Solutions Manual University Physics With Modern Physics 14th Edition Young 35g2a

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UNITS, PHYSICAL QUANTITIES, AND VECTORS

1.1.

1

IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.

⎛ 5280 ft ⎞⎛ 12 in. ⎞⎛ 2.54 cm ⎞⎛ 1 m ⎞⎛ 1 km ⎞ EXECUTE: (a) 1.00 mi = (1.00 mi) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ = 1.61 km ⎝ 1 mi ⎠⎝ 1 ft ⎠⎝ 1 in. ⎠⎝ 102 cm ⎠⎝ 103 m ⎠

1.2.

⎛ 103 m ⎞⎛ 102 cm ⎞ ⎛ 1 in. ⎞⎛ 1 ft ⎞ 3 (b) 1.00 km = (1.00 km) ⎜ ⎟⎜ ⎟ = 3.28 × 10 ft ⎜ 1 km ⎟⎜ ⎟⎜ 1 m ⎟⎟ ⎝⎜ 2.54 cm ⎠⎝ 12 in . ⎠ ⎝ ⎠⎝ ⎠ EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. IDENTIFY: Convert volume units from L to in.3. SET UP: 1 L = 1000 cm3. 1 in. = 2.54 cm

⎛ 1000 cm3 ⎞ ⎛ 1 in. ⎞3 3 EXECUTE: 0.473 L × ⎜ ⎟⎟ × ⎜ ⎟ = 28.9 in. . ⎜ 1L 2 54 cm . ⎠ ⎝ ⎠ ⎝ EVALUATE: 1 in.3 is greater than 1 cm3 , so the volume in in.3 is a smaller number than the volume in 1.3.

cm3 , which is 473 cm3. IDENTIFY: We know the speed of light in m/s. t = d/v. Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is v = 3.00 × 108 m/s. 1 ft = 0.3048 m. 1 s = 109 ns. 0.3048 m = 1.02 × 10−9 s = 1.02 ns EXECUTE: t = 3.00 × 108 m/s EVALUATE: In 1.00 s light travels 3.00 × 108 m = 3.00 × 105 km = 1.86 × 105 mi.

1.4.

IDENTIFY: Convert the units from g to kg and from cm3 to m3. SET UP: 1 kg = 1000 g. 1 m = 100 cm. EXECUTE: 19.3

3

⎛ 1 kg ⎞ ⎛ 100 cm ⎞ kg × × = 1.93 × 104 3 3 ⎜ 1000 g ⎟ ⎜ 1 m ⎟ cm ⎝ m ⎠ ⎝ ⎠ g

EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3. 1.5.

IDENTIFY: Convert volume units from in.3 to L. SET UP: 1 L = 1000 cm3. 1 in. = 2.54 cm. EXECUTE: (327 in.3 ) × (2.54 cm/in.)3 × (1 L/1000 cm3 ) = 5.36 L EVALUATE: The volume is 5360 cm3. 1 cm3 is less than 1 in.3 , so the volume in cm3 is a larger number

than the volume in in.3.

1-1

1-2 1.6.

Chapter 1 IDENTIFY: Convert ft 2 to m 2 and then to hectares. SET UP: 1.00 hectare = 1.00 × 104 m 2 . 1 ft = 0.3048 m.

⎛ 43,600 ft 2 ⎞ ⎛ 0.3048 m ⎞2 ⎛ 1.00 hectare ⎞ EXECUTE: The area is (12.0 acres) ⎜ = 4.86 hectares. ⎟⎟ ⎜ ⎟ ⎜ 4 2⎟ ⎜ ⎝ 1 acre ⎠ ⎝ 1.00 ft ⎠ ⎝ 1.00 × 10 m ⎠ EVALUATE: Since 1 ft = 0.3048 m, 1 ft 2 = (0.3048) 2 m 2 . 1.7.

IDENTIFY: Convert seconds to years. 1 gigasecond is a billion seconds. SET UP: 1 gigasecond = 1 × 109 s. 1 day = 24 h. 1 h = 3600 s.

