Module #1: Units and Vectors Revisited Introduction There are probably no concepts ore iportant in physics than the two listed in the title of this odule. In your first-year physics course, I a sure that you learned quite a lot about both of these concepts. You certainly did not learn everything, however. Whether we are talking about units or vectors, there is siply too uch inforation to possibly learn in just one year. s a result, we will take another look at both of these concepts in this first odule. This will help you war up to the task of recalling all of the things you learned in your first-year physics course, and it will help to learn both of these valuable concepts at a uch deeper level. Units Revisited lost regardless of the physics course, units should be covered first, because a great deal of physics is based on properly analyzing units. In your first-year course, you were taught how to solve probles such as the one in the following exaple: EXMPLE 1.1 saple of iron has a ass of 54.1 g. How any kg is that? In this proble, we are asked to convert fro illigras to kilogras. We cannot do this directly, because we have no relationship between g and kg. However, we do know that a illigra is the sae thing as 0.001 gras and that a kilogra is the sae thing as 1,000 gras. Thus, we can convert g into g, and then convert g into kilogras. To save space, we can do that all on one line: 541. g 0001. g 1 kg = 0. 000541 kg = 541. 10 4 1 1 g 1, 000 g kg The saple of iron has a ass of.541 x 10-4 kg. Did this exaple help dust the cobwebs out of your ind when it coes to units? It should all be review for you. I converted the units using the factor-label ethod. Because this is a conversion, I had to have the sae nuber of significant figures as I had in the beginning, and even though it was not necessary, I reported the answer in scientific notation. If you are having trouble reebering these techniques, then go back to your first-year physics book and review the. There are a couple of additional things I want you to learn about units. I a not going to show you any new techniques; I a just going to show you new ways of applying the techniques that you should already know. Consider, for exaple, the unit for speed. The standard unit for speed is.
dvanced Physics in Creation However, any distance unit over any tie unit is a legitiate unit for speed. Since that is the case, we should be able to convert fro one unit for speed to any other unit for speed. Study the following exaple to see what I ean. EXMPLE 1. s of 001, the record for the fastest lap at the Indianapolis 500 ( The greatest Spectacle in Racing ) was held by rie Luyendyk. He averaged a speed of 5. iles per hour over the entire.5-ile stretch of the Indianapolis speedway. What is that speed in eters per ond? This proble requires us to ake two conversions. To get fro iles per hour to eters per ond, we ust convert iles to eters. Then, we ust convert hours to onds. This is actually easy to do. Reeber, in iles per hour, the unit iles is in the nuerator of the fraction and the unit hours is in the denoinator. lso reeber that there are 1609 eters in a ile and that 1 hour is the sae as 3600 onds. 5. iles 1609 eters 1hr = 100. 7 1hr 1 ile 3600 eters lthough there is nothing new here, you probably haven t seen a conversion done in this way. Despite the fact that the unit for speed is a derived unit, I can still do conversions on it. I could have just converted iles to eters and gotten the unit eters/hour. I also could have just converted hours to onds and gotten iles/ond. In this case, however, I did both. That way, I ended up with eters/ond. When working with derived units, reeber that you can convert any or all units that ake up the derived unit. Thus, 5. iles per hour is the sae thing as 100.7 eters per ond. Please note that although 3600 has only significant figures, the nuber is actually infinitely precise, because there are exactly 3600 onds in an hour. Thus, it really has an infinite nuber of significant figures. This is why I say that the best rule of thub is to always end your conversion with the sae nuber of significant figures as that with which you started your conversion. Okay, we are alost done reviewing units. There is just one ore thing that you need to reeber. Soeties, units have exponents in the. You were probably taught how to deal with this fact in your first-year physics course, but we need to review it so that you really know how to deal with it. EXMPLE 1.3 One coonly used unit for volue is the cubic eter. fter all, length is easured in eters, and volue is length ties width ties height. The ore failiar unit, however, is
Module 1 Units and Vectors Revisited 3 cubic centieters (cc) which is often used in edicine. If a doctor adinisters 51 cc of edicine to a patient, how any cubic eters is that? Once again, this is a siple conversion. If, however, you do not think as you go through it, you can ess yourself up. We need to convert cubic centieters to cubic eters. Now reeber, a cubic centieter is just a c 3 and a cubic eter is just a 3. We have no relationship between these units, but we do know that 1 c = 0.01. That s all we need to know, as long as we think about it. Right now, I have the following relationship: 1 c = 0.01 This is an equation. I a allowed to do soething to one side of the equation as long as I do the exact sae thing to the other side of the equation. Okay, then, let s cube both sides of the equation: ( 1 c) = ( 001. ) 3 3 1 c = 0. 000001 3 3 Now look what we have. We have a relationship between c 3 and 3, exactly what we need to do our conversion! 51 c 1 So 51 cc s is the sae as 5.1 x 10-4 3. 0. 000001 = 51. 10 3 1c 3 3 4 3 When ost students do a conversion like the one in the exaple without thinking, they siply use the relationship between c and to do the conversion. That, of course does not work, because the c 3 unit does not cancel out, and you certainly don t get the 3 unit in the end: 51 c 1 3 0. 01 = 51. c 1 c Do you see what happened? The c unit canceled one of the c out of c 3, but that still left c. lso, since is the unit that survives fro the conversion relationship, you get the weird unit of c! When you are working with units that have exponents in the, you need to be very careful about how you convert the. t the risk of beating this to death, I want to cobine the previous two exaples into one ore exaple.
