Fisica conceptos y aplicaciones de tppens (solucionario)

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Fisica conceptos y aplicaciones de tppens (solucionario)

  1. 1. Physics, 6th Edition Chapter 2. Technical Mathematics Signed Numbers 2-1. +7 2-8. -17 2-15. +2 2-22. +12 2-2. +4 2-9. +6 2-16. -2 2-23. +8 2-3. +2 2-10. -32 2-17. -4 2-24. -4 2-4. -2 2-11. -36 2-18. -3 2-25. 0 2-5. -10 2-12. +24 2-19. +2 2-26. +220 2-6. -33 2-13. -48 2-20. -4 2-27. +32 2-7. -5 2-14. +144 2-21. -3 2-28. -32 2-29. (a) -60C; (b) -170C; (c) 360C 2-30. L = 2 mm[(-300C) – (-50C)] = 2 mm(-25) = -50 mm; Decrease in length. Algebra Review 2-31. x = (2) + (-3) + (-2) = -3; x = -3 2-32. x = (2) – (-3) – (-2) = +7; x = +7 2-33. x = (-3) + (-2) - (+2) = -7; x = -7 2-34. x = -3[(2) – (-2)] = -3(2 + 2) = -12; x = -12 2-35. x  b  c 3  ( 2) 3  2 1 2  ( 3) 2  3 1   ; x   2-36. x   ; x a 2 2 2 2 2 2 2-37. x = (-3)2 – (-2)2 = 9 – 4 = 5; x = 5 2-39. x  2-38. x  b ( 3) 3 3   ; x ac (2)( 2) 4 4 2 1 4 2  ( 2)  (4); x  2-40. x = (2)2 + (-3)2 + (-2)2; ( 3)( 2) 3 3 2-41. x  a 2  b2  c2  17 2-43. Solve for x: 2ax – b = c; x = 17 2-42. x = (2)(-3)[(-2) – (+2)]2; x = -6(-4)2 = -96 2ax = b + c; 1 x b  c 3  2 5  ; x 2a 2( 2) 4
  2. 2. Physics, 6th Edition 2-44. ax  bx  4c; (a  b) x  4c; x  2-45. 3ax  2-46. 4c 4( 2)  ; x  8 a  b 2  ( 3) 2ab 2b 2( 3) ; 3cx  2b; x   ; x  1 c 3c 3( 2) 4ac 2 x 4ac  16b 4(2)( 2)  16( 3)   16; 4ac  2 x  16b; x   ; x  32 b b 2 2 2-47. 5m – 16 = 3m – 4 2-48. 3p = 7p - 16 5m – 3m = -4 + 16 3p – 7p = -16 2m = 12; -4p = - 16; m=6 2-49. 4m = 2(m – 4) p = +4 2-50. 3(m – 6) = 6 4m = 2m – 8 3m – 18 = 6 2m = -8; 3m = 24; m = -4 m = +8 p 2 1   ; 3 6 3 p 1 2-51. x  (4)(3)  12; x  36 3 2-52. 2-53. 96 96  48; x  2 x 48 2-54. 14 = 2(b – 7); 14 = 2b – 14; b = 14 2-55. R2 = (4)2 + (3)2 = 16 + 9 R 2  25 2-56. R5 1 1 1 6p 6p 6p   ;   2 p 6 2 p 6 3p = 6 + p; p=3 V = IR; R  V I 2-58. PV  nRT ; T  2-59. F  ma; a  F m 2-60. s = vt + d; d = s – vt 2-57. 2-61. F mv 2 mv 2 ; FR  mv 2 ; R  R F PV nR 2-62. s = ½at2; 2s = at2; 2 a 2s t2
  3. 3. Physics, 6th Edition 2-63. 2as  v  v ; a  2 f 2-65. 2 0 2 v 2  v0 f 2-64. C  2s 1 1 1   ; R2 R  R1 R  R1 R2 R1 R2 R ( R1  R2 ) R  R1 R2 ; R 2-69. v = vo + at; a V R1 R2 R1  R2 mv F t 2-68. PV1 PV2 1  2 ; T1 T2 Ft  mv1 m T2  v – v0 = at 2-70. PV1T2  PV2 T1 1 2 PV2 T1 2 PV1 1 c2 = a2 + b2 ; v  v0 t Q2 2C mv t F 2-66. mv  Ft ; 2-67. mv2  mv1  Ft ; mv2  Ft  mv1 v2  Q2 ; 2V b2 = c2 - a2 b  c2  a 2 Exponents and Radicals 2-71. 212 2-72. 3523 2-73. x10 2-74. 2-75. 1/a 2-76. a/b2 2-77. 1/22 2-78. a2/b2 2-79. 2x5 2-80. 1/a2b2 2-81. m6 2-82. c4/n6 2-83. 64 x 106 2-84. (1/36) x 104 2-85. 4 2-86. 3 2-87. x3 2-88. a2b3 2-89. 2 x 102 2-90. 2 x 10-9 2-91. 2a2 2-92. x + 2 Scientific Notation 2-93. 4.00 x 104 2-94. 6.70 x 101 2-95. 4.80 x 102 2-96. 4.97 x 105 2-97. 2.10 x 10-3 2-98. 7.89 x 10-1 2-99. 8.70 x 10-2 100. 9.67 x 10-4 2-101. 4,000,000 3 x5
  4. 4. Physics, 6th Edition 2-102. 4670 2-103. 37.0 2-104. 140,000 2-105. 0.0367 2-106. 0.400 2-107. 0.006 2-108. 0.0000417 2-109. 8.00 x 106 2-110. 7.40 x 104 2-111. 8.00 x 102 2-112. 1.80 x 10-8 2-113. 2.68 x 109 2-114. 7.40 x 10-3 2-115. 1.60x 10-5 2-116. 2.70 x 1019 2-117. 1.80 x 10-3 2-118. 2.40 x 101 2-119. 2.00 x 106 2-120. 2.00 x 10-3 2-121. 2.00 x 10-9 2-122. 5.71 x 10-1 2-123. 2.30 x 105 2-124. 6.40x 102 2-125. 2.40 x 103 2-126. 5.60 x 10-5 2-127. –6.90 x 10-2 2-128. –3.30 x 10-3 2-129. 6.00 x 10-4 2-130. 6.40 x 106 2-131. –8.00x 106 2-132. -4.00 x 10-2 Graphs 2-133. Graph of speed vs. time: When t = 4.5 s, v = 144 ft/s; When v = 100 m/s, t = 3.1 s. 2-134. Graph of advance of screw vs. turns: When screw advances 2.75 in., N = 88 turns. 2-135. Graph of wavelength vs. frequency: 350 kHz  857 m; 800 kHz  375 m. 2-136. Electric Power vs. Electric Current: 3.20 A  10.4 W; 8.0 A  64.8 W. Geometry A C 2-137. 900. 1800, 2700, and 450 B D 2-138. 2-139a. A = 170, B = 350, C = 380 2-139b. A = 500 Rule 2; 2-140a. A = 500 Rule 3; 2-140b. B = 700, B = 1300 4 B = 400 Rule 2. C = 420 Rule 2
  5. 5. Physics, 6th Edition Right Triangle Trigonometry 2-141. 0.921 2-147. 19.3 2-153. 684 2-159. 54.20 2-165. 36.90 2-142. 0.669 2-148. 143 2-154. 346 2-160. 6.730 2-166. 76.00 2-143. 1.66 2-149. 267 2-155. 803 2-161. 50.20 2-167. 31.20 2-144. 0.559 2-150. 32.4 2-156. 266 2-162. 27.10 2-145. 0.875 2-151. 235 2-157. 2191 2-163. 76.80 2-146. 0.268 2-152. 2425 2-158. 1620 2-164. 6.370 Solve triangles for unknown sides and angles (Exercises 168 – 175): 2-168. tan  = 18/35,  = 35.80 ; R  182  252 R = 30.8 ft 2-169. tan  = 600/400,  = 56.30 ; R  402  802 R = 721 m. 2-170. y = 650 sin 210 = 233 m; x = 650 cos 210 = 607 m. 2-171. sin  = 200/500,  = 23.60; 5002  x 2  2002 , x = 458 km. 2-172. sin  = 210/400,  = 31.70; 5002  m2  2002 , m = 340 m. 2-173. x = 260 cos 510 = 164 in.; 2-174. tan  = 40/80,  = 26.60; y = 260 sin 510 = 202 in. R  402  802 R = 89.4 lb 2-175.  = 1800 - 1200 = 600; y = 300 sin 600 = 260 m; x = 300 cos 600 = 150 m, left 5
  6. 6. Physics, 6th Edition Challenge Problems 2-176. 30.21 – 0.59 in. = 29.62 in. 2-178. T = Tf – T0 = -150C – (290C); T = -44 C0. 2-179. Tf – T0 = -340C; Tf - 200C = -340C; Tf = -140C 2-180. Six pieces @ 4.75 in. = 6(4.75 in.) = 28.5 in.; Five cuts @ 1/16 = 5/16 = 0.3125 in. Original length = 28.5 in. + 0.3125 in. = 28.8 in. 2-181. V = r2h; Solve for h: 2-182. F  mv 2 ; R R h V r 2 mv 2 F 2-183. Solve for x and evaluate: a = 2, b = -2, c = 3, and d = -1 xb + cd = a(x + 2)  xb + cd = ax + 2a  xb – ax = 2a – cd  (b - a)x = 2a – cd x 2a  cd ; ba 2-184. c2  b2  a 2 x 2a  cd 2(2)  (3)( 1) 7 7   ; x ba ( 2)  (2) 4 4 b  c2  a 2 ; b  502  202  53.9 Gm1m2 (6.67 x 10-11 )(4 x 10-8 )(3 x 10-7 )  ; 2-185. F  R2 (4 x 10-2 ) 2 2-186. L = L0 + L0(t – t0); b = 53.9 F  5.00 x 10-22 L = 21.41 cm + (2 x 10-3/C0)( 21.41 cm )(1000C - 200C); L = 24.84 cm. 2-187. Construct graph of y = 2x and verify that x = 3.5 when y = 7 (from the graph). 2-188. (a) A + 600 = 900; A = 300. (b) D + 300 = 900; D = 600. A + C = 900; C = 600. B = 600 by rule 2. A = 600 (alt. int. angles); B = 300; C = 1200. 6
  7. 7. Physics, 6th Edition Critical Thinking Problems 2-189. A = (-8) – (-4) = -4; B = (-6) + (14) = 8; C = A – B = (-4) – (8) = -12; C = - 12 cm. B – A = (8) – (-4) = +12. There is a difference of 24 cm between B – A and A – B. 2-190. T  2 L g  T 2  4 2 Let L = 4Lo; Since L  g L gT 2 4 2 4  2 , the period will be doubled when the length is quadrupled. Let gm = ge /6, Then, T would be changed by a factor of 1  6  2.45 1/ 6 Thus, the period T on the moon would be 2(2.45) or 4.90 s. 2-191. (a) Area = LW = (3.45 x 10-4 m)(9.77 x 10-5 m); Area = 3.37 x 10-8 m2. Perimeter (P) = 2L + 2W = 2(L + W); P = 2(3.45 x 10 -4 + 9.77 x 10-5) = 8.85 x 10-4 m. (b) L = L0/2 and W = 2W0: A = (L0/2)(2W0) = L0W0 ; No change in area. P – P0 = [2(L0/2) + 2(2W0)] - [2L0 + 2W0] = 2W0 – L0 P = 2(9.77 x 10-5) – 3.45 x 10-4 P = -1.50 x 10-4 m. The area doesn’t change, but the perimeter decreases by 0.150 mm. 2-192. Graph shows when T = 420 K, P = 560 lb/in. 2; when T = 600 K, P = 798 lb/in.2 2-193. Graph shows when V = 26 V, I = 377 mA; when V = 48 V, I = 696 mA. 7
  8. 8. Physics, 6th Edition Chapter 3. Technical Measurement and Vectors Unit Conversions 3-1. A soccer field is 100 m long and 60 m across. What are the length and width of the field in feet?  100 cm  1 in.  1 ft  (100 m)      328 ft  1 m  2.54 cm  12 in.  L = 328 ft  100 cm  1 in.  1 ft  (60 m)      197 ft  1 m  2.54 cm  12 in.  W = 197 ft 3-2. A wrench has a handle 8 in. long. What is the length of the handle in centimeters?  2.54 cm  (8 in.)    20.3 cm  1 in.  L = 20.3 cm 3-3. A 19-in. computer monitor has a viewable area that measures 18 in. diagonally. Express this distance in meters. .  2.54 cm  1 m  (18 in.)     0.457 m  1 in.  100 cm  L = 0.457 m 3-4. The length of a notebook is 234.5 mm and the width is 158.4 mm. Express the surface area in square meters.  1 m  1 m  Area = (234.5 mm)(158.4 mm)     0.037  1000 mm  1000 mm  A = 0.0371 m2 3-5. A cube has 5 in. on a side. What is the volume of the cube in SI units and in fundamental USCS units? . 3 3  2.54 cm   1 m  3 V  (5 in.)3  125 in.3      0.00205 m  1 in.   100 cm    3  1 ft  3 V  (125 in. )    0.0723 ft 12 in.   3 V = 0.0723 ft3 8 V = 0.00205 m3
  9. 9. Physics, 6th Edition 3-6. The speed limit on an interstate highway is posted at 75 mi/h. (a) What is this speed in kilometers per hour? (b) In feet per second? (a) 75 mi  1.609 km     121 km/h h  1 mi  (b) 75 mi  1 h  5280 ft     = 110 ft/s h  3600 s  1 mi  3-7. A Nissan engine has a piston displacement (volume) of 1600 cm3 and a bore diameter of 84 mm. Express these measurements in cubic inches and inches. 3  1 in.  (b) 84 mm =   = 3.31 in.  25.4 mm   1 in.  3 (a) 1600 cm   = 97.6 in.  2.54 cm   3  Ans. 97.6 in. 3, 3.31 in. 3-8. An electrician must install an underground cable from the highway to a home located 1.20 mi into the woods. How many feet of cable will be needed?  5280 ft  1.2 mi    6340 ft  1 mi  L = 6340 ft 3-9. One U.S. gallon is a volume equivalent to 231 in. 3. How many gallons are needed to fill a tank that is 18 in. long, 16 in. wide, and 12 in. high? Ans. 15.0 gal. V = (18 in.)(16 in.)(12 in.) = 3456 in. 3  1 gal  V  (3456 in.3 )   15.0 gal 3   231 in.  V = 15.0 gal 3-10. The density of brass is 8.89 g/cm3. What is the density in kg/m3 ? g   1 kg   100 cm  kg    8.89 3     8890 3 cm   1000 g   1 m  m  3  = 8890 kg/m3 Addition of Vectors by Graphical Methods 3-11. A woman walks 4 km east and then 8 km north. (a) Use the polygon method to find her resultant displacement. (b) Verify the result by the parallelogram method. Let 1 cm = 1 km; Then: R = 8.94 km,  = 63.40 9 R  4 km 8 km
