# Introduction and Mathematical Concepts

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1 CHAPTER 1 Introduction and Mathematical Concepts PREVIEW In this chapter you will be introduced to the physical units most frequently encountered in physics. After completion of the chapter you will be able to convert between these systems of units and use them as an aid in problem solving. Also, after a review of trigonometry, you will become acquainted with vectors and the methods of vector addition. Upon completion of this material you should be able to add and subtract vectors graphically, decompose vectors into their components and use the components to reconstruct the vectors. Important terms QUICK REFERENCE Scalar quantity A quantity which can be described by a single number. Vector quantity A quantity which can be adequately described by a number (magnitude) and a direction. Vector components Two perpendicular vectors which added together produce the original vector. Systems of Units System Length Mass Time SI meter (m) kilogram (kg) seconds (s) BE foot (ft) slug (sl) seconds (s) CGS centimeters (cm) gram (g) seconds (s) Trigonometry sin θ = h h o (1.1) θ = sin -1 h h o (1.4) cos θ = h h a (1.2) θ = cos -1 h h a (1.5) h h tan θ = o (1.3) θ = tan -1 o ha h a (1.6)

2 h h o θ h a h 2 = ho 2 + ha 2 (1.7)

3 Graphical Addition of Vectors Vectors may be added (or subtracted) by placing them head to tail. -B A B R A R R = A + B R = A - B Components of a Vector The following expressions refer to the vector diagram shown below. A x = A cos θ A = A x 2 + Ay 2 y A y = A sin θ A y θ = tan -1 A x A y A θ A x x Addition of Vectors Using Components If C = A + B, then the components of C are: Cx = Ax + Bx Cy = Ay + By and the magnitude and direction of C are: C = C x 2 + Cy 2 C y θ = tan -1 C x where θ is measured counterclockwise from the + x axis.

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5 DISCUSSION OF SELECTED SECTIONS 1.3 The Role of Units in Problem Solving Units are very important in the study of physics in that all physical quantities have units. These may be either the base units of length (m), mass (kg), and time (s), or derived units such as the joule (kg m 2 /s 2 ). When used in algebraic expressions, the units which accompany the numbers can be used to check not only the accuracy of the calculation, but also the validity of the equation. For this reason, the units will always be displayed along with the numbers in this study guide. You are encouraged to do the same in your solutions to problems. Remember, if the units do not work out, your solution is not right either. Example 1 Manipulating Units and Converting Between Systems Convert 5.00 mi/h to m/s. Conversion factors: 1 mi = 5280 ft 1 ft = m 1 km = 1000 m 1 h = 3600 s Each of the equalities above can be used to form a fraction or conversion factor that is equal to unity (i.e., equal to 1). In multiplying by unity we do not change the value of the physical quantity; we are merely expressing the same quantity in a different set of units. One side of the equality will appear in the numerator and the other side will appear in the denominator of the fraction. The specific choice will be made so that the unwanted units cancel and the desired units appear in the final answer mi/h = 5.00 mi h 5280 ft 1 mi m 1 h = 2.24 m 1 ft 3600 s s. Notice that the first conversion factor changes miles to feet, the second changes feet to meters, and the third conversion factor changes hours to seconds. Again notice that in each conversion factor, the numerator is equal to the denominator and, therefore, the conversion factor equal to unity. Example 2 Another Look at Converting Units Convert 2.00 mi2 to ft2. We will start with the fact that 1 mi = 5280 ft. In setting up the conversion factor as a fraction equal to unity, however, we must square the fraction so that it converts square miles (mi2) to square feet (ft2) mi 2 = 2.00 mi ft 1 mi = ft 2 Notice that the conversion factor is still equal to unity (12 = 1). Also notice that the final answer is expressed in scientific notation to preserve the correct number of significant figures; three in this case.