4 dvanced Physics in Creation EXMPLE 1.4 The axiu acceleration of a certain car is 1,600 iles per hour. What is the acceleration in feet per ond? Once again, this is a derived unit, but that should not bother you. ll we have to do is convert iles into feet and hours into onds. There are 580 feet in a ile, so that conversion will be easy. We do not know a conversion between hours and onds, but we do know that: 1 hour = 3600 onds To get the conversion relationship between hours and onds, then, we just square both sides: (1 hour) = (3600 onds) 1 hour = 1.96 x 10 7 onds Once again, please note that the conversion relationship between hours and onds is exact. Thus, both nubers have an infinite nuber of significant figures. That s why I reported all digits when I squared 3600 onds. Now that I have the conversion relationships that I need, the conversion is a snap: 1600, iles 580ft 1hr ft = 880. 1hr 1ile 196. x10 7 Notice once again that had I not squared the conversion relationship between hours and onds, the units would not have worked out. In order for hr to cancel, the unit hr had to be in the proble. That s why it is iportant to watch the units and ake sure they cancel properly. Make sure that you understand how to anipulate units this way by perforing the following on your own probles. ON YOUR OWN 1.1 The speed liit on any highways in the United States is 65 iles per hour. What is the speed liit in centieters per ond? (There are 1609 eters in a ile.) 1. The size of a house is 1600 square feet. What is the square yardage of the house?
Module 1 Units and Vectors Revisited 5 1.3 The standard energy unit is the Joule ( kg a serious. It is called the erg! Convert 151 Joules into ergs. ). n alternative energy unit is the erg ( g c ). I Review of Vectors In your first-year physics course, you learned about vectors. quantity is called a vector if it contains inforation about both agnitude and direction. quantity is called a scalar if it contains only inforation about agnitude and no inforation about direction. For exaple, if I tell you that y car is oving at 55 iles per hour, I a giving you a scalar quantity. The speed of the car tells you how uch (agnitude), but it does not tell you in what direction the car is oving. If I tell you that y car is oving at 55 iles per hour due west, then I a giving you a vector quantity. Not only do you know how fast I a traveling (the agnitude), but you also know which way I a heading (the direction). You should have already learned that we call the scalar quantity in this case speed, and we call the vector quantity velocity. Since vectors contain inforation about both agnitude and direction, we use arrows to represent the. The length of the arrow represents the agnitude, while the direction in which the arrow points represents the direction. When I refer to vectors, I will ephasize that they are vectors by placing the in boldface type. For exaple, if you see in an equation, you will know that it is a vector quantity because it is in boldface type. If you see in an equation, you will know that it is a scalar quantity because it is not in boldface type. Now although vectors and scalars are quite different, they can interact atheatically. For exaple, I can ultiply a vector by a scalar. What happens if I do that? Well, a vector contains inforation about agnitude and direction, while a scalar contains inforation only about agnitude. If I ultiply a vector by a scalar, then, the ultiplication will affect the agnitude of the vector. If the scalar happens to be negative, it does affect the direction of the vector as well. Study the following figure to see what I ean. FIGURE 1.1 Scalar Multiplication ½ -
6 dvanced Physics in Creation Notice in the figure that I start with an arrow which represents vector. It has a agnitude (the length of the arrow) and a direction (the direction in which the arrow points). When I ultiply by (a scalar), what happens? The arrow points in the sae direction, but it is twice as long. That s because when I ultiply a vector by a positive scalar, I ultiply its agnitude by the scalar, but I leave the direction alone. Thus, the length of the arrow (the agnitude) changes, but the direction does not. In the sae way, when I ultiply by ½, the length of the arrow changes (it gets cut in half), but the direction does not. Now look at the last arrow in the figure. This arrow represents what happens to the vector when I ultiply by a negative scalar. When you ultiply a vector by a positive scalar, the direction of the vector does not change. However, when you ultiply by a negative scalar, the direction of the vector becoes opposite of what it once was. Thus, when I ultiply by negative, the length of the vector increases by a factor of, but the vector also points in the opposite direction. Vector points towards the upper right hand corner of the figure, while the vector - is twice as long and points to the lower left-hand corner of the figure. When vectors are ultiplied by scalars, then, the result is another vector. s you already learned in your first-year physics course, vectors can not only be ultiplied by scalars, but they can also be added to other vectors. The result in that case is a vector as well. To review how vectors are added to one another, however, we ust first review the way that vectors can be atheatically represented. two-diensional vector can be atheatically represented in one of two ways. It can be represented by its agnitude and an angle, or it can be written in ters of two perpendicular coponents. lthough any two perpendicular coponents can be used to define a two-diensional vector, we typically use horizontal and vertical coponents. These two ways of atheatically representing a vector, as well as the relationships between the, are suarized in Figure 1.. horizontal (x) axis vertical (y) axis FIGURE 1. Two-Diensional Vector θ horizontal coponent ( x ) vertical coponent ( y ) Vector can be written as: (1) (agnitude) and θ or () x and y Matheatical Relationships x = cosθ y = sinθ tanθ = y x = + x y
Module 1 Units and Vectors Revisited 7 Notice, then, that the vector can be represented with a agnitude () and a direction (θ). For exaple, if I told you that there was buried treasure 10.5 iles away fro your current location at an angle of 50.4 o, I would be giving you a vector to describe the location of the treasure. I would be giving you that vector in ters of its agnitude and angle. lternatively, I could tell you that to get to the treasure, you need to walk straight for 6.5 iles then turn left and walk another 8.1 iles. That is also a vector which represents the location of the treasure, but the vector is given in ters of two coponents rather than a agnitude and direction. Of course, since both ethods represent the sae vector, they should be related to one another. The figure suarizes the various eans by which the coponents of a vector relate to the vector s agnitude and direction. Now we can ove on to adding vectors. In order to graphically add vectors, we siply put the tail of the ond vector at the head of the first vector, and then we draw an arrow fro the tail of the first to the head of the ond. The new arrow is the graphical representation of the final vector, which is the su of the two original vectors. lternative, we can atheatically add vectors. To do this, we siply add their x-coponents to get the final vector s x-coponent, and we add their y-coponents to get the final vector s y-coponent. These processes are suarized in Figure 1.3. Two original vectors - and B: FIGURE 1.3 dding Vectors B x B x B y y dding the vectors graphically: C B B C C = + B C = B + dding the vectors atheatically: C x = x + B x C C y = y + B y On the left-hand side of the figure, two vectors ( and B) are drawn. Their horizontal (x) coponents ( x and B x ) as well as their vertical (y) coponents ( y and B y ) are shown. In the iddle of the figure, the graphical ethod for adding vectors is shown. To add vectors and B, we put the tail of B on the head of. Then, we draw an arrow fro the tail of to the head of B. The resulting arrow is the su of + B. Notice that the figure also shows the graphical representation of B +. t first glance, you ight think that the result of + B is different than the result of B +. That s not true. Reeber, a vector is deterined by its agnitude and direction. Both the result of + B and the result of B + have the sae agnitude (length) and direction (they both point in the sae direction). Thus, they each represent the sae vector. They are siply shifted relative to one another.
