. Fig. 3.14 shows the plank immediately after it was placed on a knife-edge. 83 Fig. 3.14 (a) Work out: (i) Sum of Clockwise moments (ii) Sum of Anticlockwise moments (b) Is the bar in a state of equilibrium? Give a reason for your answer. 5. A uniform metre rule is pivoted at the center. It is balanced by weights of 8 N, F and 24 N suspended at 34 cm, 43 cm and 30 cm marks respectively (Fig. 3.15). Calculate the value of F. 50 cm 60 cm 46 cm 34 cm 0 cm 100 cm 8 N F 24 N Fig. 3.15 6. Fig. 3.16 below shows a modern bar balanced by forces of 210 N, 100 N and 25 N. Calculate the distance d. 210 N 6 m 5 m d m 25 N 100 N Fig. 3.
16 3.3 Couple Couple refers to two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. Its effects is that it creates rotation without translation as shown in Fig. 3.17. 84 Fig. 3.17: A couple A couple produces a turning effect on a body. Moment of a couple Fig. 3.18 shows a couple acting on bar AB. F A d C B F Fig. 3.18: Moment of a couple Since the pivot is at C, the moment of the force F acting at point A = F × AC, in the clockwise direction. Similarly, the moment of the force, F acting at point B = F × BC, also in the clockwise direction. ∴ The moment of the couple = (F × AC) + (F × BC) = F (AC + BC), but AB = BC + BC = F × AB = F × Arm of the couple. The moment of the couple, called the torque which is defined as the total rotating effect of a couple and is given by the product one of the forces and the perpendicular distance between the forces. Hence, in Fig. 3.18, Torque = F × perpendicular distance AB. SI unit of torque is the newton-metre (Nm) 85 Example 3.6 In Fig. 3.19, each force is 4 N and the arm of the couple is 20 cm. Calculate the moment of the couple. F = 4 N 20 cm Fig. 3.19: Moment of a couple F = 4 N Solution The moment of the couple = F × perpendicular distance = 4 N × 0.20 m = 0.80 Nm. Some common real life examples of a couple are observed when: • Forces are applied by hands to turn a steering wheel of a motor car (Fig. 3.20 (a)) or the handle bars of a bicycle. • A water tap is opened or closed (Fig. 3.20 (b)). • A corkscrew is twisted into a cork in the mouth of a bottle. (Fig. 3.20 (c)). Fig. 3.20: Moment of couple 86 Exercise 3.3 1. Fig. 3.21 is a water tap in use. If the diameter of a circular path made by the tap (knob) when is open and closed is 20 cm. Calculate the moment of the couple. 2. Calculate the torque in Fig. 3.22 below
. Fig. 3.21: Moment of couple of a tap 14 cm 14 cm 8N Fig. 3.22: Torque 3. A steering wheel of a truck has a diameter of 30 cm. If the driver is holding the wheel with both hands, while negotiating a corner, calculate the force applied by the right hand if the left hand is pulling the wheel by a force equal to 200 N. 3.4 Centre of mass of a body In one of his experiments, Sir Isaac Newton showed that bodies experience a force of gravity exerted on them by the earth. This force of gravity is always directed towards the earth’s centre and is called the weight of the body. How is this weight distributed throughout the body? The answer to this question is found in the following experiment. 3.4.1 Determining centre of mass of regular objects Activity 3.6 (Work in groups) Materials To determine the weight of a beam (uniform metre rule) • Uniform metre rule • A mass, m • A knife edge (fulcrum) 87 Steps 1. Balance a uniform metre rule of mass m on a fulcrum and adjust its position until the metre rule is horizontal. Note the position P, where it is pivoted (Fig. 3.23(a)). 2. Move the fulcrum to a point A, say to the right of P. Observe what happens. Why do you think the ruler changes its state of equilibrium. 3. Place a mass M between A and 100 cm mark and adjust its position B until the metre rule is horizontal (Fig. 3.23(b)). (a) (b) Fig. 3.23: Determination of mass of a uniform metre rule 4. Find the lengths PA and BA. Repeat the experiment by changing the position of A or the mass M. 5. Record the mass of M, length PA and length BA in a table (Table 3.3). Table 3.3 Mass M(g) BA(cm) PA(cm) m = (g) M.BA PA 1 2 3 6. Take moments and determine the value of m (the mass of the metre rule). Mean m = Taking moments about A mg × PA = Mg × BA m = M × BA PA (g cancels out) Calculate the mean value for the mass of the metre rule from the last column of Table. 3.4. The weight of the metre rule = mg. 88 Example 3.7 A uniform metre rule
pivoted at the 30 cm mark is kept horizontal by placing a 50 g mass on the 80 cm mark. Calculate the mass of the metre rule (Fig. 3.24). Fig. 3.24: Determining the mass of the metre rule Solution Let the mass of the metre rule be m Force due to m = m × g where g = 10 N/kg = 10 m newtons Force due to 50 g mass = PA = 10 cm = 0.1 m AB = 20 cm = 0.2 m 50 1 000 × g = 0.05 × g = 0.50 N. By the principle of moments, taking moments about point A, 10 m × 0.1 = 0.50 × 0.2 m = 0.50 × 0.2 10 × 0.1 = 0.10 1 = 0.1 kg or 100 g Example 3.8 A coffee table of mass 22 kg and length 1.3 m long is to be lifted off the floor on one of its shorter sides to slip a carpet underneath. Calculate the maximum force needed to lift the table. 89 Solution Fig. 3.25 shows the forces acting on the table where F is the lifting force and 220 N is the weight of the table acting at the center. F A 0.8 m 0.8 m 220 N Fig. 3.25 Taking the movement about point A Sum of Clockwise moments = Sum of Anticlockwise moments 220 N × 0.8 m = F × 1.6 m ⇒ F = 220 × 0.8 1.6 = 110 N ∴ The minimum force required = 110 N Exercise 3.4 1. A uniform metre rule of uniform width 2.5 cm and thickness 0.5 cm is suspended at the 78 cm mark and kept balanced by hanging a mass of 150 g at the 100 cm mark (Fig. 3.26). Calculate, (a) the mass of the metre rule, (b) the density of the material of the metre rule, (c) the tension T in the string. Fig. 3.26: A system at balance 2. A uniform metre rule is balanced at the 20 cm mark by a mass of 240 g placed at one end. (a) Draw a diagram to show the state of balance of the metre rule (b) Determine the weight and mass of the metre rule. 3. A non-uniform plank AB shown in Fig. 3.27 is balanced when a force of 200 N is applied
at the end B. The centre of gravity, G, is shown. 90 Fig. 3.27: Centre of mass of a plank Calculate the: (a) weight and (b) mass of the plank. 3.5 Applications of moment of a force The following are some of the common examples which illustrate the turning effect of a force i.e moment of a force: 1. Opening or closing a door. 2. Opening a bottle using a bottle opener (Fig. 3.28(a)). 3. A pair of scissors or garden shears in use (Fig. 3.28(b)). 4. Children playing on a “see-saw”. 5. A wheelbarrow being used to carry some load (Fig. 3.28(c)). 6. A wheel cart being used to lift heavy loads. 7. A screwdriver being used to tighten/loosen a screw. 8. A crowbar being used to move large object (Fig. 3.28(d)). Load Pivot P Effort Effort P Load Pivot (a) Bottle opener (b) Scissors 91 Load Effort Pivot Load Pivot Effort (c) Wheelbarrow (d) Crowbar Fig. 3.28: Applications of moment of a force Note In all these tools the length of the handle determines the amount of effort to be used. Activity 3.1 will help us to understand this. Activity 3.7 (Work in groups) Materials To investigate the effect of length to the effort needed in using a tool Exercise books, 3 cm and 15 cm rulers, doors. Steps 1. Lift a book using a 15 cm ruler. Repeat the same using a 30 cm ruler. Why is it easier to lift a book using a longer ruler than the shorter one? 2. Open the classroom door with your hand near the door hinge. Now, open the same door with your hand far away from the door hinge. Of the activities above, when do you use less effort? We use less effort when using 30 cm ruler to lift a book. Several experiments have been done on several tools and machines and proved that: (a) The longer the handle, the lesser the effort used when using the machines. (b) The shorter the handle, the more the effort used while working with the tools and machines. Therefore, the manufacturers always design tools and machines such as bottle openers, see-saw, water taps, spanners and wheelbarrows with longer handles so that very little
