CHAPTER 1 INTRODUCTION AND MATHEMATICAL CONCEPTS. s K J =

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1 CHPTER 1 INTRODUCTION ND MTHEMTICL CONCEPTS CONCEPTUL QUESTIONS 1. RESONING ND SOLUTION The quantit tan is dimensionless and has no units. The units of the ratio /v are m F = m s s (m / s) H G I m K J = Thus, the units on the left side of the equation are not consistent with those on the right side, and the equation tan = /v is not a possible relationship between the variables, v, and.. SSM RESONING ND SOLUTION It is not alwas possible to add two numbers that have the same dimensions. In order to add an two phsical quantities the must be epressed in the same units. Consider the two lengths: 1.00 m and 1.00 cm. oth quantities are lengths and, therefore, have the dimension [L]. Since the units are different, however, these two numbers cannot be added. 3. RESONING ND SOLUTION a. The SI unit for is m. The SI units for the quantit vt are m s (s) = m F m I (s) m HG s K J = Therefore, the units on the left hand side of the equation are consistent with the units on the right hand side. b. s described in part a, the SI units for the quantities and vt are both m. The SI units for the quantit 1 at are F mi HG s K J = (s ) m Therefore, the units on the left hand side of the equation are consistent with the units on the right hand side. c. The SI unit for v is m/s. The SI unit for the quantit at is F mi (s) HG s K J = m s

2 INTRODUCTION ND MTHEMTICL CONCEPTS Therefore, the units on the left hand side of the equation are consistent with the units on the right hand side. d. s described in part c, the SI units of the quantities v and at are both m/s. The SI unit of the quantit 1 at3 is F mi HG s K J = 3 (s ) m s Thus, the units on the left hand side are not consistent with the units on the right hand side. In fact, the right hand side is not a valid operation because it is not possible to add phsical quantities that have different units. e. The SI unit for the quantit v 3 is m 3 /s 3. The SI unit for the quantit a is F mi m (m ) HG s K J = s Therefore, the units on the left hand side of the equation are not consistent with the units on the right hand side. 3 f. The SI unit for the quantit t is s. The SI unit for the quantit m (m / s ) = F H G I m s = s = s mkj a is Therefore, the units on the left hand side of the equation are consistent with the units on the right hand side. 4. RESONING ND SOLUTION a. The dimension of a phsical quantit describes the phsical nature of the quantit and the kind of unit that is used to epress the quantit. It is possible for two quantities to have the same dimensions but different units. ll lengths, for eample, have the dimension [L]. However, a length ma be epressed in an length unit, such as kilometers, meters, centimeters, millimeters, inches, feet or ards. s another illustration, the quantities 100 g and 1.5 kg are masses and have the dimensions [M]; however, the have different units. b. ll quantities with the same units must have the same dimensions. For eample, all quantities epressed in kilograms have the dimension [M]; all quantities epressed in meters have the dimensions [L].

3 Chapter 1 Conceptual Questions 3 5. RESONING ND SOLUTION For the equation to be valid, the dimensions of the left hand side of the equation must be the same as the dimensions on the right hand side. Since the quantit c has no dimensions, it does not contribute to the dimensions of the right hand side, regardless of the value of n. Therefore, the value of n cannot be determined from dimensional analsis. 6. RESONING ND SOLUTION The following table shows the value of sin, cos, the ratio (sin )/(cos ) and tan. sin cos (sin )/(cos ) tan From the definitions given in Equations , we have sin h / h h o o = = = tan cos h / h h a 7. RESONING ND SOLUTION a. The graph below shows sin plotted on the vertical ais and on the horizontal ais, with in 15 increments from = 0 to = 70. a Sin (degrees)

4 4 INTRODUCTION ND MTHEMTICL CONCEPTS b. The graph below shows cos plotted on the vertical ais and on the horizontal ais, with in 15 increments from = 0 to = Cos (degrees) RESONING ND SOLUTION vector quantit has both magnitude and direction. The number of people attending a football game, the number of das in a month, and the number of pages in a book can all be completel specified b giving a magnitude onl. Hence, none of these quantities can be considered a vector. 9. RESONING ND SOLUTION For two vectors to be equal, the must be equal in magnitude and have the same direction. Onl vectors,, and D have the same magnitude. These three vectors are shown below: North North North 60 D 30 East 30 East East Clearl, onl vectors and D have the same magnitude and point in the same direction; hence, and D are equal.

5 Chapter 1 Conceptual Questions RESONING ND SOLUTION For two vectors to be equal, the must be equal in magnitude and have the same direction. Thus, two vectors with the same magnitude are not necessaril equal. The must also point in the same direction. 11. SSM RESONING ND SOLUTION One can arrive back at the starting point after making eight consecutive displacements that add to zero onl if one traverses three of the edges on one face and three edges on the opposite face (si displacements; the remaining two displacements occur in going from one opposite face to the other). For an particular starting point, there are four independent was to traverse three edges on opposite faces. These are illustrated in the figures below. In () and (), one traverses three consecutive edges before moving to the opposite face; in (C) and (D), one traverses two consecutive edges, moves to the opposite face, traverses three consecutive edges, then moves back to the original face and traverses the third edge on that face ending at the starting point. () starting point (C) starting point () starting point (D) starting point For an particular starting point, one can initiall move in an one of three possible directions. Thus, there are a total of 4 (independent was) 3 (possible directions) or 1 was to arrive back at the starting point that involve eight displacement vectors. 1. RESONING ND SOLUTION (a) n vector has a zero component in the direction that is perpendicular to the direction of that vector. Thus, it is possible for one component of a vector to be zero, while the vector itself is not zero.