8 dvanced Physics in Creation That brings up an iportant thing to reeber about vectors. They can be oved around without affecting their value. s long as you do not change the length of the arrow or the direction in which the arrow points, you can ove it all over the place without changing the eaning of the vector at all. Since physics is inherently a proble-solving science, the graphical approach to adding vectors is not enough. It gives us a visual picture of the su of vectors, but it does not give us any nubers with which to work. Thus, we will ostly use the atheatical ethod for adding vectors, which is shown in the right hand portion of the figure. To add two vectors atheatically, we siply add the horizontal coponents together. This gives us the horizontal coponent of the final vector. We then add the vertical coponents together, and that gives us the vertical coponent of the vector. In the figure, then, the x-coponent of the final vector is siply x + B x, while the y-coponent is siply y + B y. Let e go through a quick exaple proble to jog your eory on all of this. EXMPLE 1.5 n explorer is given directions on how to get to a particular place. He is told to travel for 15. iles at a heading of 30.0 degrees and then to turn and travel 30.4 iles at a heading of 170. degrees. What is the vector which describes the explorer s final position? Solve this proble both graphically and atheatically. Let s start with the graphical ethod. First, we have to figure out how to draw the first vector. Well, the agnitude of the first vector (let s call it ) is 15. iles. The angle is 30.0 degrees. We can represent that vector as: 30.0 o Please realize that the drawing siply approxiates the values of the vector to give us a visual idea of what it looks like. We know that if the arrow pointed straight along the horizontal axis to the right, the angle would be 0 degrees. If it pointed up along the vertical axis, the angle would be 90.0 degrees. If it bited those two, it would be pointing at 45 degrees. n angle of 30.0 degrees, then, is close to biting the two axes, but not quite. Next, we put the tail of the ond vector at the head of the first. The ond vector has a agnitude of 30.4 iles and an angle of 170. degrees. Putting it on the head of the first vector gives us the following drawing:
Module 1 Units and Vectors Revisited 9 B 170. o 30.0 o Reeber, if the vector pointed straight up, its angle would be 90.0 degrees. If it pointed directly to the left, its angle would be 180.0 degrees. Thus, an angle of 170. degrees points alost directly to the left, but just a little up. In addition, the agnitude is twice that of the first vector, so the arrow is twice as long as the first. dding the vectors is now a snap. We just draw an arrow fro the tail of the first to the head of the ond. B 170. o C 30.0 o The arrow labeled C represents the su of the two vectors. Now, although that gives us a picture of the vector which represents the explorer s final position, it is only a picture. To get nubers which describe this vector, we ust add the two vectors atheatically. To do that, we ust get the horizontal and vertical coponents of each vector and add the together. Figure 1. gives the relationships between the coponents of a vector and its agnitude and direction, so that s not too bad: x = (15. iles) cos(30.0) = 13. iles y = (15. iles) sin(30.0) = 7.60 iles B x = (30.4 iles) cos(170.) = -30.0 iles B y = (30.4 iles) sin(170.) = 5.17 iles Notice that I used the rules of significant figures here. If you have forgotten those rules, go back to your first-year physics course (or cheistry if you took that course) and review significant figures so that you understand why I rounded the answers where I did. To get the x-coponent of the su (vector C), we just add the two x-coponents together, and to get the y-coponent of the su, we just add the two y-coponents together.
