R 2 2k 1 1 r 4 i j k 2 2 r2 2i

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Unformatted text preview: , so k = ±5/3 (b) 6 = k v = |k | v = 2 v , so v = 3 34. If k v = 0 then |k | v = 0 so either k = 0 or v = 0; in the latter case, by (9) or (10), v = 0. −→ 35. (a) Choose two points on the line, √ example P1 (0, 2) and P2 (1, 5);√then P1√2 = 1, 3 is for P √ √ parallel to the line, 1, 3 = 10, so 1/ 10, 3/ 10 and −1/ 10, −3/ 10 are unit vectors parallel to the line. −→ 4) (b) Choose two points on the line, √ example P1 (0,√ and P2 (1, 3); then P1 P2 = 1, −1 is for √ √ √ parallel to the line, 1, −1 = 2 so 1/ 2, −1/ 2 and −1/ 2, 1/ 2 are unit vectors parallel to the line. (c) Pick any line that is perpendicular to the line x + y = 4, for example y = x/5; then P1 (0, 0) −→ 1 and P2 (5, 1) are on the line, so P1 P2 = 5, 1 is perpendicular to the line, so ± √ 5, 1 are 26 unit vectors perpendicular to the line. 36. (a) ±k (b) ±j (c) ±i 37. (a) the circle of radius 1 about the origin (b) the solid disk of radius 1 about the origin (c) all points outside the solid disk of radius 1 about the origin 38. (a) the circle of radius 1 about the tip of r0 (b) the solid disk of radius 1 about the tip of r0 (c) all points outside the solid disk of radius 1 about the tip of r0 39. (a) the (hollow) sphere of radius 1 about the origin (b) the solid ball of radius 1 about the origin (c) all points outside the solid ball of radius 1 about the origin 40. The sum of the distances between (x, y ) and the points (x1 , y1 ), (x2 , y2 ) is the constant k , so the set consists of all points on the ellipse with foci at (x1 , y1 ) and (x2 , y2 ), and major axis of length k. Exercise Set 13.2 456 41. Since φ = π/2, from (13) we get F1 + F2 2 = F1 2 + F2 2 = 3600 + 900, √ 30 F2 sin φ = √ , α ≈ 26.57◦ , θ = α ≈ 26.57◦ . so F1 + F2 = 30 5 lb, and sin α = F1 + F2 30 5 1 F2 cos φ = 14,400 + 10,000 + 2(120)(100) = 36,400, so 2 √ √ 100 F2 53 sin φ = √ sin 60◦ = √ , α ≈ 27.16◦ , F1 + F2 = 20 91 N, sin α = F1 + F2 20 91 2 91 θ = α ≈ 27.16◦ . √ 3 2 2 2 , 43. F1 + F2 = F1 + F2 + 2 F1 F2 cos φ = 160,000 + 160,000 − 2(400)(400) 2 400 F2 1 sin φ ≈ , α = 75.00◦ , so F1 + F2 ≈ 207.06 N, and sin α = F1 + F2 207.06 2 θ = α − 30◦ = 45.00◦ . 42. F1 + F2 2 44. F1 + F2 + 2 F1 F2 cos φ = 16 + 4 + 2(4)(2) cos 77◦ , so 2 F2 sin φ = sin 77◦ , α ≈ 23.64◦ , θ = α − 27◦ ≈ −3.36◦ . ≈ 4.86 lb, and sin α = F1 + F2 4.86 F 1 + F2 2 = F1 = F1 2 2 + F2 + F2 2 + 2 F1 2 45. Let F1 , F2 , F3 be the forces in the diagram with magnitudes 40, 50, 75 respectively. Then F1 + F2 + F3 = (F1 + F2 ) + F3 . Following the examples, F1 + F2 has magnitude 45.83 N and makes an angle 79.11◦ with the positive x-axis. Then (F1 + F2 )+ F3 2 ≈ 45.832 +752 +2(45.83)(75) cos 79.11◦ , so F1 + F2 + F3 has magnitude ≈ 94.995 N and makes an angle θ = α ≈ 28.28◦ with the positive x-axis. 46. Let F1 , F2 , F3 be the forces in the diagram with magnitudes 150, 200, 100 respectively. Then F1 + F2 + F3 = (F1 + F2 ) + F3 . Following the examples, F1 + F2 has magnitude 279.34 N and makes an angle 91.24◦ with the positive x-axis. Then F1 + F2 + F3 2 ≈ 279.342 + 1002 + 2(279.34)(100) cos(270 − 91.24)◦ , and F1 + F2 + F3 has magnitude ≈ 179.37 N and makes an angle 91.94◦ with the positive x-axis. be 47. Let F1 , F2 √ the forces in the diagram with magnitudes 8, 10 respectively. Then F1 + F2 has √ magnitude 82 + 102 + 2 · 8 · 10 cos 120◦ = 2 21 ≈ 9.165 lb, and makes an angle F1 sin 120 ≈ 109.11◦ with the positive x-axis, so F has magnitude 9.165 lb and 60◦ + sin−1 F1 + F2 makes an angle −70.89◦ with the positive x-axis. 48. √ F1 + F2 = 1202 + 1502 + 2 · 120 · 150 cos 75◦ = 214.98 N and makes an angle 92.63◦ with the positive x-axis, and F1 + F2 + F3 = 232.90 N and makes an angle 67.23◦ with the positive x-axis, hence F has magnitude 232.90 N and makes an angle −112.77◦ with the positive x-axis. 49. F1 + F2 + F = 0, where F has magnitude 250 and makes an angle −90◦ with the positive x-axis. Thus F1 + F2 2 = F1 2 + F2 2 + 2 F1 F2 cos 105◦ = 2502 and √ F2 2 F2 −1 ◦ ◦ sin 105 , so ≈ 0.9659, F2 ≈ 183.02 lb, 45 = α = sin 250 2 250 F1 2 + 2(183.02)(−0.2588) F1 + (183.02)2 = 62,500, F1 = 224.13 lb. √ 50. Similar to Exercise 49, F1 = 100 3 N, F2 = 100 N 51. (a) c1 v1 + c2 v2 = (2c1 + 4c2 ) i + (−c1 + 2c2 ) j = −4j, so 2c1 + 4c2 = 0 and −c1 + 2c2 = −4 which gives c1 = 2, c2 = −1. (b) c1 v1 + c2 v2 = c1 − 2c2 , −3c1 + 6c2 = 3, 5 , so c1 − 2c2 = 3 and −3c1 + 6c2 = 5 which has no solution. 457 Chapter 13 52. (a) Equate corresponding components to get the system of equations c1 + 3c2 = −1, 2c2 + c3 = 1, and c1 + c3 = 5. Solve to get c1 = 2, c2 = −1, and c3 = 3. (b) Equate corresponding components to get the system of equations c1 + 3c2 + 4c3 = 2, −c1 − c3 = 1, and c2 + c3 = −1. From the second and third equations, c1 = −1 − c3 and c2 = −1 − c3 ; substitute these into the first equation to get −4 = 2, which is nonsense so the system has no solution. 53. Place u and v tip to tail so th...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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