⎛ 1 h ⎞⎛ 1 day ⎞ ⎛ 1 y ⎞ EXECUTE: 1.00 gigasecond = (1.00 × 109 s) ⎜ ⎟ = 31.7 y. ⎟⎜ ⎟⎜ ⎝ 3600 s ⎠⎝ 24 h ⎠ ⎝ 365 days ⎠ EVALUATE: The conversion 1 y = 3.156 × 107 s assumes 1 y = 365.24 d, which is the average for one 1.8.

1.9.

extra day every four years, in leap years. The problem says instead to assume a 365-day year. IDENTIFY: Apply the given conversion factors. SET UP: 1 furlong = 0.1250 mi and 1 fortnight = 14 days. 1 day = 24 h.

⎛ 0.125 mi ⎞⎛ 1 fortnight ⎞ ⎛ 1 day ⎞ EXECUTE: (180,000 furlongs/fortnight) ⎜ ⎟⎜ ⎟⎜ ⎟ = 67 mi/h ⎝ 1 furlong ⎠⎝ 14 days ⎠ ⎝ 24 h ⎠ EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number. IDENTIFY: Convert miles/gallon to km/L. SET UP: 1 mi = 1.609 km. 1 gallon = 3.788 L. ⎛ 1.609 km ⎞⎛ 1 gallon ⎞ EXECUTE: (a) 55.0 miles/gallon = (55.0 miles/gallon) ⎜ ⎟⎜ ⎟ = 23.4 km/L. ⎝ 1 mi ⎠⎝ 3.788 L ⎠ 1500 km 64.1 L = 64.1 L. = 1.4 tanks. 23.4 km/L 45 L/tank EVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly half a mile and there are roughly 4 liters in a

(b) The volume of gas required is

gallon, so 1 mi/gal ∼ 24 km/L, which is roughly our result. 1.10.

IDENTIFY: Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g = 1 kg.

ft ⎛ mi ⎞ ⎛ 1 h ⎞⎛ 5280 ft ⎞ EXECUTE: (a) ⎜ 60 ⎟ ⎜ ⎟⎜ ⎟ = 88 h ⎠ ⎝ 3600 s ⎠⎝ 1 mi ⎠ s ⎝ m ⎛ ft ⎞ ⎛ 30.48 cm ⎞ ⎛ 1 m ⎞ (b) ⎜ 32 2 ⎟ ⎜ ⎟⎜ ⎟ = 9.8 2 s ⎝ s ⎠ ⎝ 1 ft ⎠ ⎝ 100 cm ⎠ 3

g ⎞⎛ 100 cm ⎞ ⎛ 1 kg ⎞ ⎛ 3 kg (c) ⎜1.0 3 ⎟⎜ ⎟ = 10 3 ⎟ ⎜ m ⎝ cm ⎠⎝ 1 m ⎠ ⎝ 1000 g ⎠

EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The relation 1.11.

32 ft/s 2 = 9.8 m/s 2 is accurate to only two significant figures. IDENTIFY: We know the density and mass; thus we can find the volume using the relation density = mass/volume = m/V . The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: Density = 19.5 g/cm3 and mcritical = 60.0 kg. For a sphere V = 43 π r 3.

⎛ 60.0 kg ⎞ ⎛ 1000 g ⎞ 3 EXECUTE: V = mcritical /density = ⎜⎜ ⎜ ⎟ = 3080 cm . 3⎟ ⎟ 1 . 0 kg ⎠ ⎝ 19.5 g/cm ⎠ ⎝

Units, Physical Quantities, and Vectors

1-3

3V 3 3 = (3080 cm3 ) = 9.0 cm. 4π 4π EVALUATE: The density is very large, so the 130-pound sphere is small in size. IDENTIFY: Convert units. SET UP: We know the equalities 1 mg = 10−3 g, 1 µg 10−6 g, and 1 kg = 103 g. r=3

1.12.