10 dvanced Physics in Creation C x = x + B x = 13. iles + -30.0 iles = -16.8 iles C y = y + B y = 7.60 iles + 5.17 iles = 1.77 iles In this case, I had to add nubers, and the significant figure rules are different for addition and subtraction as copared to ultiplication and division. Make sure you reeber the difference! The answer, then, is that the explorer s position has an x-coponent of -16.8 iles and a y- coponent of 1.77 iles. This eans the explorer is 16.8 iles left (west) of his starting position and 1.77 iles up (north) fro his starting position. Before I leave this review, there is one ore iportant thing of which I ust reind you. When dealing with the angle associated with a vector, you need to define the angle counterclockwise fro the positive x-axis, as shown in the exaple above. If you define the angle in that way, the atheatics will always work. Thus, if you find yourself working with a vector whose angle is not defined in that way, change the angle so that it is defined properly. If you have the coponents of a vector and need to get its agnitude and its angle, you can use the equations given in Figure 1.. Reeber, however, that the angle needs to be defined counterclockwise fro the positive x-axis. How can you be sure it is defined properly when working with those equations? Well, reeber fro algebra that you can divide the Cartesian coordinate plane into four regions: II III I IV When taking the inverse tangent of a nuber, the definition of the angle that your calculator gives you depends on which of these regions the vector is in. If the vector is in region I, the angle that your calculator gives you is defined relative to the positive x-axis, just as it should be defined. Thus, if your vector is region I, the angle that your calculator gives you for the inverse tangent function will be defined properly. However, if the vector is in region II of the Cartesian coordinate plane, then the angle that your calculator gives you is defined relative to the negative x-axis and is negative. In the Cartesian coordinate syste, negative angles ean clockwise rotation while positive angles ean counterclockwise rotation. So, when a vector is in region II, your calculator gives you the nuber of degrees clockwise fro the negative x-axis. Thus, if your vector is in region II and your calculator gives you a direction of -60 o, this is what it eans:
Module 1 Units and Vectors Revisited 11 θ = -60 o This definition of angle is not proper for our purposes. s a result, we ust convert it to the proper definition. It turns out that in both regions II and III of the Cartesian coordinate plane, if you siply add 180 to the angle that your calculator gives you, you will have converted your calculator s answer to an answer in which the angle is defined properly. If the vector is in region IV, then you ust add 360 to the calculator s answer in order to get the properly defined angle. In suary, Figure 1.4 tells you what you ust do in order to change your calculator s answer into a properly defined vector angle, based on the region of the Cartesian coordinate syste: FIGURE 1.4 Converting reference angle to vector angle + 180.0 +180.0 do nothing +360.0 This figure, of course, does you no good if you can t tell what region of the Cartesian coordinate plane your vector is in. Luckily, however, this is not a difficult task. ll you have to do is look at the signs on the vector coponents. If the x-coponent and y-coponent are both positive, then the vector ust be in region I. fter all, a positive x-coponent indicates that you are to the right of the origin, while a positive y-coponent eans you are above the origin. The region that is both to the right and above the origin is region I. On the other hand, suppose that both coponents are negative. Since a negative x-coponent eans left of the origin and a negative y-coponent eans down, you ust be in region III, since that is the only region that is to the left and below the origin. See how you do this? Following the sae logic, if the x-coponent is negative and the y-coponent is positive, you ust be in region II. If the x-coponent is positive and the y-coponent is negative, however, you ust be in region IV. I want to ake sure you understand this by showing you a quick exaple. EXMPLE 1.6 What are the agnitude and direction of vector C in the previous exaple?
1 dvanced Physics in Creation In the previous exaple, C had an x-coponent of -16.8 iles and a y-coponent of 1.77 iles. Getting the agnitude is siple, given the equation in Figure 1.. C = C = C = ( 168. iles) + ( 1. 77 iles) = 11. iles x y Getting the angle is a bit ore difficult, because it ust be defined properly. When we use the equation in Figure 1., we get: 1 1. 77 iles θ = tan ( ) = 37. 168. iles Now that we have an answer, we ust figure out the region of the vector. It has a negative x- coponent (and is therefore left of the origin) and a positive y-coponent (and is therefore above the x- axis). Thus, it is in region II. ccording to Figure 1.4, then, we ust add 180.0 to it. The properlydefined angle, then, is 14.8 o. Thus vector C can also be represented as 1.1 iles at an angle of 14.8 o. o I went through all of this rather quickly, but it should be review for you. You really should have learned all of this in your first-year course. Thus, if all of this sees a bit fuzzy to you, go back and review your first-year physics course. ON YOUR OWN 1.4 The velocity vector of a car has an x-coponent of 3 / and a y-coponent of 11 /. What are the agnitude and direction of the velocity vector? 1.5 Vector has a agnitude of 3.1 / at an angle of 60.0 degrees, and vector B has a agnitude of 1.4 / at an angle of 90.0 degrees. What is the su of these two vectors? Give your answer both graphically and in ters of agnitude and direction. Unit Vectors Everything in the previous tion should be review for you, and that s why I went through it so quickly. However, I now want to introduce soe notation that ay be new for you. Since a twodiensional vector can be represented in ters of its horizontal and vertical coponents, it is nice to actually define two unit vectors one in the horizontal direction and one in the vertical direction. unit vector, as shown in the figure below, has a agnitude of 1 and a direction along either the horizontal (x) or vertical (y) axis.
Module 1 Units and Vectors Revisited 13 Vector j: j = 1, θ = 90 0 j x = 0, j y = 1 FIGURE 1.5 Two Unit Vectors Vector i: i = 1, θ = 0 0 i x = 1, i y = 0 Why do I bother to define these vectors? Think about it. I can define any vector in ters of a horizontal coponent and a vertical coponent. Well, suppose I ultiply vector i by a scalar. What will happen to the agnitude of the vector? Well, since the original agnitude is 1, the new agnitude will be equal to the value of the scalar. What about the direction? Well, unless the scalar is negative, the direction will not change. If the scalar is negative, the direction will siply be reversed. The vector 3 i, for exaple, has a agnitude of 3 and points horizontally to the right. The vector -5 j, on the other hand, has a agnitude of 5 and points vertically down. So what? Think about the fact that we can define a vector by its horizontal and vertical coponents. Suppose velocity vector has a horizontal coponent of 5 / ( x = 5 /) and a vertical coponent of -4 / ( y = -4 /). We could write that in ters of adding unit vectors: = (5 /) i + (-4 /) j fter all, if the x-coponent is 5 /, then the x-coponent points to the right and has a agnitude of 5. That s what 5 i is. If the y-coponent is -4 /, then the y-coponent points down with a agnitude of 4 /. That s what -4 j eans. In the end, then, we can also note any vector in ters of unit vectors: For any vector, = x i + y j (1.1) This is one of the standard ways in which vectors are expressed, so it is iportant that you understand it. EXMPLE 1.7 The acceleration vector of an airplane is 13 / at q = 3. o. Express the vector in ters of unit vectors. Draw a graph of the vector.