effort be used in working with them. 92 Exercise 3.5 Sketch and locate the effort, pivot, load in each of the following tools and machines. (a) (d) (g) (b) Water tap (e) (h) (c) See-saw (f) Hammer (i) Bottle opener Spanner Spade pliers Broom (j) Arm Spoon Topic summary • The moment of a force about a point is the turning effect of the force about the point. • The turning effect of the force depends upon the magnitude of the force applied and the perpendicular distance from the pivot. • The moment of a force is the product of the force applied and the perpendicular distance from the pivot to the line of action of the force. • The moment of a force is a vector quantity and its SI unit is Newton metre. • Principle of moments states that the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point, when the system is in equilibrium. Topic Test 3 1. (a) State the principle of moments. (b) Describe an experiment to prove this principle using two known masses and a uniform metre rule. 2. A uniform metre rule is suspended at the 50 cm mark and a stone at the 0 cm mark. The metre rule is balanced horizontally when a mass of 100 g is suspended at the 30 cm mark. Calculate the weight of the stone. 3. In the diagram shown (Fig. 3.29), calculate the value of the unknown mass M, when the plank is balanced horizontally. 0 M 1 m 10 kg 5 m 4 m 10 m 20 kg 8 m Fig. 3.29: A plank at balance 4. Jane and James are seated at 3 m and 2 m respectively from the centre of a seesaw on one side and Jack at 4 m from the centre on the other side. The seesaw is balanced horizontally. Find the weight of Jack, if the masses of Jane and James are 40 kg and 30 kg respectively. 93 5. Fig. 3.30 shows a uniform plank pivoted at its centre. Where should a 5 kg mass be attached if the plank is to be perfectly horizontal? 4 m 5 m 2 kg 1 kg Fig. 3.30: A uniform plank at balcon 6. A uniform metre rule is balanced horizontally at its centre. When a mass of 5 g is suspended at the 4 cm mark, the rule balances horizontally if a mass M is suspended at 30 cm mark.
Calculate M. 7. Fig. 3.32 shows a uniform metallic metre rule balanced when pivoted at the 30 cm mark under the conditions shown on the diagram. Fig. 3.31: Moments in a uniform rule (a) Redraw the diagram showing all the forces acting on the metre. (b) Calculate the weight W of the metre rule. 8. Name four importance of moments in our daily life. 9. Fig. 3.32 represents a type of safety valve that can be fitted to the boiler of a model steam engine. 60 cm 20 cm Pivot F Steam Water Heat Fig. 3.32 (a) Briefly explain how the system works.. (b) What minimum pressure should the steam have in order to escape through the plate which has a cross sectional area 10.3 cm2? (Take atmospheric pressure = 105 Pa) 94 TOPIC 4 Centre of Gravity and Equilibrium Unit outline • Definition of centre of gravity and centre of mass. • Determination of centre of gravity of regular and irregular objects. • Effects of the position of centre of gravity on stability of objects. Introduction In our day to day experiences, we may have come across statements such as ‘‘that object is not stable on the table” or “that overloaded bus is not stable on the road”. Have you ever asked yourself what factors control the stability of an object? In this topic, we will study the factors that affect stability of objects and centre of gravity. Activity 4.1 To locate the centre of gravity of a book (Work individually or in groups) Materials: Exercise book Steps 1. Take your exercise book and try to balance it horizontally on your finger as shown in Fig 4.1 below. Fig. 4.1: Balancing a book (a) What do you observe? (b) Why do you think the book balances at only one point? (c) What do you think is special about the point where the book balances? 2. Discuss with your group members your observations and thoughts in 1(a),(b) and (c). 95 By going through the following discussions, you will be able to answer questions 1(a) to (c) in Activity 4.1. 4.1 Centre of gravity and centre of mass of a body In one of his experiments, Sir Isaac Newton showed that bodies experience a force of gravity exerted on them by the earth. This force of gravity is always directed towards the earth’s centre and is called the
weight of the body. How is this weight distributed throughout the body? The answer to this question is found in the following activity. 4.1.1 To investigate where the weight of a body acts Activity 4.2 To find the centre of gravity of a regular body (Work in pairs or in groups) Materials: A table, thin rectangular card Steps 1. Place a thin rectangular cardboard near the edge of the bench top. 2. Pull the card slowly away from the bench until it is just about to topple over then released as shown in Fig. 4.2 (a). 3. Using a ruler, mark and draw the line AB along which the card balances. 4. Repeat the activity with the other side of the card, mark and draw the line CD along which the card balances. The lines AB and CD intersect at a point M (Fig. 4.2(b)). table thin rectangular card pull B M D C (a) A (b) Fig. 4.2: Location of a point where the weight of the body acts 4. Now, try to balance the card with the point M placed at the tip of your fore finger. What do you notice about the state of equilibrium of the card? Suggest a reason for this observation. From Activity 4.2 you should have observed that the cardboard balances horizontally at point M only. This shows that although the mass of the cardboard is distributed over the whole body, there is a particular point, M, where the 96 whole weight of the cardboard appears to be concentrated. When pivoted at this point the cardboard balances horizontally. This point, M, is called centre of gravity of the cardboard. The centre of gravity of a body is the point from which the whole weight of the body appears to act. The centre of gravity of an object is constant i.e. at the one location when a body in a place with uniform gravitational field strength. However, the centre of gravity of a body moves to a different location when the body is placed in a region with non-uniform gravitational field strength. Centre of mass of an object on the other hand is the point where all the mass of the object is concentrated. Since the mass of an object is constant and is not affected by pull of gravity, the location of the centre of mass of an object is constant i.e. does not change. In places like on earth where the gravitational field strength is uniform, the centre of mass and the centre of gravity coincide i.e. are at the same point.