  10. 10. Physics, 6th Edition 3-12. A land-rover, on the surface of Mars, moves a distance of 38 m at an angle of 180 0. It then turns and moves a distance of 66 m at an angle of 270 0. What is the displacement from the starting position? Choose a scale, e.g., 1 cm = 10 m 38 m, 1800 Draw each vector to scale as shown. Measure R = 7.62 cm or  67 m R = 76.2 m Measure angle  = 60.10 S of W  = 1800 + 60.10 = 240.10 R = 76.2 m, 240.10 3-13. A surveyor starts at the southeast corner of a lot and charts the following displacements: A = 600 m, N; B = 400 m, W; C = 200 m, S; and D = 100 m, E. What is the net displacement from the starting point? . Choose a scale, 1 cm = 100 m B C A Draw each vector tail to tip until all are drawn. D Construct resultant from origin to finish. R  R = 500 m,  = 53.10 N of E or  = 126.90. 3-14. A downward force of 200 N acts simultaneously with a 500-N force directed to the left. Use the polygon method to find the resultant force. Chose scale, measure: R = 539 N,  = 21.80 S. of E. 200 N 500 N  R 3-15. The following three forces act simultaneously on the same object. A = 300 N, 300 N of E; B = 600 N, 2700; and C = 100 N due east. Find the resultant force using the polygon method. A Choose a scale, draw and measure: R = 576 N,  = 51.40 S of E 10 R B C
  11. 11. Physics, 6th Edition 3-16. A boat travels west a distance of 200 m, then north for 400 m, and finally 100 m at 30 0 S C of E. What is the net displacement? (Set 1 cm = 100 N) Draw and measure: R = 368 N,  = 108.00 B R  A 3-17. Two ropes A and B are attached to a mooring hook so that an angle o 60 exists between the two ropes. The tension in rope A is 80 lb and the tension in rope B is 120 lb. Use the parallelogram method to find the resultant force on the hook. Draw and measure: R R = 174 lb B A 3-18. Two forces A and B act on the same object producing a resultant force of 50 lb at 36.9 0 N of W. The force A = 40 lb due west. Find the magnitude and direction of force B? 40 lb Draw R = 50 lb, 36.90 N of W first, then draw 40 lb, W. R F F = 30 lb, 900 0 36.9 Trigonometry and Vectors 3-19. Find the x and y-components of: (a) a displacement of 200 km, at 34 0. (b) a velocity of 40 km/h, at 1200; and (c) A force of 50 N at 330o. (a) Dx = 200 cos 340 = 166 km 300 Dy = 200 sin 340 = 112 km 340 600 (b) vx = -40 cos 600 = -20.0 km/h (a) vy = 40 sin 600 = +34.6 km/h (c) Fx = 50 cos 300 = 43.3 N; Fy = - 50 sin 300 = -25.0 N 11 (b) (c)
  12. 12. Physics, 6th Edition 3-20. A sled is pulled with a force of 540 N at an angle of 40 0 with the horizontal. What are the 540 N horizontal and vertical components of this force? 0 Fx = 540 cos 40 = 414 N 400 0 Fy = 540 sin 40 = 347 N (a) 3-21. The hammer in Fig. 3-26 applies a force of 260 N at an angle of 150 with the vertical. What is the upward component of the force on the nail? F  F = 260 lb,  = 75 ; Fxy = 260 sin   Fy = 251 N. 0  3-22. A jogger runs 2.0 mi west and then 6.0 mi north. Find the magnitude and direction of the resultant displacement. R R  (2) 2  (6) 2  6.32 mi tan = 6 ; 2  = 71.60 N of W  * 3-23. A river flows south with a velocity of 20 km/h. A boat has a maximum speed of 50 km/h in still water. In the river, at maximum throttle, the boat heads due west. What is the resultant speed and direction of the boat? R R  (50) 2  (20) 2  53.9 km / h; 20 tan = ;  = 21.80 S of W 50 20 km/h 50 km/h 0 R = 53.9 km/h, 21.8 S of E * 3-24. A rope, making an angle of 300 with the horizontal, drags a crate along the floor. What must be the tension in the rope, if a horizontal force of 40 lb is required to drag the crate? Fx = F cos 300; F Fx 40 lb  ; F = 46.2 N 0 cos30 cos300 F 300 Fx 12
  13. 13. Physics, 6th Edition * 3-25. An vertical lift of 80 N is needed to lift a window. A long pole is used to lift the window. What force must be exerted along the pole if it makes an angle of 34 0 with the wall? Fy = F sin 300; Fy F sin 34 0  40 lb ; sin 34 0 F = 96.5 N 80 N 340 F * 3-26. The resultant of two forces A and B is 400 N at 2100. If force A is 200 N at 2700, what are the magnitude and direction of force B? ( = 210 - 1800 = 300) 0 B = -400 N cos 30 = -346 N: 0 B = 346 N, 180 200 N B 300 400 N The Component Method of Vector Addition 3-27. Find the resultant of the following perpendicular forces: (a) 400 N, 0 0; (b) 820 N, 2700; and (b) 500 N, 900. Ax = +400 N; Bx = 0; Cx = 0: Rx = +400 N Ay = 0; By = -820 N; Cy = +500 N; R A  Draw each vector, then find R:  400   320 2 2 ; tan   Ry = 0 – 820 N + 500 N = -320 N 320 ; 400 R B C R = 512 N, 38.70 S of E 3-28. Four ropes, all at right angles to each other, pull on a ring. The forces are A = 40 lb, E; B = 80 lb, N; C = 70 lb, W; and D = 20 lb, S. Find the resultant force on the ring. Ax = +40 lb; Bx = 0; Cx = -70 lb Dx = 0: Rx = +40 lb – 70 lb = -30 lb Ay = 0; By = +80 lb; Cy = 0; Dy = -20 lb ; A = 40 lb, E Ry = 0 + 80 lb - 20 lb = +60 lb R  30   60 2 2 ; tan   60 ; 30 R = 67.1 N, 116.60 13
  14. 14. Physics, 6th Edition *3-29. Two forces act on the car in Fig. 3-27. Force A is 120 N, west and force B is 200 N at 600 N of W. What are the magnitude and direction of the resultant force on the car? Ax = -120; Bx = - (200 N) cos 600 = -100 N B Rx = – 120 N - 100 N; Rx = -220 N Ay = 0, By = 200 sin 600 = +173 N; tan   A R  (220)2  (173)2  280 N Thus, Rx = -100 N, Ry = +173 N and Resultant direction: 600 Ry = 0 + 173 N = 173 N; 173 ;   38.20 N of W ; 210 R = 280 N, 141.80 *3-30. Suppose the direction of force B in Problem 3-29 is reversed (+1800) and other parameters are unchanged. What is the new resultant? (This result is the vector difference A – B). The vector B will now be 600 S of E instead of N of E. Ax = -120 N; A Bx = +(200 N) cos 600 = +100 N 600 Rx = – 120 N + 100 N; Rx = -20 N Ay = 0, By = -200 sin 600 = -173 N; Thus, Rx = -20 N, Ry = -173 N and Resultant direction: tan   B Ry = 0 - 173 N = -173 N; R  (20)2  (173)2  174 N 173 ;   83.40 S of W ; 20 R = 174 N, 253.40 A = 60 lb *3-31. Determine the resultant force on the bolt in Fig. 3-28. ( Ax = 0 ) Bx = -40 cos 200 = -37.6 lb; Bx = -50 cos 600 = -25.0 lb Rx = 0 – 37.6 lb – 25.0 lb; B = 40 lb 200 Rx = -62.6 lb 600 0 Ay = +60 lb; By = 40 sin 20 = 13.7 lb; Cy = 50 sin 60 = -43.3 lb C = 50 lb Ry = 60 lb – 13.7 lb – 43.3 lb; Ry = +30.4 lb R  (62.6)2  (30.4)2 R = 69.6 lb 14 tan   30.4 ; 62.6  = 25.90 N of W
  15. 15. Physics, 6th Edition *3-32. Determine the resultant of the following forces by the component method of vector addition: A = (200 N, 300); B = (300 N, 3300; and C = (400 N, 2500). A Ax = 200 cos 300 = 173 N; Bx = 300 cos 300 = 260 N 300 300 700 Cx = -400 cos 700 = -137 N; Rx = Fx = 296 N B C Ay = 200 sin 300 = 100 N; By = 300 sin 300 = -150 N Cy = -400 sin 700 = -376 N; Ry = Fy = -430 N R  (296)2  (430)2 ; tan   426 ; 296 R = 519 N, 55.20 S of E *3-33. Three boats exert forces on a mooring hook as shown in Fig. 3-29. Find the resultant of these three forces. 420 N Ax = 420 cos 600 = +210 N; Cx = -500 cos 400 = -383 N Bx = 0; Rx = 210 N + 0 –383 N; Rx = -173 N B 500 N C 150 N A 400 600 0 Ay = 420 sin 60 = 364 N; By = 150; Cy = 500 sin 400 = 321 N R  ( 173) 2  (835) 2 ; Ry = Fy = 835 N; tan   835 ; 173 R = 853 N, 78.30 N of W Challenge Problems 3-34. Find the horizontal and vertical components of the following vectors: A = (400 N, 37 0); B = 90 m, 3200); and C = (70 km/h, 1500). Ax = 400 cos 370 = 319 N; Ay = 400 sin 370 = 241 N Bx = 90 cos 400 = 68.9 N; By = 90 sin 400 = 57.9 0 0 Cx = -70 cos 30 = -60.6 N; Cy = 70 sin 30 = 25.0 N 15 400 N 70 N C 300 A 370 400 90 N B
  16. 16. Physics, 6th Edition 3-35. A cable is attached to the end of a beam. What pull at an angle of 400 with the horizontal is needed to produce an effective horizontal force of 200 N? P 400 0 P cos 40 = 200 N; P = 261 N 3-36. A fishing dock runs north and south. What must be the speed of a boat heading at an angle of 400 E of N if its velocity component along the dock is to be 30 km/h? v cos 400 = 30 km/h; v = 39.2 km/h 3-37. Find the resultant R = A + B for the following pairs of forces: A = (520 N, south); B = 269 N, west; (b) A = 18 m/s, north; B = 15 m/s, west. 269 N (a) R  ( 269)  (520)  585 N 2 (a) A tan   520 ; R = 585 N,  = 62.60 S of W 295 (b) R  ( 15) 2  (18) 2 ; tan   18 ; 15 A R  B 2 520 N R 18 m/s  B 15 m/s (b) R = 23.4 N, 50 .20 N of W *3-38. Determine the vector difference (A – B) for the pairs of forces in Problem 3-37. Change B into the vector –B, then ADD: (a) R  (269)  (520)  585 N 2 tan   520 ; 295 2 (a)  520 N tan   18 ; 15 18 m/s R  A R = 585 N, 62.60 S of E (b) R  (15) 2  (18) 2 ; A -B = 269 N R (b) -B =15 m/s R = 23.4 N, 50.20 N of E *3-39. A traffic light is attached to the midpoint of a rope so that each segment makes an angle of 100 with the horizontal. The tension in each rope segment is 200 N. If the resultant force at the midpoint is zero, what must be the weight of the traffic light? Rx = Fx = 0; T sin 100 + T sin 100 – W = 0; 2(200) sin 100 = W: W = 69.5 N T T W 16
  17. 17. Physics, 6th Edition *3-40. Determine the resultant of the forces shown in Fig. 3-30 B = 200 lb Rx = 420 N – 200 cos 700 – 410 cos 530 = 105 lb A = 420 lb 700 Ry = 0 + 200 sin 700 – 410 sin 700 = -139.5 lb R = 175 lb;  = 306.90 2 2 R  Rx  Ry *3-41. Determine the resultant of the forces shown in Fig. 3-31. 0 0 530 C = 410 lb B = 300 N Rx = 200 cos 30 – 300 cos 45 – 155 cos 55 = 128 N 0 450 0 Ry = 0 + 200 sin 70 – 410 sin 70 = -185 N; R = 225 N;  = 124.60 2 2 R  Rx  Ry A = 200 N 0 300 550 C = 155 N *3-42. A 200-N block rests on a 300 inclined plane. If the weight of the block acts vertically downward, what are the components of the weight down the plane and perpendicular to the plane? Choose x-axis along plane and y-axis perpendicular. 300 0 Wx = 200 sin 30 ; 0 Wy = 200 sin 60 ; Wx = 173 N, down the plane. W 300 Wx = 100 N, normal to the plane. *3-43. Find the resultant of the following three displacements: A = 220 m, 600; B = 125 m, 2100; and C = 175 m, 3400. A Ax = 220 cos 600 = 110 m; Ay = 220 sin 600 = 190.5 m Bx = 125 cos 2100 = -108 m; By = 125 sin 2100 = -62.5 m Cx = 175 cos 3400 = 164.4 m; Cy = 175 sin 3400 = -59.9 m Rx = 110 m –108 m + 164.4 m; Ry = 190.5 m – 62.5 m – 59.9 m ; Rx = 166.4 m; Ry = 68.1 m tan   681 . ;   22.30 ; 166.4 R  (166.4) 2  (681) 2  180 m . R = 180 m,  = 22.30 17 30 0 B 600 200 C