14 dvanced Physics in Creation To express a vector in ters of unit vectors, we ust have its horizontal and vertical coponents. Thus, we ust convert agnitude and direction into coponents: x y o = ( 13 = ) cos( 3. ) 89. 7 o = ( 13 = ) sin( 3. ) 84. Now that we have the coponents, we just ultiply i by the x-coponent and j by the y-coponent, and we can express the vector in ters of unit vectors: = 89. 7 i + ( 84. j ) Usually, if the vector j is ultiplied by a negative, we just replace the plus sign with a inus sign, so the ore standard answer is: = 89. 7 i 84. j Now reeber, the nubers ultiplying i and j are siply the coponents of the vector. Thus, drawing the vector is a snap: -84. / -89.7 / Now reeber that to add any two vectors, you siply add the x-coponents together and the y-coponents together. Siilarly, to subtract vectors, you siply subtract the x-coponents and the y-coponents. In this notation, then, vector addition and subtraction is as follows: + B = ( x + B x ) i + ( y + B y ) j (1.) - B = ( x - B x ) i + ( y - B y ) j (1.3) We can also express scalar ultiplication in this notation. When I ultiply a vector by a scalar, I can siply ultiply each coponent by the scalar. Thus, in this notation, scalar ultiplication becoes:
Module 1 Units and Vectors Revisited 15 k = (k x ) i + (k y ) j (where k is a scalar) (1.4) Make sure you understand this new notation by solving the following probles. ON YOUR OWN 1.6 Draw the vector = (5.0 ) i - (11 ) j. Give its agnitude and direction. 1.7 What is the su of vector above with vector B = (-11 ) i + ( ) j? What is the difference - B? nswer using the notation you learned in this tion of the odule. 1.8 Write (in unit vector notation) the vector which corresponds to 6 ties the vector C = (.0 /) i + (3.0 /) j. The Dot Product When we add or subtract two vectors, the result is a vector. When we ultiply a vector by a scalar, the result is also a vector. Thus, you ight think that when you atheatically anipulate vectors, the result is always a vector. That s not true. One atheatical anipulation is called the dot product, and it is iportant in physics. When you copute the dot product of two vectors, the result is a scalar. How can two vectors produce a scalar? Well, first let s look at the echanics of taking the dot product of two vectors, and then I will tell you what the dot product really eans. B = x B x + y B y (1.5) In order to copute the dot product of two vectors, I ultiply their x-coponents together and add that result to the product of their y-coponents. Notice that there are no vectors on the right hand side of the equation, so the result is, indeed, a scalar. Given the following vectors: = 14 i + 11 j B = -.0 i + 3.0 j Copute the dot product B. EXMPLE 1.8
16 dvanced Physics in Creation Using Equation (1.5): The dot product, then, equals 5. B = x B x + y B y B = 14 (-.0) + 11 3.0 = -8 +33 = 5 Coputing the dot product between two vectors isn t too bad, is it? Of course, being able to perfor a atheatical operation is not the sae as understanding what that atheatical operation eans. Thus, you also need to know the eaning behind the dot product. When you take the dot product of two vectors, you are really ultiplying the agnitude of the first vector by the coponent of the ond vector which is parallel to the first vector. Now I know that sentence is confusing, so I want to try and explain it with a figure. FIGURE 1.6 The Meaning of the Dot Product Two vectors, and B: B θ Vector B can be split into two perpendicular coponents: one that is parallel to and one that is perpendicular to B: B θ B sinθ B = [Bcos(θ)] B cosθ Start by looking at the drawing on the left-hand side of the figure. I have two vectors there, our old friends and B. If you put their tails together (reeber, oving vectors around is okay as long as you don t change length or direction), you can define θ, the angle between the. Now we already know that any two-diensional vector can be expressed in ters of two perpendicular coponents. We usually use the horizontal and vertical coponents, but that s not necessary. We can really use any two perpendicular coponents. For the purpose of this discussion, then, let s define B in ters of a coponent that is parallel to and one that is perpendicular to. That s what is shown on the right hand side of the figure. Now look what happens when you split up B into those two coponents. One coponent lies right on vector. That s the parallel coponent of vector B, and it is calculated by taking the agnitude of the vector (B) and ultiplying by the cosine of the angle (θ). Thus, that little tion labeled B cosθ is the coponent of vector B that lies parallel to vector.