6 Example 3 Using Units to Check Equations An equation which may result from the application of the conservation of energy principle is 1 mv 2 = mgh 2 where m is a mass with the units of kg, v is a velocity with the units of m/s, g is an acceleration with the units of m/s 2 and h is a height with the units of m. This equation may be checked for validity by simply substituting the units into it and manipulating them as if they were numbers. If the units on the right and left side of the equation do not match, the equation is definitely NOT VALID. (kg) ( m s ) 2 = (kg) ( m s 2 ) (m) ( ) ( ) (kg) m2 s 2 = (kg) m2 s 2 kg m 2 s 2 = kg m 2 s 2 In this case, the units on each side of the equation do match and the equation may be a valid one. This procedure does NOT guarantee that the equation has any "true" meaning, however. By the way, the above combination of units appear so often in physics that they are given the special name of joules (J). This is the unit of ENERGY. 1.4 Trigonometry Most likely, you have already become acquainted with the basic trigonometric functions in a high school or college course. These can be found in the Quick Reference section of this book. Many times the use of the "trig" functions in physics involve finding the length of one side of a right triangle when you know one other side and one of the acute angles. This is particularly important in finding the components of a vector which will be studied in the next section. Other variations are, of course, possible and useful. For instance, if two sides of a right triangle are known, then the trigonometric functions can be used to find the angle. Example 4 Finding one side of a right triangle if one other side and an angle are known An observer, whose eyes are 6.0 ft above the ground, is standing 105 ft away from a tree. The ground is level, and the tree is growing perpendicular to it. The observer's line of sight with the tree top makes an angle of 20.0 above the horizontal. How tall is the tree? The observer's eye, the tree top, and a point on the trunk are the vertices of a right triangle. The height of the tree is H = 6.0 ft + ho

7 Now ho can be found from the right triangle by using equation (1.3) h θ h o H h a ho = ha tan θ = (105 ft) tan 20.0 = 38.2 ft. H = 6.0 ft ft = 44.2 ft. Another common need in physics is to find one side of a triangle when the other two sides are known but no acute angle is given. Then you should use the Pythagorean theorem (1.7) or one of its variants. Example 5 Using the Pythagorean theorem In the previous example, it is desired to know the straight line distance from the person's eye to the top of the tree. The Pythagorean theorem gives h = h ha 2 = (105 ft) 2 + (38.2 ft) 2 = 112 ft. 1.6 Vector Addition and Subtraction If vectors are colinear, then they may be added or subtracted by simply adding or subtracting their magnitudes. The directions of the vectors are usually specified by calling a vector pointing to the right (or up) positive and a vector pointing to the left (or down) negative. Then the sign of the resultant vector tells which way it points. Example 6 Adding and subtracting colinear vectors Find the resultant, C, of the following vectors (u is an arbitrary unit). Case 1 Case 2 A = +3.0 u A = +3.0 u B = +2.0 u B = -2.0 u In both cases the vectors are colinear. One could imagine placing the vectors A and B in a "tail-to-head" fashion. In case 1, both A and B point in the same direction. The resultant vector will have a magnitude equal to the length of A plus the length of B. Since both vectors are directed to the right, the resultant will also point to the right. In case 2, the vectors A and B point in opposite directions. If they are placed in a tail-to-head fashion it will be clear

8 that part of A is canceled by B. The resultant will have a magnitude that is equal to the magnitude of A minus the magnitude of B. Since A has the greater magnitude, the resultant will point in the direction of A. The resultant C is shown below for both cases. Case 1 Case 2 C = +5.0 u C = +1.0 u When vectors are not colinear but perpendicular, they may be added by using the Pythagorean theorem and the tangent function. This method yields the magnitude and direction of the resultant vector as a number and an angle. Example 7 Adding perpendicular vectors Find the resultant, C, of the vectors shown where A = u and B = u. The resultant is C = A + B as shown in the diagram. Notice that it was constructed by placing the vectors A and B in a "tail-to-head" fashion and then connecting the "head" of A to the "tail" of B. The length of the resultant vector, C, is given by the Pythagorean theorem to be C = A 2 + B 2 C = (2.0 u) 2 + (5.0 u) 2 C = 5.4 u. C A θ B The direction of C, as specified by the angle, θ, is given by (1.6) to be θ = tan -1 A B = tan u 5.0 u = 22 o. Now C = 5.4 u, 22 counterclockwise from B.