6 6 INTRODUCTION ND MTHEMTICL CONCEPTS (b) In order for a vector to be zero, its components in all mutuall perpendicular directions must also be zero. Therefore, it is not possible for a vector to be zero, while one of its components is not zero. 13. RESONING ND SOLUTION If two or more vectors have a resultant of zero, then, when the are arranged in a tail-to-head fashion the head of the last vector must touch the tail of the first vector. In order to satisf this requirement with two vectors, the vectors must be equal in magnitude and opposite in direction. Therefore, it is not possible to add two perpendicular vectors so that the vector sum is zero. 14. RESONING ND SOLUTION Three or more vectors with unequal magnitudes ma add together so that their vector sum is zero. To do so, the must be oriented so that, when the are placed in a tail-to-head fashion, the head of the last vector touches the tail of the first vector, as in the drawing at the right. C + + C=0 15. RESONING ND SOLUTION The vector equation + = 0, implies that =. a. The magnitude of must be equal to the magnitude of. b. The vectors and must point in opposite directions, as indicated b the minus sign in =. The same conclusions can be reached from a geometric argument. If + = 0, then, when and are placed tail-to-head, the head of must touch the tail of. Thus, the vectors and must have the same length (i.e., the same magnitude) and point in opposite directions. 16. RESONING ND SOLUTION The equation + = C tells us that the vector C is the resultant of the vectors and. The magnitudes of the vectors,, and C are related b + = C. This has the same form as the Pthagorean theorem that relates the length of the two sides of a right triangle and the length of the hpotenuse. Thus, the vectors and must be at right angles (or perpendicular) to each other. 17. RESONING ND SOLUTION The equation + = C tells us that the vector C is the resultant of the vectors and. The magnitudes of the vectors,, and C are related b + = C. In other words, the length of the vector C is equal to the combined lengths of vectors and. Therefore, the vectors and must point in the same direction. 18. RESONING ND SOLUTION If the magnitude of a vector is doubled, we ma conclude that the magnitude of each component of the vector has also doubled.

7 Chapter 1 Conceptual Questions 7 This is best eplained using the arbitrar vector,, shown at the right. The angle, that specifies the direction of, is given b = tan 1 F HG I KJ Suppose that the magnitude of is doubled. The vector is now twice as long, but its direction (and therefore the angle ) remains unchanged. In order for the angle to remain the same after the magnitude of is doubled, both and must double, so that the ratio remains the same. 19. SSM RESONING ND SOLUTION a. s the angle increases from 0 to 90, the component of the vector decreases in magnitude while the component increases in magnitude. When the vector has rotated through an angle of 90, the component has zero magnitude and the magnitude of the component is equal to the magnitude of the original vector (see figure below). b. s the angle increases from 90 to 180, the component of the vector decreases in magnitude while the component increases in magnitude. When the vector has rotated through an angle of 180, the component has zero magnitude and the magnitude of the component is equal to the magnitude of the original vector. c. s the angle increases from 180 to 70, the component of the vector decreases in magnitude while the component increases in magnitude. When the vector has rotated through an angle of 70, the component has zero magnitude and the magnitude of the component is equal to the magnitude of the original vector. d. s the angle increases from 70 to 360, the component of the vector decreases in magnitude while the component increases in magnitude. When the vector has rotated through an angle of 360, the component has zero magnitude and the magnitude of the component is equal to the magnitude of the original vector. (a) (b) (c) (d) RESONING ND SOLUTION In general, a vector that has a component of zero along the ais of a certain aes sstem does not have a component of zero along the ais of another (rotated) aes sstem. Consider the vector shown below.

8 8 INTRODUCTION ND MTHEMTICL CONCEPTS ' ' ' Figure 1 Figure Relative to the, aes shown in Figure 1, the vector has a component of zero along the ais. Figure shows the same vector along with a new aes sstem ', '. The primed aes sstem is rotated at an arbitrar angle counterclockwise relative to the unprimed, aes sstem. Clearl the component of along the ' ais is non zero for most arbitrar values of the (rotation) angle. The eceptions occur when = 0, 180 and 360.

9 Chapter 1 Problems 9 CHPTER 1 INTRODUCTION ND MTHEMTICL CONCEPTS PROLEMS 1. SSM RESONING ND SOLUTION We use the fact that 0.00 g = 1 carat and that, under the conditions stated, 1000 g has a weight of.05 lb to construct the two conversion factors: (0.00 g)/(1 carat) = 1 and (.05 lb)/(1000 g) = 1. Then, 0.00 g ( 3106 carats ) 1 carat.05 lb = 1000 g 1.37 lb. RESONING We use the facts that 1 mi = 580 ft, 1 m = 3.81 ft, and 1 d = 3 ft. With these facts we construct three conversion factors: (580 ft)/(1 mi) = 1, (1 m)/(3.81 ft) = 1, and (3 ft)/(1 d) = 1. SOLUTION multipling b the given distance d of the fall b the appropriate conversion factors we find that d = ( 6 mi ) 580 ft 1 mi 1 m d 3.81 ft ( ) 3 ft 1 d 1 m = 3.81 ft m 3. RESONING a. To convert the speed from miles per hour (mi/h) to kilometers per hour (km/h), we need to convert miles to kilometers. This conversion is achieved b using the relation km = 1 mi (see the page facing the inside of the front cover of the tet). b. To convert the speed from miles per hour (mi/h) to meters per second (m/s), we must convert miles to meters and hours to seconds. This is accomplished b using the conversions 1 mi = 1609 m and 1 h = SOLUTION a. Multipling the speed of 34.0 mi/h b a factor of unit, (1.609 km)/(1 mi) = 1, we find the speed of the bicclists is mi Speed = 34.0 () 1 = 34.0 h mi h 1.609km = 1mi km 54.7 h b. Multipling the speed of 34.0 mi/h b two factors of unit, (1609 m)/(1 mi) = 1 and (1 h)/(3600 s) = 1, the speed of the bicclists is

10 10 INTRODUCTION ND MTHEMTICL CONCEPTS mi mi 1609m 1h m Speed = 34.0 ()() 1 1 = 34.0 = 15. h h 1mi 3600s s 4. RESONING Multipling an equation b a factor of 1 does not alter the equation; this is the basis of our solution. We will use factors of 1 in the following forms: 1 gal = 1, since 1 gal = 18 oz 18 oz m = 1, since m 3 = 1 gal 1 gal 1 ml m = 1, since 1 ml = 10 6 m 3 SOLUTION The starting point for our solution is the fact that Volume = 1 oz Multipling this equation on the right b factors of 1 does not alter the equation, so it follows that 1 gal m 1 ml Volume = ( 1 oz)()()() =( 1 oz ) = 9.6 ml 18 oz 1 gal m Note that all the units on the right, ecept one, are eliminated algebraicall, leaving onl the desired units of milliliters (ml). 5. SSM RESONING When converting between units, we write down the units eplicitl in the calculations and treat them like an algebraic quantit. We construct the appropriate conversion factor (equal to unit) so that the final result has the desired units. SOLUTION a. Since grams = 1.0 kilogram, it follows that the appropriate conversion factor is ( g)/(1.0 kg) = 1. Therefore, 6 ( 5 10 kg ) b. Since milligrams = 1.0 gram, g = 1.0 kg g