Module 1 Units and Vectors Revisited 17 When you take the dot product, you are calculating the value of the agnitude of vector ultiplied by the length of the coponent of vector B which lies parallel to. That s why the last line of text in the figure says that the dot product is (the agnitude of vector ) ties Bcosθ (the length of the coponent of B which is parallel to ). This is actually another way of expressing the dot product: B = B cosθ (1.6) The dot product of two vectors, then, can be found one of two ways. If you know the coponents of the vector, you use Equation (1.5). If you know the agnitudes of each vector and the angle between the, use Equation (1.6). lternatively, if you know the dot product and agnitudes of the vectors, you can deterine the angle between the. See what I ean by studying the following exaple. Given the following vectors: = 1i + 6.0j B = -3.0i + 9.0j Deterine the angle between the vectors. EXMPLE 1.9 To get an idea of just what we are solving for, let s draw the two vectors with each tail at the origin. This is not a necessary step for solving the proble. It siply gives us a visual of the angle. B θ How do we deterine the angle? Well notice that we have the coponents of each vector. Thus, we can deterine the agnitude of each. lso, we can deterine the dot product using Equation (1.5). Once we have that, then we can solve for the angle using Equation (1.6). Let s start with the agnitudes of each vector. That s not too hard: = 1 + 6. 0 = 13 B = 30. + 9. 0 = 9. 5
18 dvanced Physics in Creation Now let s figure out the dot product. Since we have the coponents, we use Equation (1.5): B = 1 (-3.0) + 6.0 9.0 = 18 t this point, we have all of the inforation in Equation (1.6) except the angle, so we can solve for it: B = B cosθ 18 = 13 95. cosθ θ = cos 18 = cos ( 015. ) = 8 13 95. 1 1 o The angle, then, is 8 o. Look back at the drawing of the two vectors. Notice that the angle looks very close to a right angle. Thus, the answer we have akes sense based on the drawing. ON YOUR OWN 1.9 Copute the dot product of the following vectors: = (-.3 ) i - (1.) j, B = (1. ) i + (4.3 ) j. 1.10 Two vectors, and B, are defined. The angle between the is 61.0 o. Vector has a agnitude of 15.1 eters, and the coponent of B which is parallel to is 1. eters long. What is the agnitude of vector B? The Physical Significance of the Dot Product In the last tion, you learned how to copute the dot product of two vectors and what that eans atheatically. However, this is a physics course. It ight be nice to learn a new ath concept, but this is not a ath course. Thus, I would not have taught you about the dot product unless there was soe physics behind it. To learn the physics behind the dot product, you need to recall the concept of work fro your first-year physics course. Do you reeber how we define work atheatically? Try to see if this doesn t jog your eory: W = F x (1.7)
Module 1 Units and Vectors Revisited 19 In this equation, W represents work, x stands for the displaceent over which the force was applied, and F is used to indicate the coponent of the force vector that is parallel to the direction of otion. To reeber why Equation (1.7) uses only that portion of the applied force that is parallel to the otion, consider Figure 1.7. FIGURE 1.7 Force and Work F F x F y In the left side of this figure, a boy is pulling a wagon. If we reove the picture of the boy (right side of the figure) and just look at the vectors involved, we see several things. First, the wagon travels in a straight, horizontal line, as indicated by the horizontal velocity vector (v). The force that the boy is applying (F) goes in the sae direction as the wagon s handle. s a result, his pulling force is not parallel with the otion. Since the force is a two-diensional vector, we can split it into vertical (F y ) and horizontal (F x ) coponents. The horizontal coponent is parallel to the otion and, as you can see, is the only portion of the force that contributes to the otion. In contrast, there is no otion in the perpendicular direction. In other words, the wagon is not oving upwards. In the end, the perpendicular portion of the force fights gravity. Since it is not strong enough to overcoe gravity, the wagon does not ove up. Reeber, when a force is applied but there is no otion, there is no work. Thus, the perpendicular coponent of the boy s pulling force is wasted. It causes no otion, and therefore it accoplishes no work. That s why only the portion of the force parallel to the otion is considered when calculating the work done by that force. v So, when you are faced with a situation in which you ust calculate the work done by a force which takes place over a certain displaceent, you can use Equation (1.7). Wait a inute, however. Equation (1.7) tells you to take the agnitude of one vector (the displaceent vector, x), and ultiply it by the agnitude of the coponent of another vector (the force vector, F) which is parallel to the first vector. What does that sound like? It sounds like the dot product! Thus, we can re-write the definition of work in dot product notation: W = F x (1.8)
0 dvanced Physics in Creation Now reeber, work is a scalar. It doesn t tell us anything about direction. That akes sense, since the result of the dot product is a scalar. If you have been paying close attention, there ay be soething puzzling you at this point. In the dot product, we take the agnitude of the first vector, and ultiply it by the agnitude of the coponent of the ond vector which is parallel to the first vector. Thus, if I were to really write Equation (1.7) in dot product for, it should read W=x F. However, look at Equation (1.6). This equation tells us that the dot product is coutative. fter all, B = B cosθ B = B cosθ Notice that B cosθ and B cosθ are equivalent. Thus, B and B are equivalent. Thus, the order in which you take the dot product does not atter. The dot product is coutative: B is the sae as B This is iportant and worth reebering. Now that you know the physical significance of the dot product, solve the following on your own probles. ON YOUR OWN 1.11 particle undergoes a displaceent x = (1.5 ) i - (.3 ) j while being acted upon by a constant force F = (5.6 N) i - (3.4 N) j. What is the work done? 1.1 person applies a force of 16.6 Newtons to an object as the object travels 9. eters. If the work done was 14.5 J, what was the angle between the force vector and the displaceent vector? The Cross Product The dot product of two vectors produces a scalar. It only akes sense that if there is a way to ultiply two vectors to produce a scalar, there ust be a way to ultiply two vectors to produce a vector. Indeed, there is. We call it the cross product. In the cross product, we are still ultiplying the agnitude of one vector with the agnitude of another vector. In this case, however, we are ultiplying the agnitude of the first vector with the coponent of the ond vector which is perpendicular to the first vector. In addition to using a different coponent of the ond vector, the cross product also produces a vector, not a scalar. Thus, the cross product has both agnitude and direction. Let s deal with the agnitude first.