  18. 18. Physics, 6th Edition Critical Thinking Questions 3-44. Consider three vectors: A = 100 m, 00; B = 400 m, 2700; and C = 200 m, 300. Choose an appropriate scale and show graphically that the order in which these vectors is added does not matter, i.e., A + B + C = C + B + A. Is this also true for subtracting vectors? Show graphically how A – C differs from C – A. A+B+C A–C C+B+A C-A 3-45. Two forces A = 30 N and B = 90 N can act on an object in any direction desired. What is the maximum resultant force? What is the minimum resultant force? Can the resultant force be zero? Maximum resultant force occurs when A and B are in same direction. A + B = 30 N + 90 N = 120 N Minimum resultant force occurs when A and B are in opposite directions. B - A = 90 N – 30 N = 60 N No combination gives R = 0. *3-46. Consider two forces A = 40 N and B = 80 N. What must be the angle between these two forces in order to produce a resultant force of 60 N? Since R = C = 60 N is smaller than 80 N, the angle B = 40 N C = 60 N   between A and B must be > 900. Applying the law of cosines to the triangle, we can find  and then . 18 A = 80 N 
  19. 19. Physics, 6th Edition C2 = A2 + B2 – 2AB Cos ; (60)2 = (80)2 + (40)2 – 2(80)(40) Cos ;  = 46.60. The angle between the direction of A and the direction of B is  = 1800 - ;  = 133.40. *3-47. What third force F must be added to the following two forces so that the resultant force is zero: A = 120 N, 1100 and B = 60 N, 2000? Components of A: Ax = 120 Cos 1100 = -40.0 N; Ay = 120 Sin 1100 = 113 N Components of B: Bx = 60 Cos 2000 = -56.4 N; By = 60 Sin 2000 = -20.5 N Rx = 0; Rx = Ax + Bx + Fx = 0; Rx = -40.0 N –56.4 N + Fx = 0; Or Fx = +97.4 N Ry = 0; Ry = Ay + By + Fy = 0; Ry = 113 N – 20.5 N + Fy = 0; Fy = -92.2 N F  (97.4) 2  ( 92.2) 2  131 N tan   Or 92.2 ;   43.30 And  = 3600 – 43.40 97.4 Thus, the force F has a magnitude and direction of: F = 134 N,  = 316.60 *3-48. An airplane needs a resultant heading of due west. The speed of the plane is 600 km/h in still air. If the wind has a speed of 40 km/h and blows in a direction of 30 0 S of W, what direction should the aircraft be pointed and what will be its speed relative to the ground? From the diagram, Rx = R, Ry = 0, So that Ay + By = 0. A = 600 km/h Ay = 600 sin y = -40 sin 300 = -20 km/h  R 600 sin - 20 = 0; 600 sin  = 20  B = 40 km/h 20 sin   ;   1910 N of W (direction aircraft should be pointed) . 600 Noting that R = Rx and that Ax +Bx = Rx, we need only find sum of x-components. Ax = -600 cos 599.7 km/hx = 40 Cos 300 = -34.6 km/h R = -599.7 km/h –34.6 km/h; R = -634 km/h. Thus, the speed of the plane relative to the ground is 634 km/h, 00; and the plane must be pointed in a direction of 1.910 N of W. 19
  20. 20. Physics, 6th Edition *3-49. What are the magnitude F and direction  of the force needed to pull the car of Fig. 3-32 directly east with a resultant force of 400 lb? Rx = 400 lb and Ry = 0; Rx = Ax + Fx = 400 lb F 200 Cos 200 + Fx = 400 lb  0 0 Fx = 400 lb – 200 Cos 20 = 212 lb Ry = 0 = Ay + Fy; 20 Ay = -200 sin 200 = -68.4 lb A = 200 lb Fy = -Ay = +68.4 lb; So, Fx = 212 lb and Fy = +68.4 lb F  (212) 2  (68.4) 2 ; tan = 68.4 ; 212 R = 223 lb, 17.90 N of E 20 E
  21. 21. Physics, 6th Edition Chapter 4. Translational Equilibrium and Friction. Note: For all of the problems at the end of this chapter, the rigid booms or struts are considered to be of negligible weight. All forces are considered to be concurrent forces. Free-body Diagrams 4-1. Draw a free-body diagram for the arrangements shown in Fig. 3-18. Isolate a point where the important forces are acting, and represent each force as a vector. Determine the reference angle and label components. B B B 400 A By B A 600 A 300 600 Bx W W W (a) Free-body Diagram (b) Free-body with rotation of axes to simplify work. 4-2. Study each force acting at the end of the light strut in Fig. 3-19. Draw the appropriate freebody diagram. A There is no particular advantage to rotating axes. Components should also be labeled on diagram. 600 300 W B Solution of Equilibrium Problems: 4-3. Three identical bricks are strung together with cords and hung from a scale that reads a total of 24 N. What is the tension in the cord that supports the lowest brick? What is the tension in the cord between the middle brick and the top brick? Each brick must weight 8 N. The lowest cord supports only one brick, whereas the middle cord supports two bricks. 21 Ans. 8 N, 16 N.
  22. 22. Physics, 6th Edition 4-4. A single chain supports a pulley whose weight is 40 N. Two identical 80-N weights are then connected with a cord that passes over the pulley. What is the tension in the supporting chain? What is the tension in each cord? 40 N Each cord supports 80 N, but chain supports everything. T = 2(80 N) + 40 N = 200 N. T = 200 N 80 N 80 N *4-5. If the weight of the block in Fig. 4-18a is 80 N, what are the tensions in ropes A and B? By - W = 0; B sin 400 – 80 N = 0; B = 124.4 N 0 Bx – A = 0; B cos 40 = A; A = (124.4 N) cos 40 0 B A By 400 Bx W A = 95.3 N; B = 124 N. *4-6. If rope B in Fig. 4-18a will break for tensions greater than 200 lb, what is the maximum weight W that can be supported? B A 400 Fy = 0; By – W = 0; W = B sin 40 ; B = 200 N 0 W = (200 N) sin 400; Bx W W = 129 lb *4-7. If W = 600 N in Fig. 18b, what is the force exerted by the rope on the end of the boom A in Fig. 18b? What is the tension in rope B? Fx = 0; A – Wx = 0; B A = Wx = W cos 60 0 A = (600 N) cos 60 = 300 N Fy = 0; B – Wy = 0; B = Wy = W sin 600 B = (600 N) sin 600 = 520 N A = 300 N; B = 520 N 22 A 0 Wx 600 Wy W
  23. 23. Physics, 6th Edition *4-8. If the rope B in Fig. 18a will break if its tension exceeds 400 N, what is the maximum Fy = By - W = 0; By = W weight W? 0 B sin 40 = 400 N ; B A 400 By Fx = 0 B = 622 N Bx – A = 0; B cos 400 = A; A = (622 N) cos 400 Bx W A = 477 N. *4-9. What is the maximum weight W for Fig. 18b if the rope can sustain a maximum tension of B = 800 N only 800 N? (Set B = 800 N). A Draw diagram, then rotate x-y axes as shown to right. Fy = 0; 800 N – W Sin 600 = 0; 600 W = 924 N. The compression in the boom is A = 924 Cos 600 A = 462 N. W *4-10. A 70-N block rests on a 300 inclined plane. Determine the normal force and find the friction force that keeps the block from sliding. (Rotate axes as shown.) Fx = N – Wx = 0; N = Wx = (70 N) cos 300; Fx = F – Wy = 0; F = Wy = (70 N) sin 30 ; 0 N F N = 60.6 N F = 35.0 N 300 W *4-11. A wire is stretched between two poles 10 m apart. A sign is attached to the midpoint of the line causing it to sag vertically a distance of 50 cm. If the tension in each line segment is 5m 5m 2000 N, what is the weight of the sign? (h = 0.50 m)   h 0 2000 N 2000 N tan  = (0.5/5) or  = 5.71 ; 2(2000 N) sin  = W W = 4000 sin 5.71; W=? W = 398 N. *4-12. An 80-N traffic light is supported at the midpoint of a 30-m length of cable between to poles. Find the tension in each cable segment if the cable sags a vertical distance of 1 m. 15 m 15 m h = 1 m; Tan  = (1/15);  = 3.810   h T T T sin  + T sin  = 80 N; 2T sin  = 80 N W = 80 N 23
  24. 24. Physics, 6th Edition T Solution to 4-12 (Cont.): 80 N  601 N ; 2 sin 3810 . T = 601 N *4-13. The ends of three 8-ft studs are nailed together forming a tripod with an apex that is 6ft above the ground. What is the compression in each of these studs if a 100-lb weight is hung F from the apex? Three upward components Fy hold up the 100 lb weight: 3 Fy = 100 lb; sin  = (6/8); Fy = 33.3 lb F sin 48.90 = 33.3 lb; F Fy  h  = 48.90 33.3 lb  44.4 lb F = 44.4 lb, compression sin 48.9 0 *4-14. A 20-N picture is hung from a nail as in Fig. 4-20, so that the supporting cords make an angle of 600. What is the tension of each cord segment? According to Newton’s third law, the force of frame on nail (20 N) 20 N is the same as the force of the nail on the rope (20 N , up). Fy = 0; 20 N = Ty + Ty; Ty = T sin 600; 2Ty = 20 N; Ty = 10 N 600 600 T T So T sin 600 = 10 N, and T = 11.5 N. Friction 4-15. A horizontal force of 40 N will just start an empty 600-N sled moving across packed snow. After motion is begun, only 10 N is needed to keep motion at constant speed. Find the coefficients of static and kinetic friction. s  40 N  0.0667 600 N k  10 N  0.0167 600 N s = 0.0667; k = 0.016 4-16. Suppose 200-N of supplies are added the sled in Problem 4-13. What new force is needed to drag the sled at constant speed? N= 200 N + 600 N = 800 N; Fk = kN = (0.0167)(800 N); 24 Fk = 13.3 N
  25. 25. Physics, 6th Edition 4-17. Assume surfaces where s = 0.7 and k = 0.4. What horizontal force is needed to just start a 50-N block moving along a wooden floor. What force will move it at constant speed? Fs = sN = (0.7)(50 N) = 35 N ; Fk = sN = (0.4)(50 N) = 20 N 4-18. A dockworker finds that a horizontal force of 60 lb is needed to drag a 150-lb crate across the deck at constant speed. What is the coefficient of kinetic friction? k  F ; N k  60 lb  0.400 150 lb k = 0.400 4-19. The dockworker in Problem 4-16 finds that a smaller crate of similar material can be dragged at constant speed with a horizontal force of only 40 lb. What is the weight of this crate? Fk = sN = (0.4)W = 40 lb; W = (40 lb/0.4) = 100 lb; W = 100 lb. 4-20. A steel block weighing 240 N rests on level steel beam. What horizontal force will move the block at constant speed if the coefficient of kinetic friction is 0.12? Fk = sN = (0.12)(240 N) ; Fk = 28.8 N. 4-21. A 60-N toolbox is dragged horizontally at constant speed by a rope making an angle of 35 0 with the floor. The tension in the rope is 40 N. Determine the magnitude of the friction T N force and the normal force. 350 F Fx = T cos 350 – Fk = 0; Fk = (40 N) cos 350 = 32.8 N Fy = N + Ty – W = 0; N = W – Ty = 60 N – T sin 350 N = 60 N – (40 N) sin 350; N = 37.1 N Fk = 32.8 N 4-22. What is the coefficient of kinetic friction for the example in Problem 4-19? k  F 32.8 N  ; N 37.1 N 25 k = 0.884 W