⎛ 10−3 g ⎞⎛ 1 μ g ⎞ 5 EXECUTE: (a) (410 mg/day) ⎜ ⎟⎜ −6 ⎟ = 4.10 × 10 μ g/day. 1 mg 10 g ⎠ ⎝ ⎠⎝ ⎛ 10−3 g ⎞ = 0.900 g. (b) (12 mg/kg)(75 kg) = (900 mg) ⎜ ⎜ 1 mg ⎟⎟ ⎝ ⎠ ⎛ 10−3 g ⎞ −3 (c) The mass of each tablet is (2.0 mg) ⎜ ⎟ = 2.0 × 10 g. The number of tablets required each day is 1 mg ⎝ ⎠ the number of grams recommended per day divided by the number of grams per tablet: 0.0030 g/day = 1.5 tablet/day. Take 2 tablets each day. 2.0 × 10−3 g/tablet

1.13.

⎛ 1 mg ⎞ (d) (0.000070 g/day) ⎜⎜ −3 ⎟⎟ = 0.070 mg/day. ⎝ 10 g ⎠ EVALUATE: Quantities in medicine and nutrition are frequently expressed in a wide variety of units. IDENTIFY: Model the bacteria as spheres. Use the diameter to find the radius, then find the volume and surface area using the radius. SET UP: From Appendix B, the volume V of a sphere in of its radius is V = 43 π r 3 while its surface area A is A = 4π r 2 . The radius is one-half the diameter or r = d/2 = 1.0 μ m. Finally, the necessary equalities for this problem are: 1 μ m = 10−6 m; 1 cm = 10−2 m; and 1 mm = 10−3 m. 3

⎛ 10−6 m ⎞ ⎛ 1 cm ⎞3 EXECUTE: V = 43 π r 3 = 43 π (1.0 μ m)3 ⎜ = 4.2 × 10−12 cm3 and ⎜ 1 μ m ⎟⎟ ⎜⎝ 10−2 m ⎟⎠ ⎝ ⎠ 2

1.14.

⎛ 10−6 m ⎞ ⎛ 1 mm ⎞2 A = 4π r 2 = 4π (1.0 μ m)2 ⎜ = 1.3 × 10−5 mm 2 ⎜ 1 μ m ⎟⎟ ⎜⎝ 10−3 m ⎟⎠ ⎝ ⎠ EVALUATE: On a human scale, the results are extremely small. This is reasonable because bacteria are not visible without a microscope. IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of the decimal that matters. SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures. EXECUTE: (a) (12 mm) × (5.98 mm) = 72 mm 2 (two significant figures) 5.98 mm = 0.50 (also two significant figures) 12 mm (c) 36 mm (to the nearest millimeter) (d) 6 mm (e) 2.0 (two significant figures) EVALUATE: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to the nearest mm. IDENTIFY: Use your calculator to display π × 107. Compare that number to the number of seconds in a year. SET UP: 1 yr = 365.24 days, 1 day = 24 h, and 1 h = 3600 s. (b)

1.15.

1-4

1.16.

1.17.

1.18.

Chapter 1

⎛ 24 h ⎞ ⎛ 3600 s ⎞ 7 7 7 EXECUTE: (365.24 days/1 yr) ⎜ ⎟⎜ ⎟ = 3.15567…× 10 s; π × 10 s = 3.14159…× 10 s ⎝ 1 day ⎠ ⎝ 1 h ⎠ The approximate expression is accurate to two significant figures. The percent error is 0.45%. EVALUATE: The close agreement is a numerical accident. IDENTIFY: To asses the accuracy of the approximations, we must convert them to decimals. SET UP: Use a calculator to calculate the decimal equivalent of each fraction and then round the numeral to the specified number of significant figures. Compare to π rounded to the same number of significant figures. EXECUTE: (a) 22/7 = 3.14286 (b) 355/113 = 3.14159 (c) The exact value of π rounded to six significant figures is 3.14159. EVALUATE: We see that 355/113 is a much better approximation to π than is 22/7. IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express 200 months in years. SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs.1 in. = 2.54 cm. 1 y = 12 months. EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man. ⎛ 1 in. ⎞ 3 (b) 200 m = (2.00 × 104 cm) ⎜ ⎟ = 7.9 × 10 inches. This is much greater than the height of a person. ⎝ 2.54 cm ⎠ (c) 200 cm = 2.00 m = 79 inches = 6.6 ft. Some people are this tall, but not an ordinary man. (d) 200 mm = 0.200 m = 7.9 inches. This is much too short. ⎛ 1y ⎞ (e) 200 months = (200 mon) ⎜ ⎟ = 17 y. This is the age of a teenager; a middle-aged man is much ⎝ 12 mon ⎠ older than this. EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed. IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the number of gallons. SET UP: Estimate 3 × 108 people, so 2 × 108 cars. EXECUTE: (Number of cars × miles/car day)/(mi/gal) = gallons/day (2 × 108 cars × 10000 mi/yr/car × 1 yr/365 days)/(20 mi/gal) = 3 × 108 gal/day