9 1.7 The Components of a Vector As you might infer from the preceding example, any vector, C, may be expressed as the sum of two perpendicular vectors. When these perpendicular vectors are placed along the x and y axes of a Cartesian coordinate system, they are referred to as the x and y components (Cx and Cy) of the vector. The components of a vector may be found by using a vector diagram similar to the above and equations (1.1) and (1.2). Please keep in mind that if equations (1.1) and (1.2) are used to find the components of a vector, the angle, θ, is defined as being measured from the x axis. Example 8 Finding the components of a vector Find the components of the vector A = 5.0 u, θ = 30. An application of (1.2) to the vector shown in the figure gives Ax = A cos θ = (5.0 u) cos 30 Ax = 4.3 u. A similar application of (1.1) gives Ay = A sin θ = (5.0 u) sin 30 Ay = 2.5 u. y A y A θ A x x Sometimes you will encounter situations where you will need to find the x and y components of a vector, but the angle given, φ, is NOT measured from the x axis. In this case, the components may NOT be given by Ax = A cos φ and Ay = A sin φ. You now have two choices. You may either find the angle, φ, in terms of the angle, θ, or you may try to find a different right triangle in which to apply equations (1.1) and (1.2). Example 9 Finding the components of a vector when the angle is not measured from the +x axis Find the x and y components of the vector, A, whose magnitude is 10.0 u and which makes an angle of 45 with the +y axis. Refer to the figure shown. We may use Ax = A cos θ and Ay = A sin θ if we can find the angle, θ. From the figure it is seen that

10 θ = 90 + φ so that θ = = 135. A x y A y A φ θ x Now Ax = (10.0 u) cos 135 Ax = u Ay = (10.0 u) sin 135 Ay = u. We may also apply equations (1.1) and (1.2) to the right triangle shown in the figure to obtain and Ax = - A sin φ = - (10.0 u) sin 45 = u Ay = + A cos φ = + (10.0 u) cos 45 = u. 1.8 Addition of Vectors by Means of Components Since like components of two or more vectors are colinear, they may simply be added, as in example 6, to give the corresponding component of the resultant vector. In this way, all of the components of the resultant vector may be found. The magnitude and direction of the resultant vector may then be determined by equations (1.3) and (1.7). Example 10 Adding vectors by the component method Add the vectors A, B, and C shown in the figure using the component method. A = 5.0 m, B = 7.0 m and C = 4.0 m. An application of equations (1.1) and (1.2) to each of the triangles shown in the figure gives Ax = + A cos 20 Ax = + (5.0 m) cos 20 Ax = m. Bx = - B cos 40

11 Bx = - (7.0 m) cos 40 Bx = m. Cx = + C sin 25 Cx = + (4.0 m) sin 25 Cx = m. y B 40 A 20 C x 25 The x component of the resultant vector is Rx = Ax + Bx + Cx Rx = 4.7 m m m Rx = 1.0 m. Repeating the above for the y component gives Ay = + A sin 20 = + (5.0 m) sin 20 = m. By = + B sin 40 = + (7.0 m) sin 40 = m. Cy = - C cos 25 = - (4.0 m) cos 25 = m. The y component of the resultant vector is Ry = Ay + By + Cy = 1.7 m m m Ry = 2.6 m. Now the magnitude of the resultant can be found from the Pythagorean theorem R = R x 2 + Ry 2 R = (1.0 m) 2 + (2.6 m) 2 = 2.8 m The angle that R makes with the +x axis is

12 θ = tan -1 (Ry/Rx) = tan -1 (2.6 m/1.0 m) = 69.