However, the two centres are at different locations for the same object if the object is placed in a place with non-uniform gravitational field strength. 4.2 Centre of gravity (c.g) of regular lamina Activity 4.3 To locate the centre of gravity of a regular lamina (Work in groups) Materials: manila paper, ruler, pencil Instructions 1. In this activity you will conduct an investigation to locate centre of gravity of a regular lamina. 2. Using your geometrical instruments, and the other materials provided, come up with a method of determining the centre of gravity of a regular lamina. Write a brief procedure and execute it. 3. Balance them at their centres of gravity on the tip of your pencil. 4. Practically locate their centres of gravities by drawing. In the previous topic, you did Activity 3.6 on determining the weight of a beam. (uniform metre ruler). You were able to realise that the weight of the uniform metre rule tends to act at central point of the ruler that is at 50 cm mark. A thin cardboard like the one used in Activity 4.2 is a lamina. The cover of a book is a lamina. The 97 set square or protractor in your mathematical set are all examples of laminae. Experiments have shown that bodies with uniform cross-section area and density have their centres of gravity located at their geometrical centres. For example, a metre rule of uniform cross-sectional area and density has its centre of gravity located at the 50 cm mark. Fig. 4.3(a) - (d) shows the centre of gravity (c.g) of rectangular, square, triangular, and circular laminae. c.g c.g c.g c.g (a) rectangle (b) square (c) isosceles triangle (d) circle Fig. 4.3: c.g. of regular lamina 4.3 Centre of gravity (c.g) of irregular lamina Activity 4.4 (Work in groups) To determine the centre of mass of an irregular lamina using a plumbline Materials: An irregular lamina, plumbline, a drawing pin Steps 1. Guess and mark the centre of gravity of an irregular lamina. 2. Make three holes P, Q and R on an irregularly shaped lamina as close as possible to the edges and far away from each other. The holes should be large enough to allow the lamina
turn freely when supported through a drawing pin. 3. Suspend the lamina on the clamp using the drawing pin through each hole at a time. 4. Suspend a plumbline (a thin thread with a small weight at one end) from the point of support, P as shown in Fig. 4.4(a), and draw the line of the plumbline on the lamina by marking two points A and B far apart and joining them. 5. Repeat the steps with the support Q and mark the point M where the two lines intersect. 6. Check the accuracy of your method by suspending the lamina at R. What do you observe? Explain. 7. Comapre the c.g determined experimentally and that you guessed earlier. 98 irregular lamina P R A B Q smooth support e.g drawing pin hole string (plumbline) weight P A R M Q B (a) (b) Fig. 4.4: Locating centre of mass of a lamina using a plumbline 8. The plumbline pass through M (Fig. 4.4 (b)). Check the results again by balancing the lamina about point M. What do you observe? The lamina balances horizontally at point M. Point M is the centre of gravity of the lamina. This activity proves that when a body is freely suspended it rests with its centre of gravity vertically below the point of suspension. Activity 4.5 (Work in groups) To demonstrate how to locate the centre of gravity of an irregular object using a straight edge Materials: An irregular lamina, a prism Steps 1. Balance a lamina on the edge PQ of a prism as shown in Fig. 4.5(a). Mark the points A and B on the lamina and join them. 2. Repeat the activity for another position and note the points C and D on the lamina. Join C and D. knife edge P A lamina B Q A M C D B (a) (b) Fig. 4.5: Locating the centre of mass of a lamina using a straight edge 3. Label the point of intersection of lines AB and CD as point M. 4. Try balancing the lamina at point M on a sharp pointed support. What do you observe? Explain. 99 From Activiy 4.5, you should have observed that the lines AB and CD intersect at M; the centre of gravity of the lamina. 4.4 Effect of position of centre of gravity
on the state of equilibrium of a body Activity 4.6 (Work in groups) To define and describe three states of equilibrium Materials: Internet, reference books Steps 1. Conduct a research from books and the internet on the meaning of the term equilibrium in regard to forces acting on an object. 2. State the three states of equilibrium. 3. Describe each of the three states of equilibrium. 4. Compare and discuss your findings with other groups in class. In your discussion, you should have noted that, the state of balance of a body is referred to as the stability of the body. Some bodies are in a more stable (balanced) state than others. The state of balance of a body is also called its state of equilibrium. Activities 4.7, will help us distinguish between the different states of equilibrium. 4.4.1 States of equilibrium Activity 4.7 To investigate the three states of equilibrium (Work in groups) Materials: plastic thistle funnel, bench Steps 1. Place the funnel upright with the wider mouth resting on the bench (Fig 4.6). 2. Displace the funnel slightly upwards as shown in Fig 4.6 (b) and then release it. What do you observe? 100 force F C W G (a) A C A force F G W (b) Fig 4.6: To show stable state 3. Explain the behaviour of the funnel in terms of the changes in the position of centre of gravity. 4. Place the funnel upright with the narrower mouth resting on the bench as shown in Fig 4.7 (a). 5. Displace the funnel slightly with your finger. What about change of state of the funnel? G C force F G W A (a) funnel topples down force F C G W (b) Fig 4.7: To show unstable state 6. Explain the behaviour of the funnel in terms of the change in position of its centre of gravity in this activity. 7. Place the funnel horizontally as shown in Fig 4.8 (a). 8. Displace the funnel gently by tapping it with a finger. What do you observe? table G W F G W F (a) (b) Fig. 4.8: To show neutral state 101 9. Explain the behaviour of the funnel in terms of the change in position of the centre of gravity when displaced slightly. When a body is resting with its centre of gravity at the lowest point, it is very stable. When displaced slightly; its centre of gravity is raised and when it is released
, the object falls back to its original position to keep its centre of gravity as low as possible. This type of equilibrium is known as stable equilibrium. Thus, the funnel in Activity 4.7 Fig. 4.6 (a) was in stable equilibrium. Our finances and keeping the environment clean!! Note that we have used a plastic thistle funnel instead of a glass one. The latter has high chances of breaking. Any time we break a laboratory apparatus, we think of its effects on the school finances as it has to be replaced. Sometimes we may be required to pay ourselves hence affecting the finances of our parents. In case you use glass funnels, be careful when using them. As they may break and cause injuries to you or group members. If it breaks accidentally, collect the broken pieces and dispose them to keep the environment clean. When an object is resting with its centre of gravity at a very high position from the base support, it is unstable. When displaced slightly, it continues to fall up to the lowest possible position in order to lower its centre of gravity. This state of stability is known as unstable equilibrium. The funnel in Fig 4. 7(a) was in unstable equilibrium. When an object is resting such that the position of its centre of gravity remains at the same vertical position even when the object is displaces, it is said to be in neutral equilibrium. The funnel in Fig 4.8 (a) was in neutral equilibrium. 4.4.2 Relationship between position of centre of gravity and stability Activity 4.8 To investigate the relationship between position of centre of gravity of a body and its stability (Work in pairs or groups) Material: reference material Discuss with your group members why: 1. Buses sometimes carry heavy luggage at their roof tops. 2. Buses have their luggage bonets located underneath. 3. A person carrying two buckets full of water is more stable that one carrying one bucket. 102 4. Discuss and compare your explanations with your partner and report to the whole class. A body is more stable when its heavy part is as low as possible since it lowers the position of the centre of gravity. If the heavy part of the body is at high position or if the light part of the body in high position is made heavier than the lower position, the body becomes unstable and thus likely to topple over and can cause accidents like in the case of a vehicle carrying heavy luggage at its roof top. Exercise 4.1 1. Draw the figures below in your notebook and identify the centre of gravity