  26. 26. Physics, 6th Edition 4-23. The coefficient of static friction for wood on wood is 0.7. What is the maximum angle for an inclined wooden plane if a wooden block is to remain at rest on the plane? Maximum angle occurs when tan  = s; s = tan  = 0.7;  = 35.00 4-24. A roof is sloped at an angle of 400. What is the maximum coefficient of static friction between the sole of the shoe and the roof to prevent slipping? Tan  = k; k = Tan 400 =0.839; k = 0.839 *4-25. A 200 N sled is pushed along a horizontal surface at constant speed with a 50-N force that makes an angle of 280 below the horizontal. What is the coefficient of kinetic friction? Fx = T cos 280 – Fk = 0; Fk = (50 N) cos 280 = 44.1 N Fy = N - Ty – W = 0; Fk N = W + Ty = 200 N + T sin 280 N = 200 N + (50 N) sin 350; k  N F 44.1 N  N 223 N 280 P W N = 223 N k = 0.198 *4-26. What is the normal force on the block in Fig. 4-21? What is the component of the weight N acting down the plane? Fy = N - W cos 430 = 0; N = (60N) cod 430 = 43.9 N Wx = (60 N) sin 350; F P 430 W Wx = 40.9 N *4-27. What push P directed up the plane will cause the block in Fig. 4-21 to move up the plane with constant speed? [From Problem 4-23: N = 43.9 N and Wx = 40.9 N] Fk = kN = (0.3)(43.9 N); Fk = 13.2 N Fx = P - Fk – Wx = 0; P = Fk + Wx; 26 down plane. P = 13.2 N + 40.9 N; P = 54.1 N
  27. 27. Physics, 6th Edition *4-28. If the block in Fig. 4-21 is released, it will overcome static friction and slide rapidly down the plane. What push P directed up the incline will retard the downward motion until the N block moves at constant speed? (Note that F is up the plane now.) P F Magnitudes of F , Wx, and N are same as Prob. 4-25. Fx = P +Fk – Wx = 0; P = Wx - Fk; 430 W P = 40.9 N - 13.2 N P = 27.7 N directed UP the inclined plane Challlenge Problems *4-29. Determine the tension in rope A and the compression B in the strut for Fig. 4-22. Fy = 0; Fx = 0; By – 400 N = 0; B 400 N  462 N sin600 B A 600 By Bx – A = 0; A = B cos 600 0 A = (462 N) cos 60 ; A = 231 N and 400 N B = 462 N *4-30. If the breaking strength of cable A in Fig. 4-23 is 200 N, what is the maximum weight that 200 N A 400 Ay can be supported by this apparatus? Fy = 0; Ay – W = 0; W = (200 N) sin 400 = 129 N B The maximum weight that can be supported is 129 N. W *4-31. What is the minimum push P parallel to a 370 inclined plane if a 90-N wagon is to be rolled up the plane at constant speed. Ignore friction. Fx = 0; P - Wx = 0; P = (90 N) sin 370 N P 370 P = 54.2 N W = 90 N 27
  28. 28. Physics, 6th Edition 4-32. A horizontal force of only 8 lb moves a cake of ice slides with constant speed across a floor (k = 0.1). What is the weight of the ice? Fk = kN = (0.3) W; Fk = 8 lb; (0.1)W = 8 lb; W = 80 lb. *4-33. Find the tension in ropes A and B for the arrangement shown in Fig. 4-24a. Fx = B – Wx = 0; B = Wx = (340 N) cos 300; B = 294 N Fy = A – Wx = 0; A = Wy = (340 N) sin 300; A = 170 N A Fy = By – 160 N = 0; By = 160 N ; 160 N ; sin 500 W Wy 340 N B *4-34. Find the tension in ropes A and B in Fig. 4-24b. B 300 W Wy x A = 170 N; B = 294 N B sin 500 = 294 N 500 A W = 160 N B = 209 N Fx = A – Bx = 0; A = Bx = (209 N) cos 500; B A = 134 N *4-35. A cable is stretched horizontally across the top of two vertical poles 20 m apart. A 250-N sign suspended from the midpoint causes the rope to sag a vertical distance of 1.2 m. 10 m  What is the tension in each cable segment?. h = 1.2 m; tan   2Tsin 6.840 = 250 N; 12 . ; 10   6.840 T h 10 m  T W = 250 N T = 1050 N *4-36. Assume the cable in Problem 4-31 has a breaking strength of 1200 N. What is the maximum weight that can be supported at the midpoint? 2Tsin 6.840 = 250 N; 2(1200 N) sin 6.840 = W 28 W = 289 N
  29. 29. Physics, 6th Edition *4-37. Find the tension in the cable and the compression in the light boom for Fig. 4-25a. Fy = Ay – 26 lb = 0; Ay = 26 lb ; A A sin 370 = 26 lb B 370 26 lb A ; sin 37 0 A = 43.2 lb W = 26 lb Fx = B – Ax = 0; B = Ax = (43.2 lb) cos 37 ; 0 B = 34.5 lb *4-38. Find the tension in the cable and the compression in the light boom for Fig. 4-25b. First recognize that  = 900 - 420 = 480, Then W = 68 lb Fy = By – 68 lb = 0; By = 68 lb ; B 68 lb ; sin 480 B sin 480 = 68 lb 480 By W A = 915 lb Fx = Bx – A = 0; A = Bx = (91.5 lb) cos 480; B A 68 lb B = 61.2 lb *4-39. Determine the tension in the ropes A and B for Fig. 4-26a. A Fx = Bx – Ax = 0; 0 0 B cos 30 = A cos 45 ; B = 0.816 A Fy = A sin 450 – B sin 300 – 420 N = 0; 0.707 A – 0.5 B = 420 N 0.707 A – (0.5)(0.816 A) = 420 N Substituting B = 0.816A: Solving for A, we obtain: 450 300 W 420 N B A = 1406 N; and B = 0.816A = 0.816(1406) or B = 1148 N Thus the tensions are : A = 1410 N; B = 1150 N *4-40. Find the forces in the light boards of Fig. 4-26b and state whether the boards are under tension or compression. ( Note: A = 900 - 300 = 600 ) Fx = Ax – Bx = 0; A cos 600 = B cos 450; A = 1.414 B Fy = B sin 450 + A sin 600 – 46 lb = 0; 0.707 B + 0.866 A = 46 lb Substituting A = 1.414B: Solving for B: B = 23.8 lb; A = 33.7 lb, tension; A B 450 W 600 46 lb 0.707 B + (0.866)(1.414 B) = 46 lb and A = 1.414B = 01.414 (23.8 lb) or A = 33.7 lb B = 23.8 lb, compression 29
  30. 30. Physics, 6th Edition Critical Thinking Questions 4-41. Study the structure drawn in Fig. 4-27 and analyze the forces acting at the point where the rope is attached to the light poles. What is the direction of the forces acting ON the ends of the poles? What is the direction of the forces exerted BY the poles at that point? Draw the appropriate free-body diagram. Imagine that the poles are bolted together at their upper ends, then visualize the forces ON that bolt and BY that bolt. Forces ON Bolt at Ends (Action Forces): A The force W is exerted ON the bolt BY the weight. The force B is exerted ON bolt BY right pole. The force A is exerted ON bolt BY the middle pole. To understand these directions, imagine that the poles snap, then what would be the resulting motion. 600 300 W B Forces BY Bolt at Ends (Reaction Forces): The force Wr is exerted BY the bolt ON the weight. The force Br is exerted ON bolt BY right pole. The force Ar is exerted BY bolt ON the middle pole. Do not confuse action forces with the reaction forces. Br 300 600 Wr Ar *4-42. Determine the forces acting ON the ends of the poles in Fig 3-27 if W = 500 N. Fx = Bx – Ax = 0; B cos 300 = A cos 600; B = 0.577 A Fy = A sin 600 – B sin 300 – 500 N = 0; 0.866 A – 0.5 B = 500 N Substituting B = 0.577 A: Solving for A, we obtain: 0.866 A – (0.5)( 0.577 A) = 500 N A 600 300 W A = 866 N; and B = 0.577 A = 0.577(866) or B = 500 N Thus the forces are : A = 866 N; B = 500 N Can you explain why B = W? Would this be true for any weight W? Try another value, for example W = 800 N and solve again for B. 30 B
  31. 31. Physics, 6th Edition *4-43. A 2-N eraser is pressed against a vertical chalkboard with a horizontal push of 12 N. If s = 0.25, find the horizontal force required to start motion parallel to the floor? What if you want to start its motion upward or downward? Find the vertical forces required to just start motion up the board and then down the board? Ans. 3.00 N, up = 5 N, down = 1 N. For horizontal motion, P = Fs = sN F F 12 N P N P = 0.25 (12 N); P = 3.00 N 12 N N F P 2N 2N For vertical motion, P – 2 N – Fk = 0 P = 2 N + 3 N; F P 2N P = 5.00 N For down motion: P + 2 N – Fs = 0; P = - 2 N + 3 N; P = 1.00 N *4-44. It is determined experimentally that a 20-lb horizontal force will move a 60-lb lawn mower at constant speed. The handle of the mower makes an angle of 400 with the ground. What push along the handle will move the mower at constant speed? Is the normal force equal to the weight of the mower? What is the normal force? 20 lb k   0.333 60 lb Fy = N – Py - W= 0; W = 60 lb N = P sin 40 + 60 lb; 0 Fy = Px - Fk = 0; Fk 400 Fk = kN = 0.333 N W P cos 400 – 0.333N = 0 P cos 400 – 0.333 (P sin 400 + 60 lb) = 0; 0.552 P = 20 lb; N P 20 lb  36.2 lb ; 0.552 0.766 P = 0.214 P + 20 lb; P = 36.2 lb The normal force is: N = (36.2 lb) sin 400 + 60 lb 31 N = 83.3 lb P
  32. 32. Physics, 6th Edition *4-45. Suppose the lawn mower of Problem 4-40 is to be moved backward. What pull along the handle is required to move with constant speed? What is the normal force in this case? Discuss the differences between this example and the one in the previous problem. k  20 lb  0.333 60 lb N 400 N = 60 lb - P sin 400; Fy = Px - Fk = 0; Fk = kN = 0.333 N Fk W P cos 400 – 0.333N = 0 P cos 400 – 0.333 (60 lb - P sin 400) = 0; 0.980 P = 20 lb; P Fy = N + Py - W= 0; W = 60 lb P 0.766 P - 20 lb + 0.214 P = 0; 20 lb  20.4 lb ; 0.980 P = 20.4 lb The normal force is: N = 60 lb – (20.4 lb) sin 400 N = 46.9 lb *4-46. A truck is removed from the mud by attaching a line between the truck and the tree. When the angles are as shown in Fig. 4-28, a force of 40 lb is exerted at the midpoint of the line. What force is exerted on the truck? T sin 200 + T sin 200 = 40 lb   = 200 2 T sin 200 = 40 lb  h T T F T = 58.5 lb *4-47. Suppose a force of 900 N is required to remove the move the truck in Fig. 4-28. What force is required at the midpoint of the line for the angles shown?. 2 T sin 200 = F; 2(900 N) sin 200 = F; F = 616 N *4-48. A 70-N block of steel is at rest on a 400 incline. What is the static N F friction force directed up the plane? Is this necessarily the 400 maximum force of static friction? What is the normal force? F = (70 N) sin 400 = 45.0 N N = (70 N) cos 400 = 53.6 N 32 W = 70N
  33. 33. Physics, 6th Edition *4-49. Determine the compression in the center strut B and the tension in the rope A for the situation described by Fig. 4-29. Distinguish clearly the difference between the compression force in the strut and the force indicated on your free-body diagram. B Fx = Bx – Ax = 0; 0 0 B cos 50 = A cos 20 ; B = 1.46 A 500 Fy = B sin 50 – A sin 20 – 500 N = 0; 0.766 B – 0.342 A = 500 N 0 0 200 Substituting B = 1.46 A: 0.766 (1.46 A) – (0.342 A) = 500 N Solving for A, we obtain: A = 644 N; and B = 1.46 A = 1.46 (644) or B = 940 N A W Thus the tensions are : A = 644 N; B = 940 N *4-50. What horizontal push P is required to just prevent a 200 N block from slipping down a 600 inclined plane where s = 0.4? Why does it take a lesser force if P acts parallel to the plane? Is the friction force greater, less, or the same for these two cases? (a) Fy = N – Wy– Py = 0; Wy = (200 N) cos 60 = 100 N 0 0 Py = P sin 60 = 0.866 P; N F = 40 N + 0.346 P F 600 N = 100 N + 0.866 P F = N = 0.4(100 N + 0.866 P); x 600 600 P W Fx = Px – Wx + F = 0; P cos 600 - (200 N) sin 600 + (40 N + 0.346 P) = 0 0.5 P –173.2 N + 40 N + 0.346 P = 0 Solving for P gives: P = 157 N (b) If P were parallel to the plane, the normal force would be LESS, and therefore the friction force would be reduced. Since the friction force is directed UP the plane, it is actually helping to prevent slipping. You might think at first that the push P (to stop downward slipping) would then need to be GREATER than before, due to the lesser friction force. However, only half of the push is effective when exerted horizontally. If the force P were directed up the incline, a force of only 133 N is required. You should verify this value by reworking the problem. 33