11 Chapter 1 Problems ( g ) = mg 1.0 g 5 mg c. Since micrograms = 1.0 gram, 3 ( 5 10 g ) g = 1.0 g 6 µ µ g 6. RESONING This problem involves using unit conversions to determine the number of magnums in one jeroboam. The necessar relationships are 1.0 magnum = 1.5 liters 1.0 jeroboam = 0.79 U. S. gallons 1.00 U. S. gallon = m 3 = liters These relationships ma be used to construct the appropriate conversion factors. SOLUTION multipling one jeroboam b the appropriate conversion factors we can determine the number of magnums in a jeroboam as shown below: 0.79 gallons liters 1.0 magnum ( 1.0 jeroboam ) 1.0 jeroboam 1.0 gallon =.0 magnums 1.5 liters 7. RESONING ND SOLUTION a. F = [M][L]/[T] ; ma = [M][L]/[T] = [M][L]/[T] so F = ma is dimensionall correct. b. = [L]; at 3 = ([L]/[T] )[T] 3 = [L][T] so = (1/)at 3 is not dimensionall correct. c. E = [M][L] /[T] ; mv = [M][L]/[T] so E = (1/)mv is not dimensionall correct. d. E = [M][L] /[T] ; ma = [M]([L]/[T] )[L] = [M][L] /[T] so E = ma is dimensionall correct. e. v = [L]/[T]; (F/m) 1/ = {([M][L]/[T] )([L]/[M])} 1/ = {[L] /[T] } 1/ = [L]/[T] so v = (F/m) 1/ is dimensionall correct.

12 1 INTRODUCTION ND MTHEMTICL CONCEPTS 8. RESONING In the epression for the volume flow rate, the dimensions on the left side of the equals sign are [L] 3 /[T]. If the epression is to be valid, the dimensions on the right side of the equals sign must also be [L] 3 /[T]. Thus, the dimensions for the various smbols on the right must combine algebraicall to ield [L] 3 /[T]. We will substitute the dimensions for each smbol in the epression and treat the dimensions of [M], [L], and [T] as algebraic variables, solving the resulting equation for the value of the eponent n. SOLUTION We begin b noting that the smbol π and the number 8 have no dimensions. It follows, then, that n 3 ( 1) [ L] or 8η L [ T] π R P P Q = = n [ M] [ L] [ L][ T] [ M] [ L ][ ] [ L ] T n [ ] [ ] [ L][ T] n [ ] [ L][ T] L T L = = 3 [ L] [ T] n [ L] = [ ] [ L][ T] n [ L] [ L] [ ] [ ] [ ] [ ] or L = or L L = L = L n Thus, we find that n = SSM RESONING The volume of water at a depth d beneath the rectangle is equal to the area of the rectangle multiplied b d. The area of the rectangle = (1.0 nautical miles) (.60 nautical miles) = 3.1 (nautical miles). Since 6076 ft = 1 nautical mile and m = 1 ft, the conversion factor between nautical miles and meters is 6076 ft m m = 1 nautical mile 1 ft 1 nautical mile SOLUTION The area of the rectangle of water in m is, therefore, (nautical miles) m = 1 nautical mile m Since 1 fathom = 6 ft, and 1 ft = m, the depth d in meters is 6 ft ( 16.0 fathoms ) 1 fathom The volume of water beneath the rectangle is m 1 ft 1 = m ( m ) ( m) = m 3

13 Chapter 1 Problems RESONING The dimension of the spring constant k can be determined b first solving the equation T = π m/ k for k in terms of the time T and the mass m. Then, the dimensions of T and m can be substituted into this epression to ield the dimension of k. SOLUTION lgebraicall solving the epression above for k gives k = 4 π m/ T. The term 4π is a numerical factor that does not have a dimension, so it can be ignored in this analsis. Since the dimension for mass is [M] and that for time is [T], the dimension of k is Dimension of k = [ M] [ T] 11. RESONING ND SOLUTION The following figure (not drawn to scale) shows the geometr of the situation, when the observer is a distance r from the base of the arch. The angle is related to r and h b tan = h/ r. Solving for r, we find h = 19 m r = h tan = 19 m tan.0 = m = 5.5 km r 1. RESONING ND SOLUTION In the diagram below, = 14.6 and h = 830 m. We know that sin = H/h and, therefore, H = h sin = (830 m) sin 14.6 = 713 m 13. SSM WWW RESONING The shortest distance between the two towns is along the line that joins them. This distance, h, is the hpotenuse of a right triangle whose other sides are h o = 35.0 km and h a = 7.0 km, as shown in the figure below. h H SOLUTION The angle is given b tan = h o / h a so that = tan km 7.0 km = 5.9 S of W W h h o We can then use the Pthagorean theorem to find h. h a S

14 14 INTRODUCTION ND MTHEMTICL CONCEPTS h = h + h = (35.0 km) + ( 7. 0 km) = km o a 14. RESONING In both parts of the drawing the line of sight, the horizontal dashed line, and the vertical form a right triangle. The angle of the line of sight above the horizontal is known in each case, as is the horizontal distance from the building. Therefore, we can use the tangent function to calculate the vertical heights H a and H b (see the drawing). The height of the antenna is the difference Hb Ha between these heights. H a H b m m (a) (b) SOLUTION Since the tangent of an angle is the side of the right triangle opposite the angle divided b the side adjacent to the angle, we find for parts a and b that H a ( ) H ( ) = 85.0 m tan 35.0 and = 85.0 m tan 38.0 b The height of the antenna is the difference between these two values: ( ) ( ) Hb Ha = 85.0 m tan m tan 35.0 = 6.9 m 15. RESONING The drawing shows the heights of the two balloonists and the horizontal distance between them. lso shown in dashed lines is a right triangle, one angle of which is Note that the side adjacent to the 13.3 angle is the horizontal distance, while the side opposite the angle is the distance between the two heights, 61.0 m 48. m. Since we know the angle and the length of one side of the right triangle, we can use trigonometr to find the length of the other side. 48. m m SOLUTION The definition of the tangent function, Equation 1.3, can be used to find the horizontal distance, since the angle and the length of the opposite side are known:

15 Chapter 1 Problems 15 Solving for gives length of opposite side tan13.3 = length of adjacent side (= ) length of opposite side 61.0 m 48. m = = = 54.1 m tan13.3 tan RESONING ND SOLUTION Consider the following views of the cube. ottom View Na Side View Cl L a c a Na a Cl Na L Na The length, L, of the diagonal of the bottom face of the cube can be found using the Pthagorean theorem to be L = a + a = (0.81 nm) = nm or L = nm The required distance c is also found using the Pthagorean theorem. c = L + a = (0.397 nm) + (0.81 nm) = 0.37 nm Then, c = nm _ 17. RESONING Note from the drawing that the shaded right triangle contains the angle, the side opposite the angle (length = 0.81 nm), and the side adjacent to the angle (length = L). If the length L can be determined, we can use trigonometr to find. The bottom face of the cube is a square whose diagonal has a length L. This length can be found from the Pthagorean theorem, since the lengths of the two sides of the square are known nm L 0.81 nm 0.81 nm SOLUTION The angle can be obtained from the inverse tangent function, Equation 1.6, as 1 ( ) = tan 0.81 nm / L. Since L is the length of the hpotenuse of a right triangle whose sides have lengths of 0.81 nm, its value can be determined from the Pthagorean theorem:

16 16 INTRODUCTION ND MTHEMTICL CONCEPTS ( ) ( ) L = 0.81 nm nm = nm Thus, the angle is nm nm = tan = tan = 35.3 L nm 18. RESONING a. The drawing shows the person standing on the earth and looking at the horizon. Notice the right triangle, the sides of which are R, the radius of the earth, and d, the distance from the person s ees to the horizon. The length of the hpotenuse is R + h, where h is the height of the person s ees above the water. Since we know the lengths of two sides of the triangle, the Pthagorean theorem can be emploed to find the length of the third side. h R Eart d R 90 Horizon b. To convert the distance from meters to miles, we use the relation 1609 m = 1 mi (see the page facing the inside of the front cover of the tet). SOLUTION a. The Pthagorean theorem (Equation 1.7) states that the square of the hpotenuse is equal to the sum of the squares of the sides, or ( ) ields ( ) d = R+ h R = R + Rh+ h R R + h = d + R. Solving this equation for d ( )( ) ( ) 6 = Rh + h = m 1.6 m m = 4500 m b. Multipling the distance of 4500 m b a factor of unit, (1 mi)/(1609 m) = 1, the distance (in miles) from the person's ees to the horizon is 1 mi =.8 mi 1609 m d = ( 4500 m)( 1) = ( 4500 m )

17 Chapter 1 Problems SSM RESONING The drawing at the right shows the location of each deer,, and C. From the problem statement it follows that b = 6 m b α c β γ a C c = 95 m γ = = ppling the law of cosines (given in ppendi E) to the geometr in the figure, we have a ab cos γ + (b c ) = 0 which is an epression that is quadratic in a. It can be simplified to a + a + C = 0, with = 1 = b cos γ = (6 m) cos 5 = 76 m C = (b c ) = (6 m) (95 m) = 5181 m This quadratic equation can be solved for the desired quantit a. SOLUTION Suppressing units, we obtain from the quadratic formula a = ± ( 76) ( 76) 4( 1)( 5181) = m and 43 m 1 ( ) Discarding the negative root, which has no phsical significance, we conclude that the distance between deer and C is m. 0. RESONING ND SOLUTION Consider the bottom face of a tetrahedron which is an equilateral triangle. L H L H L L L 30 o 30 o

18 18 INTRODUCTION ND MTHEMTICL CONCEPTS The distance from a verte to the intersection of the perpendicular bisectors is 1 L L = = cos 30 3 Now consider an right triangle in the tetrahedron whose sides are H, L, and. The Pthagorean theorem gives L H H 1 = + = + L Then 1 H = L L = L so that H/ L= /3 3 3 _ 1. RESONING a. Since the two force vectors and have directions due west and due north, the are perpendicular. Therefore, the resultant vector F = + has a magnitude given b the Pthagorean theorem: F = +. Knowing the magnitudes of and, we can calculate the magnitude of F. The direction of the resultant can be obtained using trigonometr. b. For the vector F = we note that the subtraction can be regarded as an addition in the following sense: F = + ( ). The vector points due south, opposite the vector, so the two vectors are once again perpendicular and the magnitude of F again is given b the Pthagorean theorem. The direction again can be obtained using trigonometr. 3 SOLUTION a. The drawing shows the two vectors and the resultant vector. ccording to the Pthagorean theorem, we have F = + F = + b g b g F = 445 N + 35 N = 551 N Using trigonometr, we can see that the direction of the resultant is F North (a) North F (b) 1 35 N tan or = tan N = F H G I K J = north of west b. Referring to the drawing and following the same procedure as in part a, we find

19 b g b g b g b g Chapter 1 Problems 19 F = + or F = + = 445 N + 35 N = 551 N 1F 35 N = H G I tan or = tan K J = south of west 445 N _. RESONING For convenience, we can assign due east to be the positive direction and due west to be the negative direction. Since all the vectors point along the same east-west line, the vectors can be added just like the usual algebraic addition of positive and negative scalars. We will carr out the addition for all of the possible choices for the two vectors and identif the resultants with the smallest and largest magnitudes. SOLUTION There are si possible choices for the two vectors, leading to the following resultant vectors: F + F = 50.0 newtons newtons = newtons = 60.0 newtons, due east 1 F + F = 50.0 newtons 40.0 newtons = newtons = 10.0 newtons, due east 1 3 F + F = 50.0 newtons 30.0 newtons =+ 0.0 newtons = 0.0 newtons, due east F F F = 10.0 newtons 40.0 newtons = 30.0 newtons = 30.0 newtons, due west 3 + F = 10.0 newtons 30.0 newtons = 0.0 newtons = 0.0 newtons, due west 4 F + F = 40.0 newtons 30.0 newtons = 70.0 newtons = 70.0 newtons, due west 3 4 The resultant vector with the smallest magnitude is The resultant vector with the largest magnitude is F1+ F 3 = 10.0 newtons, due east. F3 + F 4 = 70.0 newtons, due west. 3. SSM RESONING The performer walks out on the wire a distance d, and the vertical distance to the net is h. Since these two distances are perpendicular, the magnitude of the displacement is given b the Pthagorean theorem as s = d + h. Values for s and h are given, so we can solve this epression for the distance d. The angle that the performer s displacement makes below the horizontal can be found using trigonometr. SOLUTION a. Using the Pthagorean theorem, we find that b g b g s= d + h or d = s h = 6. 7 ft 5. 0 ft = 9. 4 ft b. The angle that the performer s displacement makes below the horizontal is given b