1.19.

EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S. IDENTIFY: Estimate the number of blinks per minute. Convert minutes to years. Estimate the typical lifetime in years. SET UP: Estimate that we blink 10 times per minute.1 y = 365 days. 1 day = 24 h, 1 h = 60 min. Use 80

years for the lifetime.

⎛ 60 min ⎞ ⎛ 24 h ⎞⎛ 365 days ⎞ 8 EXECUTE: The number of blinks is (10 per min) ⎜ ⎟⎜ ⎟ (80 y/lifetime) = 4 × 10 ⎟⎜ ⎝ 1 h ⎠ ⎝ 1 day ⎠⎝ 1 y ⎠ EVALUATE: Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is surely accurate to a power of 10. 1.20.

IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and cm3 to m3 to

find the volume in m3 breathed in a year.

⎛ 24 h ⎞⎛ 60 min ⎞ 5 2 SET UP: Assume 10 breaths/min. 1 y = (365 d) ⎜ ⎟⎜ ⎟ = 5.3 × 10 min. 10 cm = 1 m so ⎝ 1 d ⎠⎝ 1 h ⎠ 106 cm3 = 1 m3. The volume of a sphere is V = 43 π r 3 = 16 π d 3 , where r is the radius and d is the diameter. Don’t forget to for four astronauts.

⎛ 5.3 × 105 min ⎞ 4 3 EXECUTE: (a) The volume is (4)(10 breaths/min)(500 × 10−6 m3 ) ⎜ ⎟⎟ = 1× 10 m /yr. ⎜ 1y ⎝ ⎠

Units, Physical Quantities, and Vectors 1/3

⎛ 6V ⎞ (b) d = ⎜ ⎟ ⎝ π ⎠

1.21.

1-5

1/3

⎛ 6[1 × 104 m3 ] ⎞ =⎜ ⎟⎟ ⎜ π ⎝ ⎠

= 27 m

EVALUATE: Our estimate assumes that each cm3 of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required. IDENTIFY: Estimation problem. SET UP: Estimate that the pile is 18 in.× 18 in.× 5 ft 8 in.. Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value. EXECUTE: The volume of gold in the pile is V = 18 in.× 18 in.× 68 in. = 22,000 in.3. Convert to cm3:

V = 22,000 in.3 (1000 cm3/61.02 in.3 ) = 3.6 × 105 cm3 .

The density of gold is 19.3 g/cm3 , so the mass of this volume of gold is m = (19.3 g/cm3 )(3.6 × 105 cm3 ) = 7 × 106 g.

The monetary value of one gram is $10, so the gold has a value of ($10/gram)(7 × 106 grams) = $7 × 107 ,

1.22.

or about $100 × 106 (one hundred million dollars). EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable. IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years. ⎛ 60 min ⎞ ⎛ 24 h ⎞⎛ 365 days ⎞⎛ 80 yr ⎞ 9 EXECUTE: N beats = (75 beats/min) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3 × 10 beats/lifespan ⎟⎜ yr ⎝ 1 h ⎠ ⎝ 1 day ⎠⎝ ⎠⎝ lifespan ⎠ 9 ⎛ 1 L ⎞⎛ 1 gal ⎞ ⎛ 3 × 10 beats ⎞ 7 Vblood = (50 cm3/beat) ⎜ ⎜ ⎟ = 4 × 10 gal/lifespan ⎟⎜ ⎟ ⎝ 1000 cm3 ⎠⎝ 3.788 L ⎠ ⎝⎜ lifespan ⎠⎟

1.23.