  34. 34. Physics, 6th Edition *4-51. Find the tension in each cord of Fig. 4-30 if the suspended weight is 476 N. Consider the knot at the bottom first since more information is given at that point. Cy + Cy = 476 N; 2C sin 600 = 476 N C 476 N C  275 N 2sin600 0 60 0 60 A B 300 0 60 476 N Fy = A sin 300 - (275 N) sin 600 = 0 A = 476 N; C C 275 N Fx = A cos 300 – C cos 600 – B = 0; 476 cos 300 – 275 cos 600 – B = 0 B = 412 N – 137 N = 275 N; Thus: A = 476 N, B = 275 N, C = 275 N *4-52. Find the force required to pull a 40-N sled horizontally at constant speed by exerting a pull along a pole that makes a 300 angle with the ground (k = 0.4). Now find the force required if you push along the pole at the same angle. What is the major factor that P changes in these cases? 300 (a) Fy = N + Py - W= 0; W = 40 N N = 40 N - P sin 300; 0.866 P – 16 N + 0.200 P = 0; (b) Fy = N - Py - W= 0; 0.866 P – 16 N - 0.200 P = 0; Fk W W P cos 400- 0.4(40 N - P sin 300) =0; P = 15.0 N N = 40 N + P sin 30 0; Fx = P cos 300 - kN = 0; Fk N 300 Fk = kN Fx = P cos 300 - kN = 0; N Fk = kN P cos 400- 0.4(40 N + P sin 300) =0; P = 24.0 N 34 Normal force is greater! P
  35. 35. Physics, 6th Edition **4-53. Two weights are hung over two frictionless pulleys as shown in Fig. 4-31. What weight W will cause the 300-lb block to just start moving to the right? Assume s = 0.3. Note: The pulleys merely change the direction of the applied forces. Fy = N + (40 lb) sin 450 + W sin 300 – 300 lb = 0 N = 300 lb – 28.3 lb – 0.5 W; 40 lb 450 F = sN 0 0.866 W – 0.3(272 lb – 0.5 W) – 28.3 lb = 0; W 300 F Fx = W cos 30 - sN – (40 lb) cos 45 = 0 0 N 300 lb W = 108 lb **4-54. Find the maximum weight than can be hung at point O in Fig. 4-32 without upsetting the equilibrium. Assume that s = 0.3 between the block and table. We first find F max for the block F B A A 200 F = sN = 0.3 (200 lb) = 60 lb W Now set A = F = 60 lb and solve for W: Fx = B cos 200 – A = 0; B cos 200 = 60 lb; B = 63.9 lb Fy = B sin 200 – W = 0; W = B sin 200 = (63.9 lb) sin 200 ; 35 W = 21.8 lb
  36. 36. Physics, 6th Edition Chapter 5. Torque and Rotational Equilibrium Unit Conversions 5-1. Draw and label the moment arm of the force F about an axis at point A in Fig. 5-11a. What is the magnitude of the moment arm? Moment arms are drawn perpendicular to action line: 250 rA A rA = (2 ft) sin 250 B 3 ft 250 2 ft rB F rA = 0.845 ft 5-2. Find the moment arm about axis B in Fig. 11a. (See figure above.) rB = (3 ft) sin 250 rB = 1.27 ft 5-3. Determine the moment arm if the axis of rotation is at point A in Fig. 5-11b. What is the magnitude of the moment arm? rB = (2 m) sin 60 0 F rB = 1.73 m 5m B 300 5-4. Find the moment arm about axis B in Fig. 5-11b. rB = (5 m) sin 300 600 2m A rA rB rB = 2.50 m Torque 5-5. If the force F in Fig. 5-11a is equal to 80 lb, what is the resultant torque about axis A neglecting the weight of the rod. What is the resultant torque about axis B? Counterclockwise torques are positive, so that A is - and B is +. (a) A = (80 lb)(0.845 ft) = -67.6 lb ft (b) B = (80 lb)(1.27 ft) = +101 lb ft 5-6. The force F in Fig. 5-11b is 400 N and the angle iron is of negligible weight. What is the resultant torque about axis A and about axis B? Counterclockwise torques are positive, so that A is + and B is -. (a) A = (400 N)(1.732 m) = +693 N m; (b) B = (400 N)(2.50 m) = -1000 N m 36
  37. 37. Physics, 6th Edition 5-7. A leather belt is wrapped around a pulley 20 cm in diameter. A force of 60 N is applied to F the belt. What is the torque at the center of the shaft? r = ½D = 10 cm;  = (60 N)(0.10 m) = +6.00 N m 5-8. The light rod in Fig. 5-12 is 60 cm long and pivoted about point A. Find the magnitude and sign of the torque due to the 200 N force if  is (a) 900, (b) 600, (c) 300, and (d) 00.  = (200 N) (0.60 m) sin  for all angles: (a)  = 120 N m 60 cm 200 N   (b)  = 104 N m (b)  = 60 N m A (d)  = 0 r 5-9. A person who weighs 650 N rides a bicycle. The pedals move in a circle of radius 40 cm. If the entire weight acts on each downward moving pedal, what is the maximum torque?  = (250 N)(0.40 m)  = 260 N m 5-10. A single belt is wrapped around two pulleys. The drive pulley has a diameter of 10 cm, and the output pulley has a diameter of 20 cm. If the top belt tension is essentially 50 N at the edge of each pulley, what are the input and output torques? Input torque = (50 N)(0.10 m) = 5 N m Output torque = (50 N)(0.20 m) = 10 N m Resultant Torque 5-11. What is the resultant torque about point A in Fig. 5-13. Neglect weight of bar.  = +(30 N)(6 m) - (15 N)(2 m) - (20 N)(3 m) 15 N 4m  = 90.0 N m, Counterclockwise. 30 N 37 3m 2m A 20 N
  38. 38. Physics, 6th Edition 5-12. Find the resultant torque in Fig. 5-13, if the axis is moved to the left end of the bar.  = +(30 N)(0) + (15 N)(4 m) - (20 N)(9 m) 15 N A 4m 3m 2m  = -120 N m, counterclockwise. 20 N 30 N 5-13. What horizontal force must be exerted at point A in Fig 5-11b to make the resultant torque F = 80 N about point B equal to zero when the force F = 80 N? 5m  = P (2 m) – (80 N)(5 m) (sin 300) = 0 300 2m 2 P = 200 N; B P = 100 N rB P 5-14. Two wheels of diameters 60 cm and 20 cm are fastened together and turn on the same axis as in Fig. 5-14. What is the resultant torque about a central axis for the shown weights? r1 = ½(60 cm) = 0.30 m ; r2 = ½(30 cm) = 0.15 m  = (200 N)(0.30 m) – (150 N)(0.15 m) = 37.5 N m;  = 37.5 N m, ccw 5-15. Suppose you remove the 150-N weight from the small wheel in Fig. 5-14. What new weight can you hang to produce zero resultant torque?  = (200 N)(0.30 m) – W (0.15 m) = 0; W = 400 N 5-16. Determine the resultant torque about the corner A for Fig. 5-15. 80 N B  = +(160 N)(0.60 m) sin 400 - (80 N)(0.20 m) 20 cm  = 61.7 N m – 16.0 N m = 45.7 N m A 160 N 60 cm 400 400 C r R = 45.7 N m 5-17. Find the resultant torque about point C in Fig. 5-15.  = - (80 N)(0.20 m) = -16 N m 160 N 20 cm 80 N r 60 cm 400 C 38
  39. 39. Physics, 6th Edition *5-18. Find the resultant torque about axis B in Fig. 5-15. Fx = 160 cos 400; Fy = 160 sin 400 B 160 N 80 N 20 cm Fy 60 cm 400  = – (123 N)(0.2 m) + (103 N)(0.6 m) = 37.2 N m Fx Equilibrium 5-19. A uniform meter stick is balanced at its midpoint with a single support. A 60-N weight is suspended at the 30 cm mark. At what point must a 40-N weight be hung to balance the system? 20 cm (The 60-N weight is 20 cm from the axis)  = 0; (60 N)(20 cm) – (40 N)x = 0 40 x = 1200 N cm or x 60 N 40 N x = 30 cm: The weight must be hung at the 80-cm mark. 5-20. Weights of 10 N, 20 N, and 30 N are placed on a meterstick at the 20 cm, 40 cm, and 60 cm marks, respectively. The meterstick is balanced by a single support at its midpoint. At what point may a 5-N weight be attached to produce equilibrium. 10 cm 30 cm  = (10 N)(30 cm) + (20 N)(10 cm) x 20 N 30 N – (30 N)(10 cm) – (5 N) x = 0 5 x = (300 + 200 –300) or 10 N x = 40 cm 5N The 5-N weight must be placed at the 90-cm mark 5-21. An 8-m board of negligible weight is supported at a point 2 m from the right end where a 50-N weight is attached. What downward force at the must be exerted at the left end to produce equilibrium? F 6m F (6 m) – (50 N)(2 m) = 0 6 F = 100 N m or F = 16.7 N 2m 50 N 39
  40. 40. Physics, 6th Edition 5-22. A 4-m pole is supported at each end by hunters carrying an 800-N deer which is hung at a point 1.5 m from the left end. What are the upward forces required by each hunter?  = A (0) – (800 N)(1.5 m) + B (4.0 m) = 0 4B = 1200 N or B = 300 N Fy = A + B – 800 lb = 0; A 1.5 m B 2.5 m Axis 800 N A = 500 N 5-23. Assume that the bar in Fig. 5-16 is of negligible weight. Find the forces F and A provided the system is in equilibrium. F  = (80 N)(1.20 m) – F (0.90 m) = 0; F = 107 N 30 cm Fy = F – A – 80 N = 0; A = 107 N – 80 N = 26.7 N Axis 80 N F = 107 N, 90 cm A A = 26.7 N 5-24. For equilibrium, what are the forces F1 and F2 in Fig. 5-17. (Neglect weight of bar.) F1  = (90 lb)(5 ft) – F2 (4 ft) – (20 lb)(5 ft) = 0; 4 ft 5 ft F2 = 87.5 lb Fy = F1 – F2 – 20 lb – 90 lb = 0 F1 = F2 +110 lb = 87.5 lb + 110 lb, F1 = 198 lb Axis 1 ft F2 90 lb 20 lb 5-25. Consider the light bar supported as shown in Fig. 5-18. What are the forces exerted by the B A supports A and B?  = B (11 m) – (60 N)(3 m) – (40 N)( 9 m) = 0; B = 49.1 N Fy = A + B – 40 N – 60 N = 0 A = 100 N – B = 100 N – 49.1 N; 2m 6m 3m Axis 60 N 40 N B = 50.9 N 5-26. A V-belt is wrapped around a pulley 16 in. in diameter. If a resultant torque of 4 lb ft is required, what force must be applied along the belt? R = ½(16 in.) = 8 in. R = (8/12 ft) = 0.667 ft F (0.667 ft) = 4 lb ft; F = 6.00 lb 40 F
  41. 41. Physics, 6th Edition 5-27. A bridge whose total weight is 4500 N is 20 m long and supported at each end. Find the forces exerted at each end when a 1600-N tractor is located 8 m from the left end. B A  = B (20 m) – (1600 N)(8 m) – (4500 N)( 10 m) = 0; 8m Fy = A + B – 1600 N – 4500 N = 0 B = 2890 N A = 6100 N – B = 6100 N – 2890 N; 2m 10 m Axis 1600 N B = 3210 N 4500 N 5-28. A 10-ft platform weighing 40 lb is supported at each end by stepladders. A 180-lb painter is located 4 ft from the right end. Find the forces exerted by the supports.  = B(10 ft) – (40 lb)(5 ft) – (180 lb)( 6 ft) = 0; B = 128 lb 5 ft Fy = A + B – 40 lb – 180 lb = 0 A = 220 lb – B = 220 lb – 128 lb; B A 1 ft 4 ft Axis 40 lb A = 92.0 lb 180 lb *5-29. A horizontal, 6-m boom weighing 400 N is hinged at the wall as shown in Fig. 5-19. A cable is attached at a point 4.5 m away from the wall, and a 1200-N weight is attached to Ty V the right end. What is the tension in the cable? 3m  = 90 – 37 = 53 ; Ty = T sin 53 0 0 0 B 1.5 m 0 Axis  = (T sin 530)(4.5 m) – (400 N)(3 m) – (1200 N)(6 m) = 0; 3.59 T = 1200 N + 7200 N; Ty 1.5 m H 400 N 1200 N T = 2340 N *5-30. What are the horizontal and vertical components of the force exerted by the wall on the boom? What is the magnitude and direction of this force? Fx = H – Tx = 0; H – T cos 530 = 0; H = (2340 N) cos 530; H = 1408 N Fy = V + T sin 530 – 400 N – 1200 N = 0; V = 1600 N – (2340 N) sin 530 = -269 N Thus, the components are: H = 1408 N and V = -269 N. The resultant of these is: R  H 2  V 2  1434 N; tan = -269  = 10.80 S of E 1408 41 R = 1434 N, 349.20