20 0 INTRODUCTION ND MTHEMTICL CONCEPTS h 1F h. tan = tan H G I 1 K J F 5 0 ft = tan H G I or = K J = 69 d d 9.4 ft 4. RESONING The triple jump consists of a double jump (assumed to end on a square that we label C) followed b a single jump. The single jump is perpendicular to the double jump, so that the length C of the double jump, the length C of the single jump, and the magnitude of the total displacement form a right triangle. Thus, we have = + C C SOLUTION The diagonal of one square on the checkerboard has a length d of b g b g d = 40. cm cm = 566. cm Since C = 4 d and C = d, it follows that b g b g b g = 4d + d = d 0 = cm 0 = 5 cm _ 5. SSM WWW RESONING When a vector is multiplied b 1, the magnitude of the vector remains the same, but the direction is reversed. Vector subtraction is carried out in the same manner as vector addition ecept that one of the vectors has been multiplied b 1. SOLUTION Since both vectors point north, the are colinear. Therefore their magnitudes ma be added b the rules of ordinar algebra. a. Taking north as the positive direction, we have = + ( ) = +.43 km + ( 7.74 km) = 5.31 km The minus sign in the answer for indicates that the direction is south so that b. Similarl, = 5.31 km, south = + ( ) = km + (.43 km) = km The plus sign in the answer for indicates that the direction is north, so that = 5.31 km, north

21 Chapter 1 Problems 1 6. RESONING t the turning point, the distance to the campground is labeled d in the drawing. Note that d is the length of the hpotenuse of a right triangle. Since we know the lengths of the other two sides of the triangle, the Pthagorean theorem can be used to find d. The direction that cclist # must head during the last part of the trip is given b the angle. It can be determined b using the inverse tangent function. Turning point 1950 m Campground 1430 m SOLUTION W E a. The two sides of the triangle have lengths of 1080 m 1080 m and 50 m (1950 m 1430 m = 50 m). The length d S of the hpotenuse can be determined from the Pthagorean theorem, Equation (1.7), as ( ) ( ) d = 1080 m + 50 m = 100 m Start N d b. Since the lengths of the sides opposite and adjacent to the angle are known, the inverse tangent function (Equation 1.6) can be used to find m = tan = 6 south of east 1080 m 7. SSM WWW RESONING ND SOLUTION The single force needed to produce the same effect is equal to the resultant of the forces provided b the two ropes. The figure below shows the force vectors drawn to scale and arranged tail to head. The magnitude and direction of the resultant can be found b direct measurement using the scale factor shown in the figure. 900 N 900 N Resultant Scale: 1000 N a. From the figure, the magnitude of the resultant is 5600 N. b. The single rope should be directed along the dashed line in the tet drawing. 8. RESONING a. Since the two displacement vectors and have directions due south and due east, the are perpendicular. Therefore, the resultant vector R = + has a magnitude given b the Pthagorean theorem: R = +. Knowing the magnitudes of R

22 INTRODUCTION ND MTHEMTICL CONCEPTS and, we can calculate the magnitude of. The direction of the resultant can be obtained using trigonometr. b. For the vector R = we note that the subtraction can be regarded as an addition in the following sense: R = + ( ). The vector points due west, opposite the vector, so the two vectors are once again perpendicular and the magnitude of R again is given b the Pthagorean theorem. The direction again can be obtained using trigonometr. SOLUTION a. The drawing shows the two vectors and the resultant vector. ccording to the Pthagorean theorem, we have R = + or = R North R R North ( ) ( ) = 3.75 km.50 km =.8 km (a) (b) Using trigonometr, we can see that the direction of the resultant is 1F.8 km tan = tan H G I or =. K J = 48 east of south 50km b. Referring to the drawing and following the same procedure as in part a, we find b g b g b g R = + or = R = 375. km 50. km = 8. km 1F.8 km = H G I tan or = tan K J = 48 west of south 50. km _ 9. RESONING a. and b. The drawing shows the two vectors and, as well as the resultant vector +. The three vectors form a right triangle, of which two of the sides are known. We can emplo the Pthagorean theorem, Equation 1.7, to find the length of the third side. The angle in the drawing can be determined b using the inverse cosine function, Equation 1.5, since the side adjacent to and the length of the hpotenuse are known. N 15.0 units + W 1.3 units E S

23 Chapter 1 Problems 3 c. and d. The drawing illustrates the two vectors and, as well as the resultant vector. The three vectors form a right triangle, which is identical to the one above, ecept for the orientation. Therefore, the lengths of the hpotenuses and the angles are equal. N W 1.3 units E 15.0 units S SOLUTION a. Let R = +. The Pthagorean theorem (Equation 1.7) states that the square of the hpotenuse is equal to the sum of the squares of the sides, so that R = +. Solving for ields ( ) ( ) = R = 15.0 units 1.3 units = 8.6 units b. The angle can be found from the inverse cosine function, Equation 1.5: units = cos = 34.9 north of west 15.0 units c. Ecept for orientation, the triangles in the two drawings are the same. Thus, the value for is the same as that determined in part (a) above: = 8.6 units d. The angle is the same as that found in part (a), ecept the resultant vector points south of west, rather than north of west: = 34.9 south of west 30. RESONING ND SOLUTION The following figure is a scale diagram of the forces drawn tail-to-head. The scale factor is shown in the figure. The head of F 3 touches the tail of F 1, because the resultant of the three forces is zero. a. From the figure, F 3 must have a magnitude of 78 N if the resultant force acting on the ball is zero. F F 1 = 50.0 N b. Measurement with a protractor indicates that the angle = 34. Scale Factor: F = 90.0 N N _