13 PRACTICE PROBLEMS The following problems are provided to give you additional practice solving single concept problems. Some work space has been left for you after or to the right of the problem. The solutions to these problems will be found at the end of this chapter. 1. How many significant figures are in the numbers a b c. 2.8 X 10 5 d e X Convert the following into the indicated units. a m to ft b. 1.2 ft/s to m/s c mi/h to km/h d m/s 2 to ft/s 2 e. 535 kg to slugs 3. A basketball coach insists that his players be at least cm tall. Would a player of height 5 ft 11.5 in tall qualify for the team? 4. A football field is yards long. Express this distance in millimeters.

14 5. Check the following equations for possible validity. a. x 2 = 1/2 gt where x is in m, g is in m/s 2, t is in s. b. v 2 = 2 ax where v is in m/s, a is in m/s 2 and x is in ft. 6. A right triangle has a side of length 3.5 m. The angle opposite the 3.5 m side is 25. Find the length of the other sides. 7. A m tall building casts a shadow m long over level ground. What is the sun's elevation angle above the horizon? 8. A right triangle has two sides of length 25 ft and 15 ft. Find the length of the hypotenuse and all angles. 9. A bridge 50.0 m long crosses a chasm. If the bridge is inclined at an angle of 20.0 to the horizontal, what is the difference in height between the two ends?

15 10. A displacement vector of magnitude 5.0 m points in an easterly direction. A second displacement vector points north and has a magnitude of 9.7 m. Find the magnitude and direction of the vector sum. 11. An electric field vector, E, has a magnitude of 1.0 newtons per coul (N/C) and makes an angle of 33 CCW from the +x axis of a Cartesian coordinate system. Find the components of E. 12. A magnetic field vector, B, is oriented 65 clockwise from the -y axis. It has a magnitude of tesla (T). Find the x and y components of B. 13. A vector, A, has a magnitude of 10.0 u and points 21.0 north of west. A second vector, B, has a magnitude of 5.2 u and points 47.0 east of south. Find the magnitude and direction of the sum by the component method. 14. A car drives 2 km west, then 8 km south, and then 10.0 km at an angle 53 north of east. Find the car's final displacement. (magnitude and direction)

17 CHAPTER QUIZ This quiz has been provided to help you diagnose possible weak areas in the understanding of the key chapter concepts. The answers can be found at the end of the chapter in this guide. 1. How many significant figures are in the answer of ? a. one b. two c. three d. four 2. How many significant figures are in the result of 1600/2.80? a. one b. two c. three d. four 3. The derived unit (kg 2 m 2 )/(kg m s 2 ) is equivalent to a. (kg 2 )/s 2 b. (kg m/s 2 ) c. (kg m)/s d. (kg m 2 )/m 4. How many feet are in 25 m? a. 82 b c d If you are given one angle and the length of the side opposite in a right triangle, which trig function would allow you, in a single step, to find the length of the hypotenuse? a. sine b. cosine c. tangent d. any 6. If you are given one angle and the length of the hypotenuse of a right triangle, which trig function would you use to find the side adjacent to the angle? a. sine b. cosine c. tangent d. any 7. Vectors may be added graphically by placing them a. tail to tail. b. parallel. c. tail to head. d. head to head. 8. Which of the following is an example of a scalar? a. force b. volume c. displacement d. velocity 9. The parts of a vector which lie in perpendicular directions are called a. magnitudes b. resultants c. components d. scalars 10. If the angle, less than 90, specifying the direction of a vector is given as measured from the y axis, then the x component will ALWAYS involve what trig function? a. sine b. cosine c. tangent d. any 11. Which of the following is NOT a vector quantity? a. time b. velocity c. force d. displacement 12. Two displacement vectors have magnitudes of 8 m and 12 m, respectively. When these two vectors are added the magnitude of the sum a. is 20 m c. is larger than 20 m. b. is 4 m. d. could be as large as 20 m or as small as 4 m. 13. A physical quantity is calculated from the formula K = 3π c (a2 + b2) where a, b, and c are all lengths. What is the dimension of K? a. [L] c. 3π [L]2 b. [L]2 d. [L]3