Module 1 Units and Vectors Revisited 1 FIGURE 1.8 The Magnitude of the Cross Product Two vectors, and B: B θ Vector B can be split into two perpendicular coponents: one that is parallel to and one that is perpendicular to B: B θ B sinθ x B = [Bsin(θ)] B cosθ Once again, we have our old friends vector and vector B. They can be drawn tail-to-tail, and the resulting angle between the is called θ. If we split vector B into two coponents: one parallel to and one perpendicular to, we find that the perpendicular coponent is B sinθ. Thus, since the cross product involves ultiplying the agnitude of the first vector and the agnitude of the coponent of the ond vector which is perpendicular to the first, the agnitude of the cross product is: x B = B sinθ (1.9) The vertical lines enclosing x B siply ean agnitude. Thus, x B eans the agnitude of the vector x B. So that s how we get the agnitude of the cross product. The cross product produces a vector, however, so there is also direction to consider. How do we coe up with the direction of the cross product? We use soething called the right hand rule. Right hand rule - To deterine the direction of the cross product x B, take your right hand and point your fingers in the direction of. Then, curl your fingers towards B, along the arc of the angle between the vectors. Your thub will then point in the direction of the cross product The right hand rule is illustrated in Figure 1.9.
dvanced Physics in Creation Illus. By Megan Whitaker FIGURE 1.9 The Right hand Rule direction of the cross product. Point the fingers of your right hand in the direction of the first vector. Curl your fingers along the arc of the angle between the vectors. Your thub points in the direction of the cross product. In this exaple, the thub pointed up out of the page. That s the direction of the cross product. When deterining the direction of the cross product, then, you take your right hand and point your fingers in the direction of the first vector. Then, you curl your fingers towards the ond vector, along the arc of the angle between the. Your thub will then point in the direction of the cross product. Think about what this eans for a oent. The direction of the cross product of two vectors will always be perpendicular to both vectors. The right hand rule has another iplication. Look at Figure 1.9 and use the right hand rule to deterine the direction of B x. Reeber, you point the fingers on your right hand in the direction of the first vector (B), and you then curl towards the ond vector (). Where does your thub point? It points down towards the paper. That s the opposite direction as that shown in the figure. Thus, unlike the dot production, the cross product is not coutative. This is iportant to reeber. The cross product is not coutative: x B = -(B x ) Okay, there is one ore thing you need to learn about the cross product. You need to know how to calculate the cross product using unit vector notation. However, before you can do that, you need to see how we define unit vectors in three-diensional space.
Module 1 Units and Vectors Revisited 3 FIGURE 1.10 Three-Diensional Space in Ters of Unit Vectors y k j i x z When we add another diension (the z-axis), we can represent that in unit vector notation with just another unit vector. Notice that i is still the unit vector in the horizontal direction and j is still the unit vector in the vertical direction. To represent three-diensional space, then, we siply add a third unit vector, k, which points out of the plane of the paper, towards you. If we ultiply by -1, the unit vector -k points behind the paper. Now that you know how three-diensional space is represented in vector notation, you can learn how we copute the cross product. First, let s start with the siple case of two-diensional vectors. For two-diensional vectors in the i/j plane: x B = ( x B y - y B x ) k (1.10) Notice the restriction placed on this equation. To use this equation, you ust be taking the cross product of two-diensional vectors which exist only in the plane defined by the horizontal (i) and vertical (j) axes. Thus, this is a pretty restrictive equation. However, ost of the physics probles that you do will involve such vectors, so it is probably what you will use ost often for calculating cross products. For copleteness sake, I will give you the total equation for calculating the cross product between any three-diensional vectors: x B = ( y B z - z B y ) i + ( z B x - x B z ) j + ( x B y - y B x ) k (1.11) Notice that Equation (1.11) reduces to Equation (1.10) for two-diensional vectors in the i/j plane. fter all, the z-coponent of such a two-diensional vector is zero. Thus, the ter ( y B z - z B y ) i is zero, as is the ter ( z B x - x B z) j. s a result, the only ter in the equation that is non-zero is ( x B y - y B x ) k, and that gives us Equation (1.10). I know that this is a lot to throw at you, but hopefully the following exaple probles will clear up any confusion that ay still be in your ind.