EVALUATE: This is a very large volume. IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in m3. Convert

m3 to L. SET UP: Estimate the diameter of a drop to be d = 2 mm. The volume of a spherical drop is V = 43 π r 3 = 16 π d 3. 103 cm3 = 1 L. EXECUTE: V = 16 π (0.2 cm)3 = 4 × 10−3 cm3. The number of drops in 1.0 L is

1.24.

1000 cm3 4 × 10−3 cm3

= 2 × 105

EVALUATE: Since V ∼ d 3 , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the number of drops is off by a factor of 8. IDENTIFY: Draw the vector addition diagram to scale. G G SET UP: The two vectors A and B are specified in the figure that accompanies the problem. G G G EXECUTE: (a) The diagram for R = A + B is given in Figure 1.24a. Measuring the length and angle of G R gives R = 9.0 m and an angle of θ = 34°. G G G G (b) The diagram for E = A − B is given in Figure 1.24b. Measuring the length and angle of E gives D = 22 m and an angle of θ = 250°. G G G G G G (c) − A − B = −( A + B ), so − A − B has a magnitude of 9.0 m (the same as A + B ) and an angle with the G G + x axis of 214° (opposite to the direction of A + B ). G G G G G G (d) B − A = −( A − B ), so B − A has a magnitude of 22 m and an angle with the + x axis of 70° (opposite G G to the direction of A − B ). G G EVALUATE: The vector − A is equal in magnitude and opposite in direction to the vector A.

1-6

Chapter 1

Figure 1.24 1.25.

IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant displacement is the single vector that points from the starting point to the stopping point. G G G G SET UP: Call the three displacements A, B, and C . The resultant displacement R is given by G G G G R = A + B + C. G EXECUTE: The vector addition diagram is given in Figure 1.25. Careful measurement gives that R is 7.8 km, 38D north of east. EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the individual displacements, 2.6 km + 4.0 km + 3.1 km.

Figure 1.25 1.26.

IDENTIFY: Since she returns to the starting point, the vector sum of the four displacements must be zero. G G G G SET UP: Call the three given displacements A, B, and C , and call the fourth displacement D. G G G G A + B + C + D = 0. G EXECUTE: The vector addition diagram is sketched in Figure 1.26. Careful measurement gives that D is 144 m, 41° south of west. G G G G EVALUATE: D is equal in magnitude and opposite in direction to the sum A + B + C .

Figure 1.26

Units, Physical Quantities, and Vectors 1.27.

1-7

G G IDENTIFY: For each vector V , use that Vx = V cosθ and V y = V sin θ , when θ is the angle V makes with the + x axis, measured counterclockwise from the axis. G G G G SET UP: For A, θ = 270.0°. For B, θ = 60.0°. For C , θ = 205.0°. For D, θ = 143.0°. EXECUTE: Ax = 0, Ay = −8.00 m. Bx = 7.50 m, B y = 13.0 m. C x = −10.9 m, C y = −5.07 m.

Dx = −7.99 m, Dy = 6.02 m. 1.28.

EVALUATE: The signs of the components correspond to the quadrant in which the vector lies. Ay IDENTIFY: tan θ = , for θ measured counterclockwise from the + x -axis. Ax G G SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies. EXECUTE: (a) tan θ = (b) tan θ = (c) tan θ = (d) tan θ =

1.29.