18 SOLUTIONS AND ANSWERS Practice Problems 1. a has THREE significant figure. b has TWO significant figures. Note that it can be written as 5.4 X10-3. c. 2.8 X 10 5 has TWO significant figures. d has TWO significant figures since both zeros are in doubt. e X 10-4 has TWO significant figures. 2. a. We have 5.00 m = (5.00 m)(3.28 ft/1 m) = 16.4 ft. b. Also 1.2 ft/s = (1.2 ft/s)(0.305 m/1 ft) = 0.37 m/s. c. We know 60.0 mi/h = (60.0 mi/1 h)(5280 ft/1 mi)(0.305 m/1 ft)(1/1000 km/1 m)= 96.6 km/h. d. In this case e. Finally 9.80 m/s 2 = (9.80 m/s 2 )(3.28 ft/1 m) = 32.1 ft/s kg = (535 kg)(6.85 X 10-2 sl/1 kg) = 36.6 sl. 3. The conversion looks like cm = (180.0 cm)/(1 in/2.54 cm) = 70.9 in = 5 ft 10.9 in. YES 4. In this case yds = (100.0 yds)(3 ft/yd)(12 in/ft)(2.54 cm/in)(10 mm/cm) = mm. 5. a. Substitute the units into the equation to see if both sides match. m 2 = (m/s 2 )(s) = m/s The units on each side do NOT match so the equation is not valid. b. Similarly, m 2 /s 2 = (m/s 2 )(m) = m 2 /s 2. The units DO match so the equation MAY be valid. 6. The hypotenuse is given by (1.1) The side adjacent is given by (1.2), h = (3.5 m)/sin 25 = 8.3 m. ha = (8.3 m)cos 25 = 7.5 m.

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20 7. The shadow, the building, and a line drawn from the top of the building to the corresponding point on the end of the shadow form a right triangle. The angle between this latter line and the horizontal is the elevation angle of the sun and is θ = tan -1 (500.0 m/800.0 m) = The length of the hypotenuse is given by (1.7), The angle opposite the 25 ft side is given by (1.6), h 2 = (25 ft) 2 + (15 ft) 2, h = 29 ft. The angle opposite the 15 ft side is θ = tan -1 (25/15). θ = 59. θ = tan -1 (15/25) = 31. Note that the angles add to 90 as they should. 9. We are given the hypotenuse and one angle in a right triangle. (1.1) gives the side opposite the angle to be ho = (50.0 m) sin 20.0 = 17.1 m. 10. The vectors are perpendicular; hence the magnitude of the vector sum can be found from the Pythagorean theorem. magnitude = 11 m. The angle the vector makes with the east-west line (+x) is θ = tan -1 (9.7/5.0) = 63 N of E. 11. The x component is The y component is Ex = (1.0 N/C) cos 33 = 0.84 N/C. Ey = (1.0 N/C) sin 33 = 0.54 N/C. 12. The angle as measured from the +x axis is θ = = 205. Then Bx = (0.010 T) cos 205 = T and By = (0.010 T) sin 205 = T. 13. The x components are: Ax = -(10.0 u) cos 21.0 = u, Bx = (5.2 u) sin 47.0 = 3.80 u, Rx = u.

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22 The y components are: Ay = (10.0 u) sin 21.0 = 3.58 u, By = -(5.2 u) cos 47.0 = u, Ry = 0.03 u. The magnitude of the resultant vector is found from the Pythagorean theorem to be The angle is R = ( u) 2 + (0.03 u) 2 = 5.54 u. θ = tan -1 (0.03/5.54) = N of W. 14. Let A be the westerly displacements, B the southerly displacement, and C be the remaining displacement. Taking east and north to be positive, the east-west (x) components are: Ax = -2 km, Bx = 0 km, Cx = +(10.0 km) cos 53 = 6 km. Rx = +4 km. The north-south (y) components are: Ay = 0 km, By = -8 km, Cy = +(10.0 km) sin 53 = +8 km. Ry = 0 km. The magnitude of the resultant vector is given by the Pythagorean theorem to be and the direction is R = (4 km) 2 + (0 km) 2 = 4 km. θ = tan -1 (0/4) = 0. That is, EAST. Quiz answers 1. b 4. a 7. c 10. a 13. d 2. b 5. a 8. b 11. a 3. b 6. b 9. c 12. d