4 dvanced Physics in Creation EXMPLE 1.10 velocity vector has a agnitude of 56.1 / and an angle of 45.0 degrees. nother velocity vector has a agnitude of 1. / and an angle of 90.1 degrees. Calculate the agnitude of the cross product and give the vector in unit vector notation. Calculating the agnitude of the cross product is not bad at all. We just use Equation (1.9). To do that, however, we need the angle between the vectors. To deterine that we will have to draw the two vectors: 90.1 o 45.0 o 360.0 o - 90.1 o = 69.9 o If the angle of the ond vector as defined fro the positive x-axis is 90.1, then the angle fro the positive x-axis down to the vector ust be 69.9 o, because the total angle ust be 360.0 o. Well, the angle fro the positive x-axis up to the first vector is 45.0 o. The angle between the two vectors, then, ust be 69.9 o + 45.0 o = 114.9 o. Now we can use Equation (1.9): x B = B sinθ = (56.1 /) (1. /) sin(114.9 o ) = 61 / The agnitude of the cross product, then, is 61 /. What about the direction? For that, we use the right hand rule. We point the fingers of our right hand in the direction of the first vector, then we curl along the arc of the 114.9 o angle in between the vectors. When we do that, our thub points down into the page. Thus, the vector is in the negative k direction. That tells us what we need to know for unit vector notation. fter all, we know the vector s agnitude (61 / ), and we know that it is pointed in the negative k direction. Thus, the vector is -(61 / ) k. Given the following vectors, calculate x B and deterine the angle between the vectors: = 3.4 i + 4.5 j B =.4 i + 1.1 j
Module 1 Units and Vectors Revisited 5 Notice that these two vectors are two-diensional and have only i and j unit vectors. Thus, they are in the i/j plane, and we can use the sipler version of the cross product equation, Equation (1.10): x B = ( x B y - y B x ) k x B = [3.4 1.1-4.5 (.4)] k = -7.1 k That s the cross product. Based solely on the unit vector notation, you know that it is pointing behind the plane of the paper. What about the angle? Well, we now know the agnitude of the vector (if the vector is coposed solely of 7.1 ties k, then the agnitude is 7.1, because k has a agnitude of 1), so we can use Equation (1.9): x B = B sinθ Reeber, x B eans agnitude of x B, so that s 7.1. We don t have the agnitudes of and B, but we can use their coponents to calculate the: = 34. + 45. = 56. B = (. 4) + 11. = 6. Now we can use Equation (1.9): xb = B sinθ 71. = ( 56. ) (. 6) sinθ θ = sin 1 71. = 9 ( 56. ) (. 6) o Given the following vectors, calculate the cross product and the angle between the: = -1.4 i + 4. j - 5.6 k B =.4 i - 1.1 j +3. k This is essentially the sae as the proble above. However, these are three-diensional vectors, so we have to use the larger forula, Equation (1.11): x B = ( y B z - z B y ) i + ( z B x - x B z ) j + ( x B y - y B x ) k x B = (4. 3.-[-5.6] [-1.1]) i + (-5.6.4 - [-1.4] 3.) j + ([-1.4] [-1.1] - 4..4) k x B = 7. i - 8.9 j - 8.6 k
6 dvanced Physics in Creation Just as we did before, we can calculate the agnitudes of the two vectors and then deterine the angle between the using Equation (1.9). The only difference now is that the agnitude of the cross product is not as easy to deterine. We ust calculate it like we calculate the agnitudes of all vectors: = ( 14. ) + 4. + ( 56. ) = 71. B =. 4 + 11. + 3. = 41. x B = 7. + ( 8. 9) + ( 8. 6) = 14 Now that we have all of the agnitudes involved, we can use Equation (1.9): x B = B sinθ 14 = ( 71. ) ( 41. ) sinθ θ = sin 1 14 = 9 ( 71. ) ( 41. ) o ON YOUR OWN 1.13 Vector is defined as 3. feet at 45.1 o, and vector B is defined as 1.1 feet at 70.1 o. What is the cross product? Give your answer in unit vector notation. 1.14 Given the following vectors, calculate the cross product and the angle between the: = -7.1 i + 4. j B = 3.4 i - 4.1 j The Physical Significance of the Cross Product s was the case with the dot product, there is physical significance to the cross product. We will apply the cross product in at least three different areas of physics, but for right now, I will concentrate on only one: the concept of torque. s you learned in your first-year physics course, when we apply a force soe distance away fro an axis of rotation, the result is a torque, which can cause rotational otion. In your first-year physics course, you learned the equation used to calculate torque. τ = F r (1.1)
Module 1 Units and Vectors Revisited 7 Where τ represents torque, r represents the agnitude of the vector drawn fro the axis of rotation to the point at which the force is applied, and F represents the coponent of the force which is perpendicular to that vector. Figure 1.11 illustrates the concept of torque. FIGURE 1.11 Torque lever ar F F F Now reeber what torque is. It is the counterpart of force when one is considering rotational otion. Reeber, force causes acceleration in a straight line. Torque causes rotational acceleration. In the figure, the wrench is going to turn the screw. To do that, it will have to give the screw rotational acceleration so that it starts to turn in a circle. Torque is the ipetus which will cause that rotational acceleration. Notice fro the figure that only a portion of the force used can generate torque. ny coponent of the force that is parallel to the vector defined fro the axis of rotation to the point at which the force is applied (the lever ar) is lost. Thus, the agnitude of the torque is given by the agnitude of the lever ar ties the coponent of the force which is perpendicular to the lever ar. Well, since the cross product takes the agnitude of a vector and ultiplies it by the agnitude of a ond vector s coponent which is perpendicular to the first vector, torque can be calculated using the cross product. t = r x F (1.13) There are two things to note about this equation. First, torque is a vector. Reeber, the cross product results in a vector. That s why the t is bold. Second, reeber that the cross product is
8 dvanced Physics in Creation NOT coutative. Thus, the order of the vectors is iportant, so one ust take the vector of the lever ar (r) and cross it with the vector representing the force (F), in that order. Since you already know how to do cross products, calculating the torque is pretty easy. Thus, I will not give you any exaples of it. However, you should perfor the following on your own probles to ake sure you understand how to use the cross product to calculate torque. ON YOUR OWN 1.15 person applies a force F = (15 N) i + (3 N) j on a lever ar r = (1. ) i + (1.1 ) j. What is the vector that represents the torque? 1.16 an is trying to turn a bolt. He exerts a force of 15. N with a pipe that creates a lever ar which is 0.36 long. If the pluber succeeds in producing 4.9 N of torque, what is the angle between the force he is exerting and the wrench? Given the diagra below, is the torque pointing up above the plane of the paper or back behind the plane of the paper? θ F r Suing Up This odule contained soe concepts which were review for you and others which were new. That will be the case with ost of the odules in this course. In each odule, I will review soe of the highlights of your first-year physics course and then go deeper into each subject. In addition, copletely new concepts will be brought in fro tie to tie. Thus, if there is soething that really baffles you in this course, you should review your first-year course to see if it is explained there.