  42. 42. Physics, 6th Edition Center of Gravity 5-31. A uniform 6-m bar has a length of 6 m and weighs 30 N. A 50-N weight is hung from the left end and a 20-N force is hung at the right end. How far from the left end will a single F upward force produce equilibrium? x Fy = F – 50 N – 30 N – 20 N = 0; F = 100 N Axis  = F x – (30 N)(3 m) – (20 N)(6 m) = 0 (100 N) x = 210 N m; 3m 3m 50 N 20 N 30 N x = 2.10 m 5-32. A 40-N sphere and a 12-N sphere are connected by a light rod 200 mm in length. How far from the middle of the 40-N sphere is the center of gravity? F x Fy = F – 40 N – 12 N = 0; F = 52 N 200 mm 40 N  = F x – (40 N)(0) – (12 N)(0.20 m) = 0 (52 N) x = 2.40 N m; x = 0.0462 m or 12 N x = 46.2 mm 5-33. Weights of 2, 5, 8, and 10 N are hung from a 10-m light rod at distances of 2, 4, 6, and 8 m from the left end. How far from the left in is the center of gravity? F x Fy = F – 10 N – 8 N – 5 N – 2 N = 0; F = 25 N 2m Fx – (2 N)(2 m) – (5 N)(4 m) – (8 N)(6 m) – (10 N)(8 m) = 0 (25 N) x = 152 N m; 2m 2m 2N x = 6.08 m 2m 2m 5N 8N 10 N 5-34. Compute the center of gravity of sledgehammer if the metal head weighs 12 lb and the 32in. supporting handle weighs 2 lb. Assume that the handle is of uniform construction and weight. x Fy = F – 2 lb – 12 lb = 0; (14 lb) x = 32 lb in.; 16 in. 16 in. F = 14 lb Fx – (12 lb)(0) – (2 lb)(16 in.) = 0; F 12 lb Fx = 32 lb in. x = 2.29 in. from head. 42 2 lb
  43. 43. Physics, 6th Edition Challenge Problems 5-35. What is the resultant torque about the hinge in Fig. 4-20? Neglect weight of the curved bar.  = (80 N)(0.6 m) – (200 N)(0.4 m) sin 400 60 cm 80 N = 48.0 N m – 51.4 N m; 400 0 40 r  = – 3.42 N m 200 N 40 cm 5-36. What horizontal force applied to the left end of the bar in Fig. 4-20 will produce rotational equilibrium? r F 80 N 500 400 60 cm From Prob. 5-33:  = - 3.42 N m. 400 200 N 40 cm Thus, if  = 0, then torque of +3.42 N m must be added. F (0.6 m) cos 400 = +3.45 N m; F = 7.45 N 5-37. Weights of 100, 200, and 500 lb are placed on a light board resting on two supports as shown in Fig. 4-21. What are the forces exerted by the supports? A  = (100 lb)(4 ft) + B(16 ft) 4 ft – (200 lb)(6 ft) – (500 lb)(12 ft) = 0; B = 425 lb Fy = A + B – 100 lb – 200 lb – 500 lb = 0 A = 800 lb – B = 800 lb – 425 lb; B 6 ft 6 ft 4 ft Axis 100 lb 200 lb 500 lb A = 375 lb The forces exerted by the supports are : A = 375 N and B = 425 N 5-38. An 8-m steel metal beam weighs 2400 N and is supported 3 m from the right end. If a 9000-N weight is placed on the right end, what force must be exerted at the left end to F balance the system? 4m  = A (5 m) + (2400 N)(1 m) – (9000 N)( 3 m) = 0; A = 4920 N Fy = A + B – 2400 N – 9000 N = 0 B = 11,400 N – A = 11,400 N – 4920 N; 43 A = 6480 N A 1m 2400 N 3m 9000 N
  44. 44. Physics, 6th Edition B *5-39. Find the resultant torque about point A in Fig. 5-22. 500  = (70 N)(0.05 m) sin 500 – (50 N)(0.16 m) sin 550 5 cm 70 N A r  = 2.68 N m – 6.55 N m = –3.87 N m 16 cm  = –3.87 N m r 50 N 550 B *5-40. Find the resultant torque about point B in Fig. 5-22.  = (70 N)(0) – (50 N)(a + b) ; 500 5 cm a 70 N First find a and b. a = (0.05 m) cos 500 = 0.0231 m; b = (0.16 m) sin 550 = 0.131 m  = – (50 N)(0.0231 m + 0.131 m) = –8.16 N m 16 cm 50 N b 550  = –8.16 N m Critical Thinking Questions *5-41. A 30-lb box and a 50-lb box are on opposite ends of a 16-ft board supported only at its midpoint. How far from the left end should a 40-lb box be placed to produce equilibrium? Would the result be different if the board weighed 90 lb? Why, or why not? F  = (30 lb)(8 ft) + (40 lb)(x) – (50 lb)(8 ft) = 0; x = 4.00 ft 8 ft Note that the weight acting at the center of the board does NOT contribute to torque about 8 ft x 30 lb 40 lb W 50 lb the center, and therefore, the balance point is not affected, regardless of the weight. 5-42. On a lab bench you have a small rock, a 4-N meterstick and a single knife-edge support. Explain how you can use these three items to find the weight of the small rock. a Measure distances a and b; determine F and then calculate the weight W from equilibrium methods. 44 F b 0.5 m 4N W
  45. 45. Physics, 6th Edition *5-43. Find the forces F1, F2, and F3 such that the system drawn in Fig. 5-23 is in equilibrium. F1 Note action-reaction forces R and R’. R 2 ft 6 ft 50 lb 2 ft First, lets work with top board: 3 ft  (about R) = 0; Force R is upward. 2 ft 5 ft 300 lb R = (300 lb)(6 ft) – (50 lb)(2 ft) – F1(8 ft) = 0 F1 = 213 lb F3 F2 R’ ’ 200 lb Now, Fy = 0 gives: 213 lb + R –300 lb – 50 lb = 0; R = 138 lb = R’ Next we sum torques about F2 with R’ = 138 lb is directed in a downward direction: F = (138 lb)(3 ft) + F3(7 ft) – (200 lb)(5 ft) = 0; From which: F3 = 83.9 lb Fy = 0 = F2 + 83.9 lb – 138 lb – 200 lb;  F2 = –254 lb The three unknown forces are: F1 = 213 lb, F2 = –254 lb, F3 = 83.9 lb *5-44. (a) What weight W will produce a tension of 400 N in the rope attached to the boom in Fig. 5-24?. (b) What would be the tension in the rope if W = 400 N? Neglect the weight 300 of the boom in each case. 2m 400 N (a) m) sin 300) – W (6 m) cos 300 = 0 4m Axis W = 154 N W 0 30 (b)  = T(4 m) sin 300 – (400 N)(6 m) cos 300 = 0 T = 600 N *5-45. Suppose the boom in Fig. 5-24 has a weight of 100 N and the suspended weight W is 300 equal to 400 N. What is the tension in the cord?  m) sin 300) – (400 N)(6 m) cos 300 – (100 N)(3 m) cos 300 = 0 T = 1169 N 45 2m T 4m Axis W 0 30 100 N
  46. 46. Physics, 6th Edition *5-46. For the conditions set in Problem 5-5, what are the horizontal and vertical components of the force exerted by the floor hinge on the base of the boom? Fx = H – 1169 N = 0; or H = 1169 N and 2m 1169 N H = 1169 N Fy = V – 100 N – 400 N = 0; or 300 V Axis V = 500 N 4m 400 N 0 100 N 30 H V = 500 N **5-47. What is the tension in the cable for Fig. 5-25. The weight of the boom is 300 N but its length is unknown. (Select axis at wall, L cancels.) 450 T L   TL sin 75  (300 N ) sin 300  546 L sin 300  0 2 0 750 L 0 T sin 75 = 75.0 N + 273 N; 300 T = 360 N r V 546 N 300 N H **5-48. What are the magnitude and direction of the force exerted by the wall on the boom in Fig. 5-25? Again assume that the weight of the board is 300 N. Refer to the figure and data given in Problem 5-7 and recall that T = 360 N. Fx = H - (360 N) cos 450 = 0; T = 360 N H = 255 N Fy = V + (360 N) sin 450 – 300 N – 546 N = 0; V = 591 N 450 600 H = 255 N and V = 591 N 300 *5-49. An car has a distance of 3.4 m between front and rear axles. If 60 percent of the weight rests on the front wheels, how far is the center of gravity located from the front axle? F Axis x  = 0.6W(0) + 0.4W(3.4 m) – F x = 0 But F = W: 1.36 W – W x = 0 0.4W x = 1.36 m from front axle 46 3.4 m 0.6W
  47. 47. Physics, 6th Edition Chapter 6. Uniform Acceleration Problems: Speed and Velocity 6-1. A car travels a distance of 86 km at an average speed of 8 m/s. How many hours were required for the trip? t s  vt 86, 000 m  1h   10, 750 s   8 m/s  3600 s  t = 2.99 h 6-2. Sound travels at an average speed of 340 m/s. Lightning from a distant thundercloud is seen almost immediately. If the sound of thunder reaches the ear 3 s later, how far away is the storm? t s 20 m   0.0588 s t 340 m / s t = 58.8 ms 6-3. A small rocket leaves its pad and travels a distance of 40 m vertically upward before returning to the earth five seconds after it was launched. What was the average velocity for the trip? v s 40 m + 40 m 80 m   t 5s 5s v = 16.0 m/s 6-4. A car travels along a U-shaped curve for a distance of 400 m in 30 s. It’s final location, however is only 40 m from the starting position. What is the average speed and what is the magnitude of the average velocity? Average speed: v Average velocity: v  s 400 m  t 30 s D 40 m  t 30 s D = 40 m v = 13.3 m/s v = 1.33 m/s, E s = 400 m 47
  48. 48. Physics, 6th Edition 6-5. A woman walks for 4 min directly north with a average velocity of 6 km/h; then she walks eastward at 4 km/h for 10 min. What is her average speed for the trip? t1 = 4 min = 0.0667 h; t2 = 10 min = 0.167 h 4 km/h, 10 min B  s2 s1 = v1t1 = (6 km/h)(0.0667 h) = 0.400 km 6 km/h, 4 min s1 D s1 = v2t2 = (4 km/h)(0.167 h) = 0.667 km  A s s 0.4 km + 0.667 km v 1 2  t1  t2 0.0667 h + 0.167 h C E v = 4.57 km/h 6-6. What is the average velocity for the entire trip described in Problem 6-5? D  (0.667 km) 2 + (0.400 km) 2 ; v tan   0.4 km 0.667 km 0.778 km  3.33 km / h 0.0667 h + 0.167 h D = 0.778 km, 31.00 v = 3.33 km/h, 31.00 6-7. A car travels at an average speed of 60 mi/h for 3 h and 20 min. What was the distance? t = 3 h + 0.333 h = 3.33 h; s = vt = (60 mi/h)(3.33 h); s = 200 mi 6.8 How long will it take to travel 400 km if the average speed is 90 km/h? t s 400 km  t 90 km / h t = 4.44 h *6-9. A marble rolls up an inclined ramp a distance of 5 m, then stops and returns to a point 5 m below its starting point. The entire trip took only 2 s. What was the average speed and what was the average velocity? 5 m + 10 m speed = 2s velocity = D 5 m - 10 m  t 2s (s1 = 5 m, s2 = -10 m) v = 7.50 m/s s2 D v = – 2.5 m/s, down plane. 48 s1
  49. 49. Physics, 6th Edition Uniform Acceleration 6-10. The tip of a robot arm is moving to the right at 8 m/s. Four seconds later, it is moving to the left at 2 m/s. What is the change in velocity and what is the acceleration. v = vf - vo = (–2 m/s) – (8 m/s) a v 10 m / s  t 4s v = –10 m/s a = –2.50 m/s2 6-11. An arrow accelerates from zero to 40 m/s in the 0.5 s it is in contact with the bow string. What is the average acceleration? a v f  vo t  40 m / s - 0 0.5 s a = 80.0 m/s2 6-12. A car traveling initially at 50 km/h accelerates at a rate of 4 m/s 2 for 3 s. What is the final speed? vo = 50 km/h = 13.9 m/s; vf = vo + at vf = (13.9 m/s) + (4 m/s2)(3 s) = 25.9 m/s; vf = 25.9 m/s 6-13. A truck traveling at 60 mi/h brakes to a stop in 180 ft. What was the average acceleration and stopping time? vo = 60 mi/h = 88.0 ft/s a 2 v 2  vo f  v0  v f x  2 2s   t;   0  (88.0 ft/s)2 2(180 ft) t 2as = vf2 – vo2 a = – 21.5 ft/s2 2x  2(180 ft)    v0  v f  88.0 ft/s + 0  49 t = 4.09 s