of each. (a) (d) (b) (e) (g) (h) (c) (f) (i) Fig 4.9: Various types of figures 2. Fig 4.10 shows a Bunsen burner at different states of equilibrium. (i) (ii) (iii) Fig 4.10: A bunsen burner at different states of equilibrium 103 (a) Name the states in which the Bunsen burner is at in (i), (ii) and (iii). (b) Describe each state named in (a) above. 3. With aid of a diagram, describe how you can determine the centre of gravity of an irregular plane sheet of metal. 4. State and explain the states of equilibrium in Fig 4.11. Fig 4.11: A sphere 4.5 Factors affecting the stability of a body Activity 4.9 (Work in groups) To design and investigate factors that affect the stability of a body Materials: Plastic thistle funnels, Benches Instructions 1. By modifying the set-up, we used in Activity 4.7, using the materials provided, conduct an investigation on factors affecting stability of a body. 2. Sketch the step-up and write a brief procedure to investigate the factors. 3. Execute the procedure and answer the following questions (a) What happens to the funnel when the vertical line through the centre of gravity falls outside the base of the funnel? Deduce the factors that affect stability of the funnel. 4. Write a report and present it in a class discussions. 5. What are some of the sources of errors in the experiment and how can they be minimised? The funnel is more stable when its c.g is at a very low position and vice varsa. In addition, the activities show that the wider the base the more stable a body is. Activity 4.9 further shows that the funnel becomes unstable when the vertical line drawn through the centre of gravity falls outside the base that supports the body. In summary, a body is more stable if: 1. the centre of gravity is as low as possible. 2. the area of the base is as large as possible, and 3. the vertical line drawn from the centre of gravity falls within its base. 104 4.6 Applications of the position of centre of gravity Activity 4.10 (Work in groups) To describe the applications of the position centre of gravity Materials: reference books, Internet Steps 1. Conduct a research from Internet and reference books on the
applications of position of centre of gravity. 2. In your research also find out: (a) Why a bird toy balances on its beak? (b) Why it is not advisable to stand on a small boat on the surface of the water? (c) Why one leans to the opposite direction when carrying a load? (d) Why the bus chassis is made heavier than the other parts of the bus? (e) How a tight-rope walker balances himself/herself? 3. Discuss your findings with other groups in class. Do you have the same explanations? 4. Have a class presentation on your findings from your research. In your research and discussion, you should have learnt the following: 1. The balancing bird is a toy that has its centre of gravity located at the tip of the beak. The bird balances with its beak resting on one finger or any other support placed underneath the beak, and the rest of the body in the air. This is because it is designed with its centre of gravity at that point (Fig 4.12). Fig. 4.12: Balancing bird toy 105 2. People in a small boat are advised neither to stand up nor lean over the sides while in the boat. This is because when they stand, they raise the position of the centre of gravity making the boat unstable and more likely to tip over (See Fig 4.13). Fig. 4.13: People in a boat 3. A person normally leans to the opposite direction when carrying heavy loads with one hand e.g. a bucket full water. This helps to maintain the position of the c.g to within the base of the person in order to maintain stability (See Fig 4.14) Fig. 4.14: Leaning while carrying heavy load 4. Most buses have their cargo in compartment in the basement instead of the roof rack in order to keep the centre of gravity of the buses as low as possible (Fig. 4.15). 106 Fig. 4.15: Buses carry their cargo below passengers’ level. 5. A tight-rope walker carries a pole to maintain stability. By swaying from side to side, he/she ensures that the vertical line drawn from his/her centre of gravity falls within the feet on the rope in order to maintain stability. (Fig. 4.16). tight-rope Fig. 4.16: A tight-rope walker carries a pole to maintain balance. Topic summary • The centre of
gravity, c. g, of a body is the point where the whole weight of the body appears to act from. • Centre of mass of an objects is the point where all the mass of the object is concentrated. • The centre of gravity of a regular lamina or object is at its geometric centre. • The centre of gravity of a lamina can be found using a plumbline or by balancing it on a knife edge. • A body is said to be in stable equilibrium if it returns to its original position after being displaced slightly. • A body is in unstable equilibrium if on being slightly displaced, it does not return to its original position. • A body is said to be in neutral equilibrium it moves to a new position • but maintains the position of the c.o.g above its base support. Bodies can be made more stable if their centres of gravity are made as low as possible and the bases are made as broad as possible. 107 Topic Test 4 1. Define the term centre of gravity?. 2. Differentiate between centre of mass and centre of gravity. 3. Redraw the figures shown in Fig 4.17 below and indicate their centres of gravity. 4. Describe how you can determine the centre of gravity of the lamina shown in Fig 4.18. Fig. 4.17: Solids Fig. 4.18: Irregular shape 5. Fig. 4.19 shows a marble in three types of equilibrium. State and explain the type of equilibrium in each case. Fig. 4.19: Marble in three state of equilibrium 6. What is stability? 7. One vehicle which was travelling from Juba to Gulu was seen carrying heavy goods on its roof top and some of its passengers in the vehicle were standing. Discuss why the vehicle is likely to topple if it negotiates a corner at high speed. 108 8. Explain why a three-legged stool design is less stable than a four legged one. 9. Explain the following: (a) The passengers of a double-decker bus are not allowed to stand on the upper deck. (b) A racing car is made of a heavy chassis in its lower parts. (c) When one is alighting from a moving vehicle, it is advisable to spread out his/her legs. My safety Do not stand in a moving vehicle. Let us observe traffic rules. 109 110 UNIT 3 Work, Energy and Power Topics in the unit Topic 5: Work, Energy and Power Learning outcomes Knowledge and Understanding •
Understand the concepts of work, energy and power. Skills • Design tests to relate work done to the magnitude of a force and the distance moved, power to work done and time taken, using appropriate examples. • Observe carefully. • Predict what might happen. • Use appropriate measures. • Collect and present results appropriate in writing or drawing. • Interpret results accurately and derive kinetic and potential energy formula. • Report findings appropriately and relate work, energy and power. Attitudes • Appreciate that food eaten is energy. Key inquiry questions • How does the world get its energy? • Why is energy not destroyed? • Why work done is energy? 111 TOPIC 5 Work, Energy and Power Unit outline • Forms of energy • Transformation of kinetic energy to potential energy and vice verse. • • Different ways to conserve energy • Law of conservation of mechanical energy Sources of energy Introduction Everyday, we do many types of work. We work in the offices, in the farms, in the factories etc. To make our work easier, we use machines ranging from simple tools to sophisticated machinery. Different machines or people do work at different rates (known as power). The ability and the rate of doing certain amount of work depends on how much energy is used. In this topic, we will seek to understand these three terms i.e work, energy and power from the science point of view. 5.1 Work Activity 5.1 To distinguish cases when work as defined in science is done or not (Work individually or in groups) Materials: a chart showing people carrying out different activities, pieces of chalk, pen, chair, desk. Steps 1. Conduct research from books on the scientific definition of work. 2. Walk from your chair to the chalkboard and write the word ‘work’ on the chalkboard. 3. Collect any litter in your classroom. Be responsible Always keep where you live clean. It is good for your health. 112 4. Carry your chair to the front of you classroom and sit on it. 5. Push against a rigid wall of your classroom. 6. Discuss with your colleagues whether scientifically speaking work, is done in steps 1, 2, 3 and 4. What do think is the meaning of ‘work’? 7. Now, look at the activities being performed by the people in Fig. 5.1 below. (a) (b) (c) (d) Fig. 5.1: People performing different tasks 8. According to the scientific definition of work, in which of the diagrams above
is the person doing work? Explain. 9. Give other examples of doing work. Work is only said to have been done when an applied force moves the object through some distance in the direction of force. Therefore in Activity 5.1, work was done in steps 2, 3 and part of 4 (when carrying the desk). However, no work was done when you sat on your chair without moving in step 4 and pushing the wall without moving it in step 5. 113 Similary, in Fig 5.1, work is being done in (a) and (d) only. When the girl applies a force to a wall in (b) and even becomes exhausted, she is not doing any work because the wall is not displaced. When the woman carries the basket on the head, she is not doing any work. This is because she exerts an upward force on the basket which is balanced by the weight hence there is no motion of the basket in the direction of the applied force. Definition of work Work is defined as the product of force and distance moved in the direction of the force. i.e Work = force × distance moved in the direction of the force W = F × d The SI unit of work is joule (J). A joule is the work done when a force of 1 newton moves a body through a distance of 1 metre. 1 joule =1 newton × 1 metre Bigger units used are kilojoules (1 kJ) = 1 000 J Megajoule (1 MJ) = 1 000 000 J Note: Whenever work is done, energy is transferred. Example 5.1 Find the work done in lifting a mass of 2 kg vertically upwards through 10 m. (g = 10 m/s2) Solution To lift the mass upwards against gravity, a force equal to its own weight is exerted. Applied force = weight = mg = 2kg × 10N/kg = 20 N Work done = F × d = 20 N × 10 m = 200 Nm = 200 J 5.1.1 Work done in pulling an object along a horizontal surface Activity 5.2 To design an investigation to determine the work done in pulling an object along a horizontal surface (Work in groups) Materials: a block, a weighing scale, and a tape measure/metre ruler, string. 114 Instructions 1. In this experiment you will design and carry out an investigation to determine work done in pulling an object on a horizontal surface. 2. Using