Ay Ax Ay Ax Ay Ax Ay Ax

=

−1.00 m = −0.500. θ = tan −1 (−0.500) = 360° − 26.6° = 333°. 2.00 m

=

1.00 m = 0.500. θ = tan −1 (0.500) = 26.6°. 2.00 m

=

1.00 m = −0.500. θ = tan −1 (−0.500) = 180° − 26.6° = 153°. −2.00 m

=

−1.00 m = 0.500. θ = tan −1 (0.500) = 180° + 26.6° = 207° −2.00 m

EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value of θ . IDENTIFY: Given the direction and one component of a vector, find the other component and the magnitude. SET UP: Use the tangent of the given angle and the definition of vector magnitude. A EXECUTE: (a) tan 32.0° = x Ay

Ax = (9.60 m)tan 32.0° = 6.00 m. Ax = −6.00 m. (b) A = Ax2 + Ay2 = 11.3 m. 1.30.

EVALUATE: The magnitude is greater than either of the components. IDENTIFY: Given the direction and one component of a vector, find the other component and the magnitude. SET UP: Use the tangent of the given angle and the definition of vector magnitude. A EXECUTE: (a) tan 34.0° = x Ay Ay =

Ax tan 34.0°

=

16.0 m = 23.72 m tan 34.0°

Ay = −23.7 m. (b) A = Ax2 + Ay2 = 28.6 m. 1.31.

EVALUATE: The magnitude is greater than either of the components. G G G IDENTIFY: If C = A + B, then C x = Ax + Bx and C y = Ay + B y . Use C x and C y to find the magnitude and G direction of C .

1-8

Chapter 1 SET UP: From Figure E1.24 in the textbook, Ax = 0, Ay = −8.00 m and Bx = + B sin 30.0° = 7.50 m,

B y = + B cos30.0° = 13.0 m. G G G EXECUTE: (a) C = A + B so C x = Ax + Bx = 7.50 m and C y = Ay + B y = +5.00 m. C = 9.01 m. Cy

5.00 m = and θ = 33.7°. C x 7.50 m G G G G G G (b) B + A = A + B, so B + A has magnitude 9.01 m and direction specified by 33.7°. G G G (c) D = A − B so Dx = Ax − Bx = −7.50 m and D y = Ay − B y = −21.0 m. D = 22.3 m. tan θ =

tan φ =

Dy Dx

=

G −21.0 m and φ = 70.3°. D is in the 3rd quadrant and the angle θ counterclockwise from the −7.50 m

+ x axis is 180° + 70.3° = 250.3°. G G G G G G (d) B − A = −( A − B ), so B − A has magnitude 22.3 m and direction specified by θ = 70.3°. 1.32.

EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.24. IDENTIFY: Find the vector sum of the three given displacements. SET UP: Use coordinates for which + x is east and + y is north. The driver’s vector displacements are: K K K A = 2.6 km, 0° of north; B = 4.0 km, 0° of east; C = 3.1 km, 45° north of east. EXECUTE: Rx = Ax + Bx + C x = 0 + 4.0 km + (3.1 km)cos(45°) = 6.2 km; Ry = Ay + By + C y =

2.6 km + 0 + (3.1 km)(sin 45°) = 4.8 km; R = Rx2 + Ry2 = 7.8 km; θ = tan −1[(4.8 km)/(6.2 km)] = 38°; K R = 7.8 km, 38° north of east. This result is confirmed by the sketch in Figure 1.32. G EVALUATE: Both Rx and R y are positive and R is in the first quadrant.

1.33.

Figure 1.32 IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are asked to find their sum. SET UP:

A = 3.25 km B = 2.20 km C = 1.50 km

Figure 1.33a

Units, Physical Quantities, and Vectors

1-9

G G G Select a coordinate system where + x is east and + y is north. Let A, B, and C be the three G G G G G displacements of the professor. Then the resultant displacement R is given by R = A + B + C . By the method of components, Rx = Ax + Bx + C x and Ry = Ay + By + C y . Find the x and y components of each vector; add them to find the components of the resultant. Then the magnitude and direction of the resultant can be found from its x and y components that we have calculated. As always it is essential to draw a sketch. EXECUTE:

Ax = 0, Ay = +3.25 km Bx = −2.20 km, By = 0 Cx = 0, C y = −1.50 km Rx = Ax + Bx + C x Rx = 0 − 2.20 km + 0 = −2.20 km