24 4 INTRODUCTION ND MTHEMTICL CONCEPTS 31. SSM RESONING ND SOLUTION In order to determine which vector has the largest and components, we calculate the magnitude of the and components eplicitl and compare them. In the calculations, the smbol u denotes the units of the vectors. = (100.0 u) cos 90.0 = 0.00 u = (00.0 u) cos 60.0 = u C = (150.0 u) cos 0.00 = u = (100.0 u) sin 90.0 = u = (00.0 u) sin 60.0 = 173 u C = (150.0 u) sin 0.00 = 0.00 u a. C has the largest component. b. has the largest component. 3. RESONING ND SOLUTION a. From the Pthagorean theorem, we have F = (150 N) + (130 N) =.0 10 N b. The angle is given b F 130 N 150 N 1 F 130 N = H G I tan K J = N _ 33. RESONING ND SOLUTION The east and north components are, respectivel a. e = cos = (155 km) cos 18.0 = 147 km b. n = sin = (155 km) sin 18.0 = 47.9 km _ 34. RESONING The triangle in the drawing is a right triangle. We know one of its angles is 30.0, and the length of the hpotenuse is 8.6 m. Therefore, the sine and cosine functions can be used to find the magnitudes of and. The directions of these vectors can be found b eamining the diagram.

25 Chapter 1 Problems 5 SOLUTION a. The magnitude of the displacement vector is related to the length of the hpotenuse and the 30.0 angle b the sine function (Equation 1.1). The drawing shows that the direction of is due east. #1 W N S E ( ) = sin 30.0 = 8.6 m sin 30.0 = 4.3 m, due east b. In a similar manner, the magnitude of can be found b using the cosine function (Equation 1.). Its direction is due south. ( ) = cos 30.0 = 8.6 m cos30.0 = 7.4 m, due south # m Net 35. SSM RESONING The and components of r are mutuall perpendicular; therefore, the magnitude of r can be found using the Pthagorean theorem. The direction of r can be found using the definition of the tangent function. SOLUTION ccording to the Pthagorean theorem, we have r = + = ( 15 m ) + ( 184 m) = m The angle is = tan m 15 m = RESONING Two vectors that are equal must have the same magnitude and direction. Equivalentl, the must have identical components and identical components. We will begin b eamining the given information with respect to these criteria, in order to see if there are obvious reasons wh some of the vectors could not be equal. Then we will compare our choices for the equal vectors b calculating the magnitude and direction and the scalar components, as needed. SOLUTION Vectors and cannot possibl be equal, because the have different scalar components of = 80.0 m and = 60.0 m. Furthermore, vectors and C cannot possibl be equal, because the have different magnitudes of = 75.0 m and C = m. Therefore, we conclude that vectors and C are the equal vectors. To verif that this is indeed the case we have two choices. We can either calculate the magnitude and direction of (and compare it to the given magnitude and direction of C) or determine the scalar components of C (and compare them to the given components of ). Either choice will do, although both are shown below. r

26 6 INTRODUCTION ND MTHEMTICL CONCEPTS The magnitude and direction of are ( ) ( ) = + = 80.0 m m = m m = tan = tan = m 1 1 These results are identical to those given for C. + The scalar components of C are + C C ( ) = C cos 36.9 = m cos 36.9 = 80.0 m ( ) = Csin 36.9 = m sin 36.9 = 60.0 m These results are identical to the components given for. C 36.9 C C RESONING The force F and its two components form a right triangle. The hpotenuse is 8.3 newtons, and the side parallel to the + ais is F = 74.6 newtons. Therefore, we can use the trigonometric cosine and sine functions to determine the angle of F relative to the + ais and the component F of F along the + ais. SOLUTION a. The direction of F relative to the + ais is specified b the angle as b. The component of F along the + ais is F newtons = cos = 5.0 (1.5) 8.3 newtons ( ) = F sin 5.0 = 8.3 newtons sin 5.0 = 34.8 newtons (1.4) 38. RESONING ND SOLUTION a. = cos 30.0 = (750 units)(0.866) = 650 units = sin 30.0 = (750 units)(0.500) = 380 units b. ' = cos 40.0 = (750 units)(0.643) = 570 units

27 Chapter 1 Problems 7 ' = sin 40.0 = (750 units)(0.766) = 480 units 39. SSM WWW RESONING ND SOLUTION The force F can be first resolved into two components; the z component F z and the projection onto the - plane, F p as shown in the figure below on the left. ccording to that figure, F p = F sin 54.0 = (475 N) sin 54.0 = 384 N The projection onto the - plane, F p, can then be resolved into and components. z F F z F p 33.0 F F p F a. From the figure on the right, F = F p cos 33.0 = (384 N) cos 33.0 = 3 N b. lso from the figure on the right, F = F p sin 33.0 = (384 N) sin 33.0 = 09 N c. From the figure on the left, F z = F cos 54.0 = (475 N) cos 54.0 = 79 N 40. RESONING Using the component method for vector addition, we will find the component of the resultant force vector b adding the components of the individual vectors. Then we will find the component of the resultant vector b adding the components of the individual vectors. Once the and components of the resultant are known, we will use the Pthagorean theorem to find the magnitude of the resultant and trigonometr to find its direction. We will take east as the + direction and north as the + direction. SOLUTION The component of the resultant force F is

28 8 INTRODUCTION ND MTHEMTICL CONCEPTS F b g b g b g = 40 N cos N cos = 40 N cos F F The component of the resultant force F is F = 40 N sin N F F b g b g Using the Pthagorean theorem, we find that the magnitude of the resultant force is F = F + F = 40 N cos N sin N = 4790 N b g b g Using trigonometr, we find that the direction of the resultant force is Lb N g 40 N sin N 1 = tan M P = 67. south of east 40 N cos b g 41. SSM RESONING The individual displacements of the golf ball,,, and C are shown in the figure. Their resultant, R, is the displacement that would have been needed to "hole the ball" on the ver first putt. We will use the component method to find R. O Q N R C 0.0 o E SOLUTION The components of each displacement vector are given in the table below. Vector Components Components (5.0 m) cos 0 = 5.0 m (5.0 m) sin 0 = 0 C (.1 m) cos 0.0 =.0 m (0.50 m) cos 90.0 = 0 (.1 m) sin 0.0 = 0.7 m (0.50 m) sin 90.0 = 0.50 m R = + + C 7.0 m 1. m The resultant vector R has magnitude