the materials provided think of a set-up to do the investigation. Set-up the apparatus and sketch the set-up. 3. Write a brief procedure to execute in doing the activity. 4. Correctly execute the procedure and answer the following questions. 5. Using relevant formula, calculate the work done in pulling the block. What assumption did you make? Explain. 6. Compare and discuss your findings with other groups in class. 7. Explain how tractors pull large loads. Since the block was on a smooth surface, we assume that friction force is negligible hence the force applied is constant along the distance of motion, d. Work done in moving the block is given by: Work = force × distance W = F × d = Fd Example 5.2 A horizontal pulling force of 60 N is applied through a spring to a block on a frictionless table, causing the block to move by a distance of 3 m in the direction of the force. Find the work done by the force. Solution The work done = F × d = 60 N × 3 m = 180 Nm = 180 J Example 5.3 A horizontal force of 75 N is applied on a body on a frictionless surface. The body moves a horizontal distance of 9.6 m. Calculate the work done on the body. Solution Work = force × distance = 75 N × 9.6 m = 75 × 9.6 Nm = 720 J 115 Example 5.4 A towing truck was used to tow a broken car through a distance of 30 m. The tension in the towing chain was 2 000 N. If the total friction is 150 N, determine. (a) Work done by the pulling force. (b) Work done against friction. (c) Useful work done. Solution Fig. 5.2 shows the forces acting on the two cars. Fr = 150 N 2 000 N Fig. 5.2: Diagram of cars (a) Work done by the pulling force (b) Work done against friction W = F × d W = Fr × d (Fr is the frictional force) = 2 000 N × 30 m = 150 N × 30 m = 60 000 J = 4 500 J (c) Useful work done Useful work done = Fd – Frd = (60 000 – 4 500) J = 55 500 J Exercise 5.1 1. Explain why in trying to push a rigid wall, a person is said to be doing no work. 2. Define the term work and state its
SI unit. 3. How much work is required to lift a 2 kilogram mass to a height of 10 metres (Take g=10 m/s2). 4. A garden tractor drags a plough with a force of 500 N at a distance of 2 metres in 20 seconds. How much work is done? 116 5.1.2 Work done against the force of gravity Activity 5.3 (Work in pairs) To determine the work done against the force of gravity Materials: Masses, meter rule (tape measure), a single fixed pulley, string, retort stand, newtonmeter. Steps 1. Hang a mass from a newtonmeter and record its weight. 2. Lift a mass from the ground or bench vertically upwards at a constant speed up to a certain point. For a pulley, tie the string on a mass and pass it around the pulley / clamped on a retort stand pull the other end of the thread. Explain why you must use some work in lifting a mass from the ground. 3. Have a friend to measure the height through which the mass has been raised using the meter ruler or tape measure. 4. The force needed to lift the mass is equal to its weight mg. 5. The work done on the mass is then w = Fd = mgh. 6. Repeat the steps with different masses and calculate the work. 7. When an object is thrown upwards, it raises upto a certain point and then drops. Explain why it raises and drops back. You can do this practically by throwing some objects (of different masses). Approximate amount of work done. The gravitational force (weight) acting on a body of mass m is equal to the product of mass and acceleration due to gravity, g, i.e. w = mg. Thus, to lift a body, work has to be done against the force of gravity (Fig. 5.3). – – – – – – – – – • – – – – – – – – mg – – – – – x = h • mass, m mg Ground Fig 5.3: Work done against gravity Work done against gravity to lift a body through height is given by: Work = Force × vertical height = mg × h = mgh 117 Example 5.5 Calculate the work done by a weight lifter in raising a weight of 400 N through a vertical distance of 1.4 m. Solution Work done against gravity = Force × displacement = mg × h = 400 N ×
1.4 m = 560 J Example 5.6 A force of 200 N was applied to move a log of wood through a distance of 10 m. Calculate the work done on the log. Solution W = F × d = 200 N × 10 m = 2 000 J Exercise 5.2 1. Define work done and give its SI unit. 2. Calculate the work done by a force of 12 N when it moves a body through a distance of 15 m in the direction of that force. 3. Determine the work done by a person pulling a bucket of mass 10 kg steadily from the well through a distance of 15 m. 4 A car moves with uniform speed through a distance of 40 m and the net resistive force acting on the car is 3 000 N. (a) What is the forward driving force acting on the car? Explain your answer. (b) Calculate the work done by the driving force. (c) State the useful work done. 5. A student of mass 50 kg climbs a staircase of vertical height 6 m. Calculate the work done by the student. 6. A block was pushed by a force of 20 N through 9 m. Calculate the work done. 118 5.1.3 Work done along an inclined plane Activity 5.4 (Work in groups) Materials To determine the work done along an inclined plane Spring balance, one piece of wood of about 10 cm, a wedge, ruler, trolley/ piece of wood/mass hanger/stone. Steps 1. Make an inclined plane by putting a piece of wood on a wedge. 2. Attach the mass hanger/stone/trolley to a spring balance (calibrated in Newtons). What happens to the spring immediately when the mass is hanged on it? 3. Measure the length of the incline and record it down l=..... cm. 4. Pull the spring balance with its object on from the bottom of the incline and note down the force used in pulling.............N 5. Change the length in cm to m and find the work done using the formula, work = Fd = (J) 6. Using the above skills, approximate the amount of work you do when climbing a slope of 100 m long. Consider an inclined plane as shown in Fig. 5.4 below. A body of mass m moved up by a force f through a distance d li e d p F ( a θ Fig. 5.4: Work done along inclined plane.
Work done by the applied force is given by Work done = F × d The work done against the gravitional force is given by: Work done = weight of the object × vertical height Work = mgh In case the inclined plane is frictionless force: Work done by the applied force = work done against gravity 119 In case there is some frictional force opposing the sliding of the object along the plane: Work done by the applied force > Work done against gravity Work done against friction = Work done by applied force – work done against gravity Example 5.7 A box of mass 100 kg is pushed by a force of 920 N up an inclined plane of length 10 m. The box is raised through a vertical distance of 6 m (Fig. 5.5). m = 100 kg Fig 5.5: Inclined plane (a) Calculate: (i) the work done by the applied force, (ii) the work done against the gravitational force. (iii) the difference in work done. (b) Why do the answers to (i) and (ii) in part (a) differ? Solution (a) (i) Work done by the applied force = F × d = 920 N × 10 m = 9 200 J (ii) Work done against gravity = F × d = mg × h = 100 kg × 10 N/kg × 6 m = 6 000 J (iii) The difference in work done = 9 200 J – 6 000 J = 3 200 J (b) This work done is used to overcome the friction between the box and surface of the incline plane. The useful work done is 6 000 N. 120 Exercise 5.3 1. A box of mass 50 kg is pushed with a uniform speed by a force of 200 N up an inclined plane of length 20 m to a vertical height for 8 m (Fig. 5.6). 0 N 0 F = 2 8 m Fig. 5.6: Work done along an inclined plane Calculate the: (a) Work done to move the box up the inclined plane. (b) Work done if the box was lifted vertically upwards. 2. A body of mass 85 kg is raised through a vertical height 6 m through an inclined plane as shown in Fig. 5.7. Calculate the: F = 150 N (a) Slant distance. (b) Work done by the force 150 N. 6 m (c) Work done, if the body was lifted vertically upwards. (d) Work done against friction. 85