Ry = Ay + By + C y Ry = 3.25 km + 0 − 1.50 km = 1.75 km

Figure 1.33b

R = Rx2 + Ry2 = ( −2.20 km) 2 + (1.75 km) 2 R = 2.81 km R 1.75 km = −0.800 tan θ = y = Rx −2.20 km θ = 141.5° Figure 1.33c

The angle θ measured counterclockwise from the + x -axis. In of com directions, the resultant displacement is 38.5° N of W. G EVALUATE: Rx < 0 and Ry > 0, so R is in the 2nd quadrant. This agrees with the vector addition diagram. 1.34.

IDENTIFY: Use A =

Ax + Ay and tan θ = 2

2

Ay Ax

to calculate the magnitude and direction of each of the

given vectors. G G SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies. EXECUTE: (a) (b)

⎛ 5.20 ⎞ (−8.60 cm)2 + (5.20 cm)2 = 10.0 cm, arctan ⎜ ⎟ = 148.8° (which is180° − 31.2° ). ⎝ −8.60 ⎠

⎛ −2.45 ⎞ (−9.7 m) 2 + (−2.45 m) 2 = 10.0 m, arctan ⎜ ⎟ = 14° + 180° = 194°. ⎝ −9.7 ⎠

⎛ −2.7 ⎞ (7.75 km) 2 + (−2.70 km)2 = 8.21 km, arctan ⎜ ⎟ = 340.8° (which is 360° − 19.2° ). ⎝ 7.75 ⎠ EVALUATE: In each case the angle is measured counterclockwise from the + x axis. Our results for θ agree with our sketches.

(c)

1-10 1.35.

Chapter 1

G G G G IDENTIFY: Vector addition problem. A − B = A + (− B ). G G SET UP: Find the x- and y-components of A and B. Then the x- and y-components of the vector sum are G G calculated from the x- and y-components of A and B. EXECUTE: Ax = A cos(60.0°) Ax = (2.80 cm)cos(60.0°) = +1.40 cm Ay = A sin (60.0°)

Ay = (2.80 cm)sin (60.0°) = +2.425 cm Bx = B cos(−60.0°) Bx = (1.90 cm)cos(−60.0°) = +0.95 cm B y = B sin ( −60.0°) B y = (1.90 cm)sin (−60.0°) = −1.645 cm Note that the signs of the components correspond to the directions of the component vectors. Figure 1.35a G G G (a) Now let R = A + B. Rx = Ax + Bx = +1.40 cm + 0.95 cm = +2.35 cm.

Ry = Ay + By = +2.425 cm − 1.645 cm = +0.78 cm. R = Rx2 + Ry2 = (2.35 cm)2 + (0.78 cm) 2 R = 2.48 cm R y +0.78 cm tan θ = = = +0.3319 Rx +2.35 cm

θ = 18.4° Figure 1.35b

G G G EVALUATE: The vector addition diagram for R = A + B is G R is in the 1st quadrant, with | Ry | < | Rx | , in agreement with our calculation.

Figure 1.35c

Units, Physical Quantities, and Vectors

G G G (b) EXECUTE: Now let R = A − B. Rx = Ax − Bx = +1.40 cm − 0.95 cm = +0.45 cm. R y = Ay − By = +2.425 cm + 1.645 cm = +4.070 cm. R = Rx2 + R y2 = (0.45 cm)2 + (4.070 cm) 2 R = 4.09 cm Ry 4.070 cm tan θ = = = +9.044 0.45 cm Rx

θ = 83.7°

Figure 1.35d

G G G EVALUATE: The vector addition diagram for R = A + ( − B ) is G R is in the 1st quadrant, with | Rx | < | R y |, in agreement with our calculation.

Figure 1.35e (c) EXECUTE:

G G G G B − A = −( A − B ) G G G G B − A and A − B are equal in magnitude and opposite in direction. R = 4.09 cm and θ = 83.7° + 180° = 264°

Figure 1.35f

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