  50. 50. Physics, 6th Edition 6-14. An arresting device on a carrier deck stops an airplane in 1.5 s. The average acceleration was 49 m/s2. What was the stopping distance? What was the initial speed? 0 = vo + (– 49 m/s2)(1.5 s); vf = vo + at; s = vf t - ½at2 ; vo = 73.5 m/s s = (0)(1.5 s) – ½(-49 m/s2)(1.5 s)2; s = 55.1 m 6-15. In a braking test, a car traveling at 60 km/h is stopped in a time of 3 s. What was the acceleration and stopping distance? ( vo = 60 km/h = 16.7 m/s) (0) = (16.7 m/s) + a (3 s); vf = vo + at; v0  v f s 2 a = – 5.56 m/s2  16.6 m/s + 0  t   3 s  ; 2   s = 25.0 m 6-16. A bullet leaves a 28-in. rifle barrel at 2700 ft/s. What was its acceleration and time in the barrel? (s = 28 in. = 2.33 ft) a 2as = vo2 - vf2 ; s v0  v f 2 2 v 2  v0 f 2s t; t =  (2700 ft / s) 2  0 ; 2(2.33 ft) 2s 2(2.33 ft) ;  v0  v f 0 + 2700 ft / s a = 1.56 x 106 m/s2 t = 1.73 ms 6-17. The ball in Fig. 6-13 is given an initial velocity of 16 m/s at the bottom of an inclined plane. Two seconds later it is still moving up the plane, but with a velocity of only 4 m/s. What is the acceleration? vf = vo + at; a v f  v0 t  4 m / s - (16 m / s) ; 2s a = -6.00 m/s2 6-18. For Problem 6-17, what is the maximum displacement from the bottom and what is the velocity 4 s after leaving the bottom? (Maximum displacement occurs when vf = 0) 2 2as = vo - vf2; s 2 v 2  v0 f 2a 0  (16 m / s) 2  ; 2(-6 m / s2 ) vf = vo + at = 16 m/s = (-6 m/s2)(4 s); 50 s = +21.3 m vf = - 8.00 m/s, down plane
  51. 51. Physics, 6th Edition 6-19. A monorail train traveling at 80 km/h must be stopped in a distance of 40 m. What average acceleration is required and what is the stopping time? 2 2as = vo - s a vf2; v0  v f 2 2 v 2  v0 f 2s t; t =  ( vo = 80 km/h = 22.2 m/s) 0  (22.2 m / s) 2 ; 2(40 m) 2s 2(40 m) ;  v0  v f 22.2 m / s + 0 a = -6.17 m/s2 t = 3.60 m/s Gravity and Free-Falling Bodies 6-20. A ball is dropped from rest and falls for 5 s. What are its position and velocity? s = vot + ½at2; s = (0)(5 s) + ½(-9.8 m/s2)(5 s)2 ; vf = vo + at = 0 + (-9.8 m/s2)(5 s); s = -122.5 m v = -49.0 m/s 6-21. A rock is dropped from rest. When will its displacement be 18 m below the point of release? What is its velocity at that time? s = vot + ½at2; (-18 m) = (0)(t) + ½(-9.8 m/s2)t2 ; vf = vo + at = 0 + (-9.8 m/s2)(1.92 s); t = 1.92 s vf = -18.8 m/s 6-22. A woman drops a weight from the top of a bridge while a friend below measures the time to strike the water below. What is the height of the bridge if the time is 3 s? s = vot + ½at2 = (0) + ½(-9.8 m/s2)(3 s)2; s = -44.1 m 6-23. A brick is given an initial downward velocity of 6 m/s. What is its final velocity after falling a distance of 40 m? 2as = vo2 - vf2 ; 2 v f  v0  2as  (-6 m / s) 2  2(-9.8 m / s2 )( 40 m) ; v = 28.6 m/s; Since velocity is downward, 51 v = - 28.6 m/s
  52. 52. Physics, 6th Edition 6-24. A projectile is thrown vertically upward and returns to its starting position in 5 s. What was its initial velocity and how high did it rise? s = vot + ½at2; 0 = vo(5 s) + ½(-9.8 m/s2)(5 s)2 ; It rises until vf = 0; 2as = vo2 - vf2 ; s vo = 24.5 m/s 0  (24.5 m / s) 2 ; 2(-9.8 m / s2 ) s = 30.6 m 6-25. An arrow is shot vertically upward with an initial velocity of 80 ft/s. What is its maximum height? (At maximum height, vf = 0; a = g = -32 ft/s2) 2 2as = vo - vf2; s 2 v 2  v0 f 2a 0 - (80 ft / s) 2  ; 2(-32 ft / s2 ) s = 100 ft 6-26. In Problem 6-25, what are the position and velocity of the arrow after 2 s and after 6 s? s = vot + ½at2 = (80 ft/s)(2 s) + ½(-32 ft/s2)(2 s)2 ; vf = vo + at = (80 ft/s) + (-32 ft/s2)(2 s); vf = 16 ft/s s = vot + ½at2 = (80 ft/s)(6 s) + ½(-32 ft/s2)(6 s)2 ; vf = vo + at = (80 ft/s) + (-32 ft/s2)(6 s); s = 96 ft s = -96 ft vf = -112 ft/s 6-27. A hammer is thrown vertically upward to the top of a roof 16 m high. What minimum initial velocity was required? 2as = vo2 - vf2 ; v0  v 2  2as  (0) 2  2(-9.8 m / s2 )(16 m) ; f vo = 17.7 m/s Horizontal Projection 6-28. A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, how far will it have traveled horizontally and how far has it fallen vertically? x = vox t = (20 m/s)(2.5 s) ; x = 50.0 m y = voy + ½gt2 = (0)(2.5 s) + ½(-9.8 m/s2)(0.25 s)2 52 y = -0.306 m
  53. 53. Physics, 6th Edition 6-29. An airplane traveling at 70 m/s drops a box of supplies. What horizontal distance will the box travel before striking the ground 340 m below? 0 y = voy t + ½gt2 First we find the time to fall: t = 8.33 s ; t 2y 2( 340 m)  g 9.8 m / s2 x = vox t = (70 m/s)(8.33 s) ; x = 583 m 6-30. At a lumber mill, logs are discharged horizontally at 15 m/s from a greased chute that is 20 m above a mill pond. How far do the logs travel horizontally? t y = ½gt2; 2y  g 2( 20 m) ; 9.8 m / s2 x = vox t = (15 m/s)(8.33 s) ; t = 2.02 s x = 30.3 m 6-31. A steel ball rolls off the edge of a table top 4 ft above the floor. If it strikes the floor 5 ft from the base of the table, what was its initial horizontal speed? First find time to drop 4 ft: x = vox t ; v0 x  t 2y  g x 5 ft ;  t 0.5 s 2( 4 ft) ; 32 ft / s2 t = 0.500 s vox = 10.0 ft/s 6-32. A bullet leaves the barrel of a weapon with an initial horizontal velocity of 400 m/s. Find the horizontal and vertical displacements after 3 s. x = vox t = (400 m/s)(3 s) ; x = 1200 m y = voy + ½gt2 = (0)(3 s) + ½(-9.8 m/s2)(3 s)2 y = -44.1 m 6-33. A projectile has an initial horizontal velocity of 40 m/s at the edge of a roof top. Find the horizontal and vertical components of its velocity after 3 s. vx = vox = 40 m/s vy = voy t + gt = 0 + (-9.8 m/s2)(3s); 53 vy = -29.4 m/s
  54. 54. Physics, 6th Edition The More General Problem of Trajectories 6-34. A stone is given an initial velocity of 20 m/s at an angle of 58 0. What are its horizontal and vertical displacements after 3 s? vox = (20 m/s) cos 580 = 10.6 m/s; voy = (20 m/s) sin 580 = 17.0 m/s x = voxt = (10.6 m/s)(3 s); x = 31.8 m y = voyt + ½gt2 = (17.0 m/s)(3 s) +½(-9.8 m/s2)(3 s)2; y = 6.78 m 6-35. A baseball leaves the bat with a velocity of 30 m/s at an angle of 30 0. What are the horizontal and vertical components of its velocity after 3 s? vox = (30 m/s) cos 300 = 26.0 m/s; voy = (30 m/s) sin 300 = 15.0 m/s vx = vox = 26.0 m/s ; vx = 26.0 m/s vy = voy + gt = (15 m/s) + (-9.8 m/s2)(3 s) ; vy = -14.4 m/s 6-36. For the baseball in Problem 6-33, what is the maximum height and what is the range? ymax occurs when vy = 0, or when: vy = voy + gt = 0 and t = - voy/g voy 30 sin 300 t  ; g 9.8 m / s t  153 s ; Now we find ymax using this time. . ymax = voyt + ½gt2 = (15 m/s)(1.53 s) + ½(-9.8 m/s2)(1.53 s)2; ymax = 11.5 m The range will be reached when the time is t’ = 2(1.53 s) or t’ = 3.06 s, thus R = voxt’= (30 m/s) cos 300 (3.06 s); R = 79.5 m 6-37. An arrow leaves the bow with an initial velocity of 120 ft/s at an angle of 37 0 with the horizontal. What are the horizontal and vertical components of is displacement two seconds later? vox = (120 ft/s) cos 370 = 104 ft/s; voy = (120 ft/s) sin 300 = 60.0 ft/s 54
  55. 55. Physics, 6th Edition 6-37. (Cont.) The components of the initial velocity are: vox = 104 ft/s; x = voxt = (104 ft/s)(2 s); voy = 60.0 ft/s x = 208 ft y = voyt + ½gt2 = (60.0 m/s)(2 s) +½(-32 ft/s2)(2 s)2; y = 56.0 ft *6-38. In Problem 6-37, what are the magnitude and direction of arrow’s velocity after 2 s? vx = vox = 104 ft/s ; vx = 104 ft/s vy = voy + gt = (60 m/s) + (-32 ft/s2)(2 s) ; vy = -4.00 ft/s *6-39. A golf ball in Fig. 6-14 leaves the tee with a velocity of 40 m/s at 65 0. If it lands on a green located 10 m higher than the tee, what was the time of flight, and what was the horizontal distance to the tee? vox = (40 m/s) cos 650 = 16.9 m/s; voy = (40 m/s) sin 650 = 36.25 m/s y = voyt + ½gt2: 10 ft = (36.25 m/s) t + ½(-9.8 m/s2)t2 Solving quadratic (4.9t2 – 36.25t + 10 = 0) yields: t1 = 0.287 s and t2 = 7.11 s The first time is for y = +10 m on the way up, the second is y = +10 m on the way down. Thus, the time from tee to green was: t = 7.11 s Horizontal distance to tee: x = voxt = (16.9 m/s)(7.11 s); x = 120 m *6-40. A projectile leaves the ground with a velocity of 35 m/s at an angle of 32 0. What is the maximum height attained. vox = (35 m/s) cos 320 = 29.7 m/s; voy = (35 m/s) sin 320 = 18.55 m/s ymax occurs when vy = 0, or when: vy = voy + gt = 0 and t = - voy/g t voy g  18.550 ; 9.8 m / s2 t  189 s ; Now we find ymax using this time. . ymax = voyt + ½gt2 = (18.55 m/s)(1.89 s) + ½(-9.8 m/s2)(1.89 s)2; 55 ymax = 17.5 m
  56. 56. Physics, 6th Edition *6-41. The projectile in Problem 6-40 rises and falls, striking a billboard at a point 8 m above the ground. What was the time of flight and how far did it travel horizontally. vox = (35 m/s) cos 320 = 29.7 m/s; voy = (35 m/s) sin 320 = 18.55 m/s y = voyt + ½gt2: 8 m = (18.55 m/s) t + ½(-9.8 m/s2)t2 Solving quadratic (4.9t2 – 18.55t + 8 = 0) yields: t1 = 0.497 s and t2 = 3.36 s The first time is for y = +8 m on the way up, the second is y = +8 m on the way down. Thus, the time from tee to green was: t = 3.29 s Horizontal distance to tee: x = voxt = (29.7 m/s)(3.29 s); x = 97.7 m Challenge Problems 6-42. A rocket travels in space at 60 m/s before it is given a sudden acceleration. It’s velocity increases to 140 m/s in 8 s, what was its average acceleration and how far did it travel in this time? a s v f  v0 t v0  v f 2  t (140 m / s) - (60 m / s) ; 8s a = 10 m/s2 140 F m / s + 60 m / sIbsg G 2 J8 ; H K t = 800 s 6-43. A railroad car starts from rest and coasts freely down an incline. With an average acceleration of 4 ft/s2, what will be the velocity after 5 s? What distance does it travel? vf = vo + at = 0 + (4 ft/s2)(5 s); s = vot + ½at2 = 0 + ½(4 ft/s2)(5 s)2; 56 vf = 20 ft/s s = 50 ft