29 Chapter 1 Problems 9 and the angle is R = ( 70. m) + ( 1. m) = 71. m = tan 1 1. m 7.0 m = 9.9 Thus, the required direction is 9.9 north of east. 4. RESONING We will use the scalar and components of the resultant vector to obtain its magnitude and direction. To obtain the component of the resultant we will add together the components of each of the vectors. To obtain the component of the resultant we will add together the components of each of the vectors. + SOLUTION The and components of the resultant vector R are R and R, respectivel. In terms of these components, the magnitude R and the directional angle (with respect to the ais) for the resultant are R 1 R= R + R and = tan R (1) The following table summarizes the components of the individual vectors shown in the drawing: Vector component component C 50.0 D + ( ) ( ) = 16.0 m cos 0.0 = 15.0 m = 16.0 m sin 0.0 = 5.47 m C D C D = 0 m = 11.0 m ( ) C ( ) = 1.0 m cos 35.0 = 9.83 m = 1.0 m sin 35.0 = 6.88 m ( ) D ( ) = 6.0 m cos50.0 = 16.7 m = 6.0 m sin 50.0 = 19.9 m R R = 15.0 m + 0 m 9.83 m m R = 5.47 m m 6.88 m 19.9 m = 8.1 m = 10.3 m

30 30 INTRODUCTION ND MTHEMTICL CONCEPTS Note that the component is zero, because points along the ais. Note also that the components C and C are both negative, since C points between the and aes. Finall, note that the component D is negative since D points below the + ais. Using equation (1), we find that ( ) ( ) R= R + R = 8.1 m m = 13 m R 10.3 m = tan = tan = 5 R 8.1 m Since both R and R are negative, the resultant points between the and aes. R R R RESONING ND SOLUTION The horizontal component of the resultant vector is R h = h + h +C h = (0.00 m) + (15.0 m) + (18.0 m) cos 35.0 = 9.7 m Similarl, the vertical component is R v = v + v + C v = (5.00 m) + (0.00 m) + ( 18.0 m) sin 35.0 = 5.3 m The magnitude of the resultant vector is The angle is obtained from R = R + R = 9. 7 m m = 30. m h v b g b g = tan 1 [(5.3 m)/(9.7 m)] = 10. _ 44. RESONING The magnitude of the resultant force in part (b) of the drawing is given b F + F cos F C cos 0.0 = F + F cos 0.0 where, since F sin 0.0 = F C sin 0.0, the components add to zero. s stated in the problem, the magnitude of the resultant force acting on the elephant in part (b) of the drawing is twice that in part (a); therefore, F + F cos 0.0 = F or F cos 0.0 = F SOLUTION Solving for the ratio F / F, we have

31 Chapter 1 Problems 31 F 1 = F (cos 0.0 ) = 0.53 _ 45. RESONING If we let the directions due east and due north be the positive directions, then the desired displacement has components E = ( 4.8 km) cos 4 = 3.57 km N = ( 4.8 km) sin 4 = 3.1 km while the actual displacement has components E = (.4 km) cos =.3 km N = (.4 km) sin = 0.90 km Therefore, to reach the research station, the research team must go and 3.57 km.3 km = 1.34 km, eastward 3.1 km 0.90 km =.31 km, northward SOLUTION a. From the Pthagorean theorem, we find that the magnitude of the displacement vector required to bring the team to the research station is N R = (1.34 km) + (.31 km) =.7 km R.31 km b. The angle is given b 1.34 km = tan 1.31 km = km 1 degrees, north of east _ 46. RESONING We are given that the vector sum of the three forces is zero, so F 1 + F + F 3 = 0 N. Since F 1 and F are known, F 3 can be found from the relation F 3 = (F 1 + F ). We will use the - and -components of this equation to find the magnitude and direction of F 3. SOLUTION The - and -components of the equation F 3 = ( F 1 + F ) are: -component F ( F F ) = + (1) 3 1

32 3 INTRODUCTION ND MTHEMTICL CONCEPTS -component F3 ( F1 F ) The table below gives the - and -components of F 1 and F : = + () Vector component component F 1 F F 1 = (1.0 N) sin 30.0 = 10.5 N F = N F 1 = (1.0 N) cos 30.0 = +18. N F = 0 N Substituting the values for F 1 and F into Equation (1) gives ( ) ( ) F3 = F1 + F = 10.5 N N = 4.5 N Substituting F 1 and F into Equation () gives ( ) ( ) F3 = F1 + F = N + 0 N = 18. N The magnitude of F 3 can now be obtained b emploing the Pthagorean theorem: ( ) ( ) F3 = F3 + F3 = 4.5 N N = 18.7 N The angle that F 3 makes with respect to the ais can be determined from the inverse tangent function (Equation 1.6), 1 F N = tan = tan = 76 F N 47. SSM RESONING Since the finish line is coincident with the starting line, the net displacement of the sailboat is zero. Hence the sum of the components of the displacement vectors of the individual legs must be zero. In the drawing in the tet, the directions to the right and upward are taken as positive. SOLUTION In the horizontal direction R h = h + h + C h + D h = 0 R h = (3.0 km) cos 40.0 (5.10 km) cos 35.0 (4.80 km) cos D cos = 0 D cos = 6.14 km. (1)