kg 8 m Fig. 5.7: Work done on an inclined (e) Frictional force between the body and the track. 3. A block of mass 60 kg was raised through a vertical height of 7 m. If the slant height of a frictionless track is 21 m, and the force used to push the block up the plane is 800 N, calculate the work done in pushing the block. 4. A car engine offers a thrust of 2 500 N to ascend a sloppy road for 1.1 km. At the top of the slope, the driver realized that the attitude change was 200 m. If the mass of the car is 1.2 tonnes, calculate the; (a) Work done by the car engine. (b) Work done against resistance. 121 5.2 Power Activity 5.5 (Work in groups) To compare the time taken to do a piece of work by a person and a machine Materials: writing material, stopwatch and scientific calculator. Steps 1. By timing yourself, start solving the following problems without using calculator: (a) 37 6998 J × 276 J (b) 35 264 J × 469 J 2. Repeat step 1, but now with a scientific calculator and compare the time taken. Which one takes longer or shorter time to complete the task? Explain your answer. 3. Now, think of a man ploughing a square piece of land that measure 100 m by 100 m (a) by hand, (b) using a tractor. Which task do you think takes longer or shorter time to complete the activity? Suggest a reason. 4. What is power? In your groups discuss the meaning of power. In your discussion, you might have noted that sometimes work is done very quickly and sometimes very slowly. For instance, it takes a longer time to multiply the problems without a calculator in step 1 than with a calculator in step 2. Similarly, in step 3, a tractor will take few hours ploughing a piece of land while a man will take more hours ploughing the same piece of land. The person and the tractor are doing the same work but the tractor is doing it at a faster rate than the person does. This is because they have different power ratings. Different machines and engines have different power ratings. Engines with bigger power ratings are said to be powerful and operate very fast. Definition of power Power is the rate of doing work. i.e. Power = work done time taken = force × distance time 122 SI units of power are
Watts. 1watt = 1 joule second Large units used are kilowatt and megawatt. 1 kilowatt = 1 000 W 1 megawatt = 1 000 000 W Example 5.8 What is the power of a boy lifting a 300 N block through 10 m in 10 s? Solution Force = 300 N, Distance = 10 m, Time = 10 s Work done by the boy = F ×d = 300 ×10 Power = = 3 000 J work time = 3 000 J 10 s = 300 W 5.2.1 Estimating the power of an individual climbing a flight of stairs Activity 5.6 To estimate the power of an individual climbing a flight of stairs (Work in groups) Materials: stopwatch, weighing machine, tape measure Steps 1. Find a set of stairs that you can safely walk and run up. If there are no stairs in your school, create some at garden. 2. Count their number, measure the vertical height of each stair and then find the total height of the stairs in metres. 3. Let one member weigh himself/herself on a weighing machine and record the weight down. 4. Let him/her walk then run up the stairs. Using a stopwatch, record the time taken in seconds to walk and running up the stairs (Fig 5.8). 123 Fig. 5.8: Measuring one’s own power output 5. Calculate the work done in walking and running up the stairs. Let each group member do the activity. Is the work done by different members in walking and running up the stairs same? Explain the disparity of work done by various group members. 6. Calculate the power developed by each individual in walking and running up the stairs. Which one required more power, walking or running up the flight of stairs? Why? Note: (i) The disabled should be the ones to time others. Care must be taken on the stairs not to injure yourselves. (ii) Incase of lack of stairs, learners can perform other activities like lifting measured weights. From your discussion, you should have established that: Height moved up (h) = Number of steps (n) × height of one step (x) h = n × x = n x Time taken to move height (h) = t P = = P = Work done against gravity time W × n × x t Wnx t = mgh t = W × h t where, P = power, W = weight, x = height of first step
, n = number of steps If x is in metres, W in newtons and t in seconds then power is in watts. 124 Example 5.9 A girl whose mass is 60 kg can run up a flight of 35 steps each of 10 cm high in 4 seconds. Find the power of the girl. (Take g = 10 m/s2). Solution Force overcome (weight) = mg = 60 kg × 10 N/kg = 600 N Total distance = 10 × 35 = 350 cm = 3.5 m Work done by the girl = F × d = 600 × 3.5 Power = = 2 100 J work time = 2 100 J 4 s The power of the girl is 525 W My health Do you know that regular exercises are good for your health. Do exercises regularly to keep your body healthy Exercise 5.4 1. Define power and give its SI unit. 2. To lift a baby from a crib, 100 J of work is done. How much power is needed 3. if the baby is lifted in 0.5 seconds. If a runner’s power is 250 W, how much work is done by the runner in 30 minutes? 4. The power produced by an electric motor is 700 W. How long will it take the motor to do 10,000 J of work? 5. Find the force a person exerts in pulling a wagon 20 m if 1 500 J of work is done. 6. A car’s engine produces 100 kw of power. How much work does the engine do in 5 seconds? 7. A color TV uses 120 W of power. How much energy does the TV use in 1 hour? 8. A machine is able to do 30 joules of work in 6.0 seconds. What is the power developed by the machine? 125 9. Rebecca is 42 kg. She takes 10 seconds to run up two flights of stairs to a landing, a total of 5.0 metres vertically above her starting point. What power does the girl develop during her run? 10. Student A lifts a 50 newton box from the floor to a height of 0.40 metres in 2.0 seconds. Student B lifts a 40 newton box from the floor to a height of 0.50 metres in 1.0 second. Which student has more power than the other? 11. Four machines do the amounts of work listed in Table 5.1 shown below. The time they take to do the work is also listed. Which machine develops the most power?
Machine A B C D Work 1 000 joules 1 000 joules 2 000 joules 2 000 joules Table 5.1 Time 5 sec 10 sec 5 sec 10 sec 5.3 Energy Activity 5.7 (Work in pairs) To brainstorm about energy Materials: Reference materials, a pen, desk or slopy platform Steps 1. What enables your body to perform various functions besides keeping warm. Discuss this with your group member. 2. Define the term energy. 3. Push a pen on your desk such that it rolls for a small distance. What makes the pen roll? 4. State some examples of objects that posses energy in our environment. 5. Discuss the importance of energy in our lives. 6. With a heavy bag on your back, climb the stairs to the top most floor or the steep ground to the top most point. How do you feel? Have you done work? Did you use energy to do so? 7. What happens to a car that has been moving if it runs out of fuel (petrol or diesel)? Explain. 8. Discuss the observations and findings from the above activities in a class discussion. 126 Energy is one of the most fundamental requirement of our universe. It moves motorcycles, cars along roads, airplanes through air, and boats over water. It warms and lights our homes, makes our bodies grow and allows our minds to think. A person is able to push a wheelbarrow, a stretched catapult when released is able to make a stone in it move, wind mills are turned by a strong wind and cooking using electricity in a cooker. All these are possible because of energy. Therefore, for any work to be done, energy must be provided. But what is energy? Definition of energy Energy is the ability to do work. Work done = energy transferred SI unit of energy is joules (J). Relationship between energy and work In your discussion, you should have noted that you got exhausted because you did a lot of work against gravity to carry your body and the heavy bag to the top of the building. The work you did led to the loss of energy (chemical energy from the food) from your body. 5.4 Forms of energy Energy is not visible, it occupies no space and has neither mass nor any other physical property that can describe it. However, it exists in many forms, some of these forms include: 5.4.1 Solar energy Activity 5.8 (Work in groups) To investigate the effect of solar energy Materials: plastic
basin, water, convex lens, thin piece of paper Steps 1. On a bright sunny day, fill a plastic basin with cold water and place it in an open place with no shade. Dip your hand into the water after 2 hours. What is the temperature difference of the water initially and after 2 hours? 2. Get a convex lens on the same day and put it horizontally with one surface facing the sun and another surface facing down. Place a thin paper below the lens. What do you observe after 5 minutes or more? 3. Discuss your observations in steps 1 and 2. 127 The water becomes hot in case 1 and in case 2, the paper burns because of the heat from the sun. These are some of the effects of solar energy. This energy from the sun is in form of radiant heat and light. In some countries where the sun shines throughout, large concave mirrors have been set to collect energy from the sun by focusing its rays on special boilers which provide power for running electric generators. 5.4.2 Sound energy Activity 5.9 (Work in pairs) To design and investigate the production of sound energy Materials: Two pens, a stone Instructions 1. In this activity, design and carry out an investigation on production of sound energy using the materials provided. 2. Write a brief procedure and execute it to produce sound energy. Then write a report about your investigation and then discuss it in a class presentation. In what form is the energy released by the pen and the stone? Discuss with your class partner. 3. From your discussion, you should have heard sound in steps 1 and 2. In each case, kinetic energy has been converted to sound and heat energy. Sound energy is the energy associated with the vibration or disturbance of bodies or matter. 5.4.3 Heat energy Activity 5.10 (Work in groups) To demonstrate heat energy Materials: Bunsen burner/candle, matchbox, a retort stand, a nail/metallic rod Steps 1. Light a Bunsen burner or a candle using a lighter (matchbox). 2 Clamp a nail (metallic rod) on a retort stand and bring it near the flame. 3. Carefully touch the other end of the nail after sometime. What do you feel? Explain. 128 The other end of the nail is felt to be hot after sometime. The hotness is due to heat energy that has been transferred from the hot part to the cold part of the nail. Therefore, heat energy only travels from a hot