  57. 57. Physics, 6th Edition *6-44. An object is projected horizontally at 20 m/s. At the same time, another object located 12 m down range is dropped from rest. When will they collide and how far are they B A located below the release point? y A: vox = 20 m/s, voy = 0; B: vox = voy = 0 12 m Ball B will have fallen the distance y at the same time t as ball A. Thus, x = voxt and (20 m/s)t = 12 m; y = ½at2 = ½(-9.8 m/s2)(0.6 s)2 ; t = 0.600 s y = -1.76 m 6-45. A truck moving at an initial velocity of 30 m/s is brought to a stop in 10 s. What was the acceleration of the car and what was the stopping distance? a s v f  v0 t v0  v f 2  t 0 - 30 m / s ; 10 s a = -3.00 m/s2 30 F m / s + 0Ib sg G 2 J10 ; H K s = 150 m 6-46. A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2s, after 4 s, and after 8 s? Apply s = vot + ½at2 and vf = vo + at for time of 2, 4, and 8 s: (a) s = (23 m/s)(2 s) + ½(-9.8 m/s2)(2 s)2 ; vf = (23 m/s) + (-9.8 m/s2)(2 s) ; s = 26.4 m vf = 3.40 m/s (b) s = (23 m/s)(4 s) + ½(-9.8 m/s2)(4 s)2 ; vf = (23 m/s) + (-9.8 m/s2)(4 s) ; s = 13.6 m vf = -16.2 m/s (c) s = (23 m/s)(8 s) + ½(-9.8 m/s2)(8 s)2 ; vf = (23 m/s) + (-9.8 m/s2)(8 s) ; s = -130 m vf = -55.4 m/s 57
  58. 58. Physics, 6th Edition 6-47. A stone is thrown vertically downward from the top of a bridge. Four seconds later it strikes the water below. If the final velocity was 60 m/s. What was the initial velocity of the stone and how high was the bridge? vf = vo + at; v0 = vf – at = (-60 m/s) - (-9.8 m/s)(4 s); s = vot + ½at2 = (-20.8 m/s)(4 s) + ½(-9.8 m/s)(4 s)2; vo = -20.8 m/s s = 162 m 6-48. A ball is thrown vertically upward with an initial velocity of 80 ft/s. What are its position and velocity after (a) 1 s; (b) 3 s; and (c) 6 s Apply s = vot + ½at2 and vf = vo + at for time of 2, 4, and 8 s: (a) s = (80 ft/s)(1 s) + ½(-32 ft/s2)(1 s)2 ; s = 64.0 ft vf = (80 ft/s) + (-32 ft/s2)(2 s) ; vf = 16.0 ft/s (b) s = (80 ft/s)(3 s) + ½(-32 ft/s2)(3 s)2 ; s = 96.0 ft vf = (80 ft/s) + (-32 ft/s2)(3 s) ; vf = -16.0 ft/s (c) s = (80 ft/s)(6 s) + ½(-32 ft/s2)(6 s)2 ; s = 64.0 ft vf = (80 ft/s) + (-32 ft/s2)(6 s) ; vf = -96.0 ft/s 6-49. An aircraft flying horizontally at 500 mi/h releases a package. Four seconds later, the package strikes the ground below. What was the altitude of the plane? y = ½gt2 = ½(-32 ft/s2)(4 s)2; y = -256 ft *6-50. In Problem 6-49, what was the horizontal range of the package and what are the components of its final velocity? vo = 500 mi/h = 733 ft/s; vx = vox = 733 ft/s; x = vxt = (733 ft/s)(4 s); vy = voy + at = 0 + (-32 ft/s)(4 s); 58 voy = 0; t=4s x = 2930 ft vy = -128 ft/s; vx = 733 m/s
  59. 59. Physics, 6th Edition *6-51. A putting green is located 240 ft horizontally and 64 ft vertically from the tee. What must be the magnitude and direction of the initial velocity if a ball is to strike the green at this location after a time of 4 s? x = voxt; s = vot + ½at2; 240 ft = vox (4 s); vox = 60 m/s 64 ft = voy(4 s) + ½(-32 ft/s2)(4 s)2; tan   2 2 v  vx  v y  (60 ft / s) 2  (80 ft / s) 2 ; 80 ft / s 60 ft / s voy = 80 ft/s v = 100 ft/s,  = 53.10 Critical Thinking Questions 6-52. A long strip of pavement is marked off in 100-m intervals. Students use stopwatches to record the times a car passes each mark. The following data is listed: Distance, m Time, s 0 10 m 20 m 30 m 40 m 50 m 0 2.1 s 4.3 s 6.4 s 8.4 s 10.5 s Plot a graph with distance along the y-axis and time along the x-axis. What is the significance of the slope of this curve? What is the average speed of the car? At what instant in time is the distance equal to 34 m? What is the acceleration of the car? Data taken directly from the graph (not drawn): Ans. Slope is v, 4.76 m/s, 7.14 s, 0. 6-53. An astronaut tests gravity on the moon by dropping a tool from a height of 5 m. The following data are recorded electronically. Height, m Time, s 5.00 m 4.00 m 3.00 m 2.00 m 1.00 m 0m 0 1.11 s 1.56 s 1.92 s 2.21 s 2.47 s 59
  60. 60. Physics, 6th Edition 6-53. (Cont.) Plot the graph of this data. Is it a straight line? What is the average speed for the entire fall? What is the acceleration? How would you compare this with gravity on earth? Data taken directly from the graph (not drawn): Ans. Slope is v, 4.76 m/s, 7.14 s, 0. *6-54. A car is traveling initially North at 20 m/s. After traveling a distance of 6 m, the car passes point A where it's velocity is still northward but is reduced to 5 m/s. (a) What are the magnitude and direction of the acceleration of the car? (b) What time was required? (c) If the acceleration is held constant, what will be the velocity of the car (a) vo = 20 m/s, vf = 5 m/s, x = 6 m 2  t; t 2as = vo - (b) s  v0  v f 2 2 v 2  v0 f vf2; a 2s A v = 5 m/s v = 20 m/s when it returns to point A? x=6m x=0 (5 m/s)2  (20 m/s)2  ; 2(6 m) 2s 2(6 m)   ;  v0  v f  20 m/s + 5 m/s   a = -31.2 m/s2 t = 0.480 s (c) Starts at A with vo = + 5 m/s then returns to A with zero net displacement (s = 0): 2as = vo2 - vf2; 0 = (5 m/s)2 – vf2; v f  (5 m / s) 2  5 m / s ; vf = - 5 m/s *6-55. A ball moving up an incline is initially located 6 m from the bottom of an incline and has a velocity of 4 m/s. Five seconds later, it is located 3 m from the bottom. Assuming constant acceleration, what was the average velocity? What is the meaning of a negative average velocity? What is the average acceleration and final velocity? vo = + 4 m/s; s = -3 m; t = 5 s Find vavg s = vavg t; v  3 m ; vavg = -0.600 m/s 5s 6m 4 m/s 3m s=0 Negative average velocity means that the velocity was down the plane most of the time. 60
  61. 61. Physics, 6th Edition *6-55. (Cont.) s = vot + ½at2; -3 m = (4 m/s)(5 s) + ½a (5 s)2; vf = vo + at = 4 m/s + (-1.84 m/s2)(5 s); a = -1.84 m/s2 vf = -5.20 m/s *6-56. The acceleration due to gravity on an distant planet is determined to be one-fourth its value on the earth. Does this mean that a ball dropped from a height of 4 m above this planet will strike the ground in one-fourth the time? What are the times required on the planet and on earth? The distance as a function of time is given by: s = ½at2 so that one-fourth the acceleration should result in twice the drop time. te  2s 2(4 m)  ge 9.8 m / s2 tp  te = 0.904 s 2s 2(4 m)  gp 2.45 m / s2 tp = 1.81 s *6-57. Consider the two balls A and B shown in Fig. 6-15. Ball A has a constant acceleration of 4 m/s2 directed to the right, and ball B has a constant acceleration of 2 m/s 2 directed to the left. Ball A is initially traveling to the left at 2 m/s, while ball B is traveling to the left initially at 5 m/s. Find the time t at which the balls collide. Also, assuming x = 0 at the initial position of ball A, what is their common displacement when they collide? Equations of displacement for A and B: s = so + vot + ½at2 (watch signs) aa = +4 m/s2 ab = -2 m/s2 2 v = - 5 m/s A B v = - 2 m/s x=0 For A: sA = 0 + (-2 m/s)t + ½(+4 m/s2) t2 For B: sB = 18 m + (-5 m/s)t + ½(-2 m/s2) t2; - 2t + 2t2 = 18 – 5t - t2  3t2 + 3t – 18 = 0  + Next simplify and set sA = sB t1 = - 3 s, t2 = +2 s Accept t = +3 s as meaningful answer, then substitute to find either sA or sB: sA = -2(2 s) + 2(2 s)2; 61 x=+4m x = 18 m
  62. 62. Physics, 6th Edition *6-58. Initially, a truck with a velocity of 40 ft/s is located distance of 500 ft to the right of a car. If the car begins at rest and accelerates at 10 ft/s2, when will it overtake the truck? How far is the point from the initial position of the car? v = 40 ft/s v=0 + Equations of displacement for car and truck: s = 500 ft s=0 s = so + vot + ½at 2 (watch signs) For car: sC = 0 + ½(+10 ft/s2) t2 ; Truck: sT = 500 ft + (40 ft/s)t + 0; 5t2 = 500 + 40t or t2 – 8t –100 = 0; Set sC = sT Solve for either distance: t1 = -6.77 s; sC = ½(10 ft/s2)(14.8 s)2; t2 = +14.8 s s = 1092 ft *6-59. A ball is dropped from rest at the top of a 100-m tall building. At the same instant a second ball is thrown upward from the base of the building with an initial velocity of 50 m/s. When will the two balls collide and at what distance above the street? For A: sA = 100 m + v0At + ½gt2 = 100 m + 0 + ½(-9.8 m/s2) t2 For B: sB = 0 + (50 m/s)t + ½(-9.8 m/s2) t2 100 – 4.9 t2 = 50 t – 4.9 t2; 50 t = 100; s = 100 m A Set sA = sB t = 2.00 s B Solve for s: sA = 100 m – (4.9 m/s2)(2 s)2; s = 80.4 m s=0 *6-60. A balloonist rising vertically with a velocity of 4 m/s releases a sandbag at the instant when the balloon is 16 m above the ground. Compute the position and velocity of the sandbag relative to the ground after 0.3 s and 2 s. How many seconds after its release will it strike the ground? The initial velocity of the bag is that of the balloon: voB = + 4 m/s From ground: s = soB + voBt + ½gt2; s = 18 m + (4 m/s)t + ½(-9.8 m/s2)t2 s = 18 m + (4 m/s)(0.3 s) – (4.9 m/s2)(0.3 s)2 ; s = 16.8 m 62
  63. 63. Physics, 6th Edition *6-61. An arrow is shot upward with a velocity of 40 m/s. Three seconds later, another arrow is shot upward with a velocity of 60 m/s. At what time and position will they meet? Let t1 = t be time for first arrow, then t2 = t - 3 for second arrow. s1 = (40 m/s)t1 + ½(-9.8 m/s2)t12 ; s1 = 40t – 4.9t2 s2 = (60 m/s)t2 + ½(-9.8 m/s2)t22 ; s2 = 60(t – 3) - 4.9(t – 3)2 60 m/s s1 = s2 s1 = s2; 40t – 4.9t2 = 60t – 180 – 4.9(t2 – 6t + 9) 40 m/s The solution for t gives: t = 4.54 s Now find position: s1 = s2 = (40 m/s)(4.54 s) – (4.9 m/s2)(4.54 s)2; s = 80.6 m *6-62. Someone wishes to strike a target, whose horizontal range is 12 km. What must be the velocity of an object projected at an angle of 35 0 if it is to strike the target. What is the time of flight? y = voyt + ½gt2 = 0; ( vo sin 350)t = (4.9 m/s2)t2 or R = voxt = 12 km; t (vo cos 350)t = 12,000 m; 0.574v0 14,649 ; From which  4.9 v0 t vo = 354 m/s 0.574v0 ; t  0117v0 . 4.9 14,649 v0 Set t = t and t = 41.4 s *6-63. A wild boar charges directly toward a hunter with a constant speed of 60 ft/s. At the instant the boar is 100 yd away, the hunter fires an arrow at 300 with the ground. What must be the velocity of the arrow if it is to strike its target? y = 0 = (v0 sin 300)t + ½(-32 ft/s2)t2; Solve for t 0.5(2)v0 t  0.03125v0 ; 32 t = 0.03125 vo vo 300 s=0 s1 =( v0 cos 300) t = (0.866 vo)(0.03125 vo); 63 s1 = 0.0271 vo2 v = -60 ft/s s1 = s2 s = 300 ft
  64. 64