33 Chapter 1 Problems 33 In the vertical direction R v = v + v + C v + D v = 0. R v = (3.0 km) sin (5.10 km) sin 35.0 (4.80 km) sin 3.0 D sin = 0. Dividing () b (1) gives D sin = 3.11 km () Solving (1) gives tan = (3.11 km)/(6.14 km) or = 6.9 D = (6.14 km)/cos 6.9 = 6.88 km 48. RESONING We know that the three displacement vectors have a resultant of zero, so that + + C = 0. This means that the sum of the components of the vectors and the sum of the components of the vectors are separatel equal to zero. From these two equations we will be able to determine the magnitudes of vectors and C. The directions east and north are, respectivel, the + and + directions. SOLUTION Setting the sum of the components of the vectors and the sum of the components of the vectors separatel equal to zero, we have b g b g 1550 m cos sin C cos = 0 C b g b g 1550 m sin cos C sin = 0 C These two equations contain two unknown variables, and C. The can be solved simultaneousl to show that a. = 5550 m and b. C = 6160 m 49. RESONING The drawing shows the vectors,, and C. Since these vectors add to give a resultant that is zero, we can write that + + C = 0. This addition will be carried out b the component method. This means that the -component of this equation must be zero ( + + C = 0) and the -component must be zero ( + + C = 0). These two equations will allow us to find the magnitudes of and C. 145 units 35.0 C + (north) (east)

34 34 INTRODUCTION ND MTHEMTICL CONCEPTS SOLUTION The - and -components of,, and C are given in the table below. The plus and minus signs indicate whether the components point along the positive or negative aes. Vector Component Component (145 units) cos 35.0 = 119 units +(145 units) sin 35.0 = +83. units C + sin 65.0 = + (0.906) C sin 15.0 = C (0.59) + cos 65.0 = + (0.43) C cos 15.0 = C (0.966) + + C 119 units + (0.906) C (0.59) +83. units + (0.43) C (0.966) Setting the separate - and - components of + + C equal to zero gives -component ( 119 units) + (0.906) C (0.59) = 0 -component (+83. units) + (0.43) C (0.966) = 0 Solving these two equations simultaneousl, we find that a. = 178 units b. C = 164 units 50. RESONING The following table shows the components of the individual displacements and the components of the resultant. The directions due east and due north are taken as the positive directions. Displacement (1) () (3) (4) East/West Component 7.0 cm (3.0 cm) cos 35.0 = cm (8.0 cm) cos 55.0 = cm (35.0 cm) cos 63.0 = cm North/South Component 0 (3.0 cm) sin 35.0 = cm (8.0 cm) sin 55.0 =.94 cm (35.0 cm) sin 63.0 = cm Resultant cm 4.94 cm SOLUTION a. From the Pthagorean theorem, we find that the magnitude of the resultant displacement vector is R = (13.89 cm) + (4.94 cm) = 14.7 cm b. The angle is given b 4.94 cm cm 1 F 4.94 cm = H G I tan K J = 19. 6, south of west cm _ R

35 Chapter 1 Problems RESONING The shortest distance between the tree and the termite mound is equal to the magnitude of the chimpanzee's displacement r. SOLUTION a. From the Pthagorean theorem, we have r = (51 m) + (39 m) = 64 m 51 m r 39 m b. The angle is given b 1F 39 m = H G I tan K J = 37 south of east 51 m _ 5. RESONING ND SOLUTION The observer's ee, the treetop and a point on the tree trunk 1.83 m above the ground are the vertices of a right triangle. The height of the tree, H is H = 1.83 m + (3.0 m) tan 0.0 = 13.5 m _ 53. SSM RESONING The ostrich's velocit vector v and the desired components are shown in the figure at the right. The components of the velocit in the directions due west and due north are v W and v N, respectivel. The sine and cosine functions can be used to find the components. SOLUTION a. ccording to the definition of the sine function, we have for the vectors in the figure v N v 68.0 v W vn sin = or vn = vsin = (17.0 m/s) sin 68 = 15.8 m/s v b. Similarl, vw cos = or vw = vcos = v (17.0 m/s) cos 68.0 = 6.37 m/s _ 54. RESONING ND SOLUTION has the dimensions of [L], v has the dimensions of [L]/[T], and a has the dimensions of [L]/[T]. The equation under consideration is v n = a.

36 36 INTRODUCTION ND MTHEMTICL CONCEPTS The dimensions of the right hand side are [ ] L hand side are = [ T] n [ L] [ T] n n L T L L =, while the dimensions of the left T. The right side will equal the left side onl when n =. 55. RESONING ND SOLUTION The first three rows of the following table give the components of each of the three individual displacements. The fourth row gives the components of the resultant displacement. The directions due east and due north have been taken as the positive directions. Displacement East/West Component North/South Component 5 paces 0 C (4 paces) cos 30.0 = 36 paces 0 (4 paces) sin 30.0 = 1 paces 5 paces R = + + C 88 paces 46 paces a. Therefore, the magnitude of the displacement in the direction due north is 46 paces. b. Similarl, the magnitude of the displacement in the direction due west is 88 paces. _ 56. RESONING Using the component method, we find the components of the resultant R that are due east and due north. The magnitude and direction of the resultant R can be determined from its components, the Pthagorean theorem, and the tangent function. SOLUTION The first four rows of the table below give the components of the vectors,, C, and D. Note that east and north have been taken as the positive directions; hence vectors pointing due west and due south will appear with a negative sign. Vector East/West Component North/South Component +.00 km km C.50 km 0 D km R = + + C + D 0.50 km km

37 Chapter 1 Problems 37 The fifth row in the table gives the components of R. The magnitude of R is given b the Pthagorean theorem as R = ( 0.50 km) + ( km) = 090. km The angle that R makes with the direction due west is = tan km 0.50 km = 56 north of west R R N R E _ 57. SSM RESONING ND SOLUTION single rope must suppl the resultant of the two forces. Since the forces are perpendicular, the magnitude of the resultant can be found from the Pthagorean theorem. a. ppling the Pthagorean theorem, F = ( 475 N ) + (315 N) = N W 475 N b. The angle that the resultant makes with the westward direction is N 475 N F = H G I tan K J = F 315 N S Thus, the rope must make an angle of 33.6 south of west. _ 58. RESONING The force vector F points at an angle of above the + ais. Therefore, its and components are given b F = F cos and F = F sin. SOLUTION follows: a. The magnitude of the vector can be obtained from the component as F F = F sin 90 N or F = = sin sin = 370 N 5 b. Now that the magnitude of the vector is known, the component of the vector can be calculated as follows: b g F = F cos = 370 N cos 5 = + 30 N _