Physics Solution Manual for 1100 and 2101

# 9 south of west 33

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Unformatted text preview: f west ______________________________________________________________________________ 33. REASONING AND SOLUTION The figures below are scale diagrams of the forces drawn tail-to-head. The scale factor is shown in the figure. a. From the figure on the left, we see that FA − FB = 142 N, θ = 67° south of east . b. Similarly, from the figure on the right, FB − FA = 142 N, θ = 67° north of west . Chapter 1 Problems –FA = 90.0 N FA = 90.0 N θ 21 75° –FB = 135 N FA – FB FB = 135 N FB – FA Scale Factor: 20.0 N θ 75° _____________________________________________________________________________________________ 34. REASONING The magnitude of the x-component of the force vector is the product of the magnitude of the force times the cosine of the angle between the vector and the x axis. Since the x-component points in the +x direction, it is positive. Likewise, the magnitude of the y-component of the force vector is the product of the magnitude of the force times the sine of the angle between the vector and the x axis. Since the vector points 36.0º below the positive x axis, the y component of the vector points in the −y direction; thus, a minus sign must be assigned to the y-component to indicate this direction. SOLUTION The x and y scalar components are a. Fx = (575 newtons) cos 36.0° = 465 newtons b. Fy = –(575 newtons) sin 36.0° = –338 newtons 22 INTRODUCTION AND MATHEMATICAL CONCEPTS 35. SSM REASONING The ostrich's velocity vector v and the desired components are shown in the figure at the right. The components of the velocity in the directions due west and due north are v W and v N , respectively. The sine and cosine functions can be used to find the components. v vN 68.0 ° vW SOLUTION a. According to the definition of the sine function, we have for the vectors in the figure sin θ = vN v or vN = v sin θ = (17.0 m/s) sin 68° = 15.8 m/s or vW = v cos θ = (17.0 m/s) cos 68.0° = 6.37 m/s b. Similarly, cos θ = vW v _____________________________________________________________________________________________ 36. REASONING The triangle in the drawing is a right triangle. We know one of its angles is 30.0°, and the length of the hypotenuse is 8.6 m. Therefore, the sine and cosine functions can be used to find the magnitudes of Ax and Ay. The directions of these vectors can be found by examining the diagram. SOLUTION a. The magnitude Ax of the displacement vector Ax is related to the length of the hypotenuse and the 30.0° angle by the sine function (Equation 1.1). The drawing shows that the direction of Ax is due east. Ax = A sin 30.0° = ( 8.6 m ) sin 30.0° = 4.3 m, due east N E S 30.0° 8.6 m Ay b. In a similar manner, the magnitude Ay of Ay can be found by using the cosine function (Equation 1.2). Its direction is due south. Ay = A cos 30.0° = ( 8.6 m ) cos 30.0° = 7.4 m, due south W #1 A 90° #2 Net Ax ______________________________________________________________________________ 37. REASONING AND 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. Chapter 1 Problems Displacement East/West Component North/South Component A –52 paces 0 B C –(42 paces) cos 30.0° = –36 paces 0 (42 paces) sin 30.0° = 21 paces 25 paces R=A+B+C –88 paces 23 46 paces a. Therefore, the magnitude of the displacement in the direction due north is 46 paces . b. Similarly, the magnitude of the displacement in the direction due west is 88 paces . _____________________________________________________________________________________________ 38. REASONING AND SOLUTION a. From the Pythagorean theorem, we have F= (150 N) 2 + (130 N) 2 = 2.0 × 10 2 N 130 F NI= G NJ HK 150 130 N θ b. The angle θ is given by θ = tan −1 F 41° 150 N _____________________________________________________________________________________________ 39. SSM REASONING The x and y components of r are mutually perpendicular; therefore, the magnitude of r can be found using the Pythagorean theorem. The direction of r can be found using the definition of the tangent function. SOLUTION According to the Pythagorean theorem, we have r= x x 2 + y 2 = ( −125 m ) 2 + ( −184 m) 2 = 222 m θ y The angle θ is r 184 m = 55.8° 125 m θ = tan −1 ____________________________________________________________________________________________ 24 INTRODUCTION AND MATHEMATICAL CONCEPTS 40. REASONING Using trigonometry, we can determine the angle θ from the relation tan θ = Ay /Ax: SOLUTION Ay −1 12 m o = tan = 45 12 m Ax A 12 m o θ = tan −1 y = tan −1 = 35 Ax 17 m θ = tan −1 a. b. Ay −1 17 m o = tan = 55 Ax 12 m ______________________________________________________________________________ θ = tan −1 c. 41. REASONING Two vectors that are equal must have the same magnitude and direction. Equivalently, they must have identical x components and identic...
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