object to a cooler one. Heat energy is a form of energy that is transferred from one body to another due to the difference in temperature. 5.4.4 Electrical energy Activity 5.11 To demonstrate production of light by electrical energy (Work in groups) Materials: bulb, electric wire, cells (battery), switch, bulb holder and cell holder Steps 1. Fix the battery/cells in their holders and the bulb too. 2. Connect one wire from one end of the cell holder to the bulb holder. Then connect the same wire from the bulb holder to the switch holder and then connect another wire from the other part/side of cell holder to the switch. Make sure the switch is open and the cells are fixed into their holder. 3. What do you observe after the connection? Explain. 4. Complete the circuit then close the switch and observe what happens. The bulb lights when the circuit is complete. Electrical energy is the energy produced by the flow of electric charges (electrons). Work is done when electrons move from one point to another in an electric circuit with electrical appliances such as bulbs. 5.4.5 Nuclear energy Activity 5.12 (Work in groups) To find out what is nuclear energy 1. Conduct a research from internet and reference books on the meaning of 2. nuclear energy. In your research, also find out advantages and disadvantages of nuclear energy. 3. Compare and discuss your findings with those of other groups in your class. You may consult your teacher for more guidance on your discussion. 129 In your discussion, you should have noted that nuclear energy is the energy that results from nuclear reactions in the nucleus of an atom. It is released when the nuclei are combined or split. 5.4.6 Chemical energy Activity 5.13 To investigate and demonstrate chemical energy (Work in groups) Materials: glass beaker, a small bowl, steel wool, white vinegar, thermometer Steps 1. Place the steel wool in the bowl and soak it in white vinegar for a couple of minutes. 2. Squeeze out excess vinegar and wrap the steel wool around the thermometer in a way that you are still able to read the temperature. 3. Put the steel wool in the beaker, then place a cover with a paper or small book on the top. 4. Record the temperature immediately, then again in a minute or so, and again every minute for about five minutes. What did you observe? 5. Discuss your observation with other groups in a class discussion. The thermometer records a higher
temperature reading. The chemical reaction of vinegar and steel wool generates energy in form of heat. This causes temperature to rise as shown by the thermometer. Chemical energy is a type of energy stored in the bonds of the atoms and molecules that make up a substance. Once chemical energy is released by a substance, it is transferred into a new substance. Food and fuels like coal, oil, and gas are stores of chemical energy. Fuels release their chemical energy when they are burnt in the engine (e.g in a car engine). 5.4.7 Mechanical energy Activity 5.14 (Work in groups) To describe mechanical energy Materials: pen, a piece of chalk Steps 1. Raise a piece of chalk or pen from the ground to a position above your head and release it to fall to the ground. What do you observe with the change in height and the speed of the piece of chalk as it falls. 130 2. Throw two full pieces of chalk on the wall one at a time using different forces or at different speeds (one should move faster and another one slowly). Note the sound created by the piece of chalk after colliding with the wall. Which one makes more noise after collision? Mechanical energy is the energy possessed by a body due to its motion or due to its position. It can either be kinetic energy or potential energy of both. When an object is falling down through the air, it posses both potential energy (PE) due to its position above the ground, and kinetic energy (KE) due to its speed as it falls. The sum of its PE and KE is its mechanical energy. Mechanical energy = kinetic energy + potential energy. (a) Potential energy Activity 5.15 (Work in pairs) To demonstrate the forms of potential energy Materials: a catapult or a spring, a small stone. Steps 1. Raise a small stone from the ground or any other resting position upwards to a particular height above its resting surface. What kind of energy do you think it attains? 2. Now, release the stone and observe what happens. Explain your observations. 3. Compress a spring to a particular size. What kind of energy do you think it attains? Explain. 4. Release the spring and observe what happens to the spring. 5. Why does the spring move in such a manner? Discuss your observations in steps 1, 2, 3 and 4 with your colleagues and identify two types of potential energy from the activities. You should have observed that when the stone was released it moved down to the ground.
This implies that the stone had stored energy due to its position that makes it to start moving down after it has been released. The energy possessed by a body (e.g. a stone) due to its position above the ground is called gravitational potential energy. In other words potential energy is energy by virtue of position. Similarly, when the compressed spring was released, it relaxed to a bigger size. This implies that the spring had stored energy due to compression. The energy possessed by a body due to compression (e.g. a spring) or stretch (e.g catapult) is called elastic potential energy. 131 Therefore, potential energy is in two forms; gravitational potential energy and elastic potential energy. (i) Gravitational potential energy Bodies which are at a given height above the ground posses gravitational potential energy. This energy depends on the position of objects above the ground. The following activity will help us to understand how to calculate potential energy of a body at a particular position above the ground. Activity 5.16 (Work in pairs) To determine gravitational potential energy of a raised object Materials: three bricks, meter ruler, beam balance, soft board bridge Steps 1. Conduct research from the Internet and books on the mathematical expression of potential energy. 2. Support the soft board on two bricks. 3. Measure the mass of the third brick by using a beam balance then place it on the soft board. 4. Now lift the third brick to a height h1. Let your partner measure the height h1, in metres. 5. Allow the brick to drop gently onto the soft board. Describe the energy possessed by the brick as it drops. 6. Calculate the potential energy gained by the stone using the expression of potential energy you got from the research. 7. Repeat the activity with the other two different heights h2 and h3. 8. Compare and discuss your observations and values of PE in the three cases and deduce a general conclusion from your discussion. If a stone is lifted upwards through a height h; and placed on a table (Fig 5.9), work is done against gravity. h F mg Fig. 5.9: Potential energy depends on height, h. 132 The work done to overcome gravity is equal to the gravitational potential energy gained by the stone. But work done = F × h ; F = mg ∴ work done = mg × h But, potential energy = work done. Therefore: P.E = mgh Example 5.10 A crane is used to lift a body of mass
30 kg through a vertical distance of 6.0 m. (a) How much work is done on the body? (b) What is the P.E stored in the body? (c) Comment on the two answers. Solution (a) Work done = F × d = mg × d = 30 × 10 × 6 = 300 × 6 = 1 800 J (b) P.E = mgh = 300 × 6 = 1 800 J (c) The work done against gravity is stored as P.E in the body. Caution A stone dropped from the roof of a building will cause more pain if it falls on someone’s foot than when the same stone falls from a table. This is because the one on the roof has more gravitation potential energy due to its greater height (position) above the ground. (ii) Elastic potential energy In Activity 5.15, we saw that a stretched catapult or compressed (Fig. 5.10(a)) has energy stored inform of elastic potential energy. When the stretched spring catapult is released it releases the energy that can be used to do work e.g. to throw a stone. (a) (b) Fig. 5.10: A compressed and a stretched spring 133 pull – – – – – – – – – – – – – – – – – – force = Work done in stretching the spring = Elastic P.E gained by the spring Fig.5.11: Elastic potential energy = average force × extension ) × e = ( 0 + F 2 1 2 Fe Work done is stored as elastic potential energy. = Note: Since the force is not uniform (F increases from 0 to F) we should use the average force in calculating the work done. Example 5.11 Calculate the elastic gravitational p.e stored in a spring when stretched through 4 cm by a force of 2 N. Solution Elastic P.E = 1 We will learn more about elastic potential energy and how to determine it later. 2 × 2 × (0.04) = 0.04 J 2 Fe = 1 (b) Kinetic energy Activity 5.17 (Work in pairs) To demonstrate kinetic energy Material: trolley, table Steps 1. Place a trolley on the table and give it a slight push. Observe what happens to it. Explain your observations. 2. Now, observe any moving objects or things around you. Which energy do you think they possess when they are in motion? Explain your answer. 134 From your discussion in activity