23 MCAT REVIEW PROBLEMS When an airplane flies, its total velocity with respect to the ground is vtotal = vplane + vwind, where vplane denotes the plane s velocity through motionless air, and vwind denotes the wind s velocity. Crucially, all the quantities in this equation are vectors. The magnitude of a velocity vector is often called the speed. Consider an airplane whose speed through motionless air is 100 meters per second (m/s). To reach its destination, the plane must fly east. The heading of a plane is the direction in which the nose of the plane points. So, it is the direction in which the engines propel the plane. 1. If the plane has an eastward heading, and a 20 m/s wind blows towards the southwest, then the plane s speed is: A. 80 m/s B. more than 80 m/s but less than 100 m/s C. 100 m/s D. more than 100 m/s 2. The pilot maintains an eastward heading while a 20 m/s wind blows northward. The plane s velocity is deflected from due east by what angle? A. sin B. cos C. tan D. none of the above 3. Let φ denote the answer to question 2. The plane in question 2 has what speed with respect to the ground? A. (100 m/s)sin φ B. (100 m/s)cos φ 100 m/s 100 m/s C. sin φ D. cos φ 4. Because the 20 m/s northward wind persists, the pilot adjusts the heading so that the plane s total velocity is eastward. By what angle does the new heading differ from due east? A. sin B. cos C. tan D. none of the above 5. Let θ denote the answer to question 4. What is the total speed, with respect to the ground, of the plane in

24 question 4? A. (100 m/s)sin θ B. (100 m/s)cos θ C. 100 m/s sin θ D. 100 m/s cos θ ANSWERS TO MCAT REVIEW PROBLEMS 1. B. In all of these problems, a good diagram gets you half way to the answer. For this plane, v plane = 100 m/s eastward, and v wind = 20 m/s towards the southwest. To represent the total velocity, add these two vectors: FIGURE 1 Just by looking at this rough diagram, you can see that the wind slows down the plane. So, the answer must be A or B. Since the wind does not blow westward, it does not directly oppose the plane s motion. Therefore, the wind cannot slow the plane down by a full 20 m/s from the plane s default velocity of 100 m/s. 2. C. Because the pilot maintains an eastward heading, the engines still push the plane due east. So, v plane = 100 m/s eastward. But the wind blows at v wind = 20 m/s northward, thereby knocking the plane off course by angle φ. FIGURE 2 In this triangle, we know the two legs but not the hypotenuse. Specifically, in units of m/s, the opposite leg has length h o = 20, while the adjacent leg has length h a = 100. Therefore, by the definition of tangent,

25 tan φ = Take the inverse tangent of both sides to get φ = tan -1 h o h a = D. By looking at figure 2, and remembering the definition of cosine, we get h v ah plane 100 m/s cos φ = = v total = v total. Multiply through by vtotal, and then divide through by cos φ, to get vtotal = 100 m/s cos φ. 4. A. The pilot now adjusts the heading so that v total points due east. Since the wind knocks the plane northward, the plane must head slightly southward, to offset the wind:

26 [--- Unable To Translate Graphic ---]

27 FIGURE 3 Here, v plane is the hypotenuse. From figure 3, we get sin θ = h o h = v wind v plane = 20 m / s 100 m / s, and hence, θ = sin B. Again from figure 3, the hypotenuse has length h = v plane = 100 m/s, while the adjacent leg has length h a = v total. So, cos θ = h a h = v total v plane = v total 100 m / s. Multiply through by 100 m/s to get (100 m/s)cos θ = v total.

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