above, you should have observed that the trolley starts to move once given a slight push. It possess energy as it moves. The energy which is possessed by a moving object due to its speed is called kinetic energy (KE). In other words we can define kinetic energy as energy by virtue of motion. Examples of objects that posses KE include moving air, rotating windmills, falling water, rotating turbines and a moving stone. In general, any moving body possesses energy called kinetic energy. The kinetic energy of a moving body is given by: Kinetic energy = 1 2 mv2, where m and v are the mass and velocity of the body respectively. Example 5.12 A body of mass 400 g falls freely from a tower and reaches the ground after 4 s. Calculate the kinetic energy of the mass as it hits the ground. (Take g = 10 m/s2) Solution The final velocity of the body as it hits the ground given by the equation of motion, v = u + at = u + gt = 0 + 10 × 4 = 40 m/s K.E = 1 2 mv2 (as the body hits the ground) 2 × 0.400 × 402 = 320 J = 1 Example 5.13 A car of mass 1 000 kg travelling at 36 km/h is brought to rest by applying brakes. Calculate the distance travelled by the car before coming to rest, if the frictional force between the wheels and the road is 2 000 N. Solution V = 36 km/h to m/s = 36 × 1 000 m/s 60 × 60 = 10 m/s K.E = work done against friction 135 1 2 mv2 = F × d 2 × 1000 × 102 = 2 000 × d ⇒ 1 ⇒ 50 000 = 2 000 d ∴ d = 50 000 2 000 = 25 m The stopping distance is 25 m. 5.5 Work and energy relationship When work is done a transfer of energy always occurs. For example carrying a box up the stairs/lifting something heavy from the ground, you transfer energy to the box which is stored as gravitational energy. Therefore its energy increases. Doing work is a way of transferring energy from one object to another. Just as power is the rate at which work is done when energy is transferred the power involved can be calculated by dividing the energy transferred by time needed for the transfer to occur. Power(in watts) = Energy transferred (in joules) time (in watts) P =
E t or Energy = pt For example, when the light bulb is connected to an electric circuit, energy is transferred from the circuit to the light bulb filament. The filament converts the electrical energy supplied to the light bulb into heat and light. The power used by the bulb is the amount of electrical energy transferred to the light bulb each second. Exercise 5.5 1. Define the term energy. 2. State and explain briefly six forms of energy. 3. Differentiate between: (a) Potential energy and kinetic energy. (b) Gravitational potential energy and elastic potential energy. 4. A brick of mass 0.5 kg is lifted through a distance of 100 m to the top of a building. Calculate the potentials energy attained by the brick. 136 5. Explain what is meant by gravitational potential energy. 6. A force of 8.5 N stretches a certain spring by 6 cm. How much work is done in strecthing the spring by 10 cm. 7. A body is acted on by a varying force, F over a distance of 7 cm as shown in Fig. 5.121 –2 –3 –4 –5 –6 1 2 3 4 5 6 7 Distance (m) Calculate the total work done by the force. Fig.5.12 8. An object of mass 3.5 kg is released from a height of 7.0 m above the ground. (a) Calculate the gravitational potential energy of the object release. (b) Calculate the velocity of the object just before it strikes the ground. What assumption have you made in your calculation? 5.6 Sources of energy Activity 5.18 To find out sources of energy (Working in groups) Materials: Internet, reference books, a stream of water or a water tap Steps 1. Tell your group members the meaning of the terms ‘source’ and ‘energy source’. 137 2. Now, think of plants, animals, vehicles and so on. Where do you think their energy comes from? What of electricity used in your school and at home, where does it come from? 3. Compare and discuss your findings in step 1 and 2 with other groups in your class. 4. What is the meaning of primary sources of energy? 5. What is the meaning of secondary sources of energy? 6. Conduct a research from the Internet and reference books on primary and secondary sources of energy. 7. In your research find out: (a) The types of primary and secondary sources of energy.
(b) The generation of energy from each source. 8. Let your secretary write down a summary of your discussion and present it to the whole class. 9. State and explain various types of primary and secondary sources of energy? The word ‘source’ means the beginning of something e.g. the stream begins from the mountain or hills around your school. Hey!! Are you aware that cutting down trees will lead to the loss of our forests in our country and consequently the loss of water sources? Let us protect our water sources by planting more trees. You should have also established that the energy source is a system which produces energy in a certain way. For instance, a hydroelectric station uses the motion of the water of the river to turn the turbines and thus producing electricity. There are two kinds of energy sources; 1. Primary sources 2. Secondary sources 138 5.6.1 Primary sources of energy From your research and discussion in activity 5.18, you should have established the following that Primary Sources are from sources which can be used directly as they occur in the natural environment. They include. 1. Flowing water 2. Nuclear 3. Sun 4. Wind 5. Geothermal ( interior of the earth) 6. Fuels 7. Minerals 8. Biomass (living things and their waste materials 1. Water (a) Hydropower - the flowing water from dams rotate turbines at the bottom of the dam which turn the generator resulting in generation of electricity. This water is kept behind a dam (reservoir) and released at a controlled rate downwards where it meets the turbines and turns them. An example is the falling water in the Fulla rapids on the River Nile in Nimule in South Sudan. (b) Waves - energy from water waves (generated by winds) is also used in generating electricity using sea wave converters. An example is pelamis wave energy converter, a technology that uses the motion of ocean surface waves to create electricity. 2. Nuclear energy Nuclear energy is created through reactions that involve the splitting or merging of the atoms of nuclei together. The process of splitting of large atoms such as those of uranium into smaller atoms is called fission. Fusion on the other hand, is the combining of two smaller atoms such as hydrogen or helium to produce a heavier atom. All these reactions release heat which is turned into electricity in nuclear power plants (Fig 5.13). An atomic bomb derives its energy from these kinds of reactions. 139 3. The sun Fig 5.13:
Nuclear plant The sun is the biggest source of energy and has played an important role in shaping our life on earth since the dawn of time. The sun gives off radiant energy in form of electromagnetic waves. The light energy (visible spectrum) part of the spectrum can be converted directly into electricity in a single process using the photovoltaic (PV) cell otherwise known as the solar cell. The solar thermal energy is used for heating swimming pools, heating water for domestic use (solar heater) and heating of building. Solar thermal electricity generation is where the sun’s rays are used to heat a fluid for the production of high pressure and high temperature steam. The steam is in turn converted into mechanical energy in a turbine to generate electricity. (a) Solar panel (b) Solar heater Fig. 5.14: Solar panel and solar heater 140 4. Wind Wind is caused by the sun heating the earth unevenly. The air is heated differently causing hotter air to expand rise, and the colder one to condense and sink. This results to the movement of air and hence formation of wind. Modern wind turbines placed on the top of steel tubular towers harness the natural wind in our atmosphere and convert it into mechanical energy and then to electricity. Wind mills (Fig. 5.15) are also be used to pump water from the underground and do some other work. 5. Geothermal energy Fig 5.15: Wind mills Geothermal gradient is the difference in temperature between the core (interior) of the earth (planet) and its surface brings about conduction of heat from the core to the surface. The earth’s internal heat is generated from radioactive decay and continual heat loss from the earth’s formation. From hot springs, geothermal energy has been used for bathing to heal some diseases as in some cultures. Geothermal energy is also used to generate electricity at geothermal power stations where heat is used to heat water to get steam which in turn is used to turn the turbines to generate electricity. 141 Fig. 5.16 Geothermal power station 6. Thermal energy Thermal energy is the internal energy in a system by virtue of its temperature. It is defined as the average translational kinetic energy possessed by free particles in a system of free particles in a thermodynamic equilibrium. It can also include the potential energy of a system’s particle which may be an electron or an atom. Thermal (heat) energy is transferred of heat across the system boundaries. Thermal energy is important in our daily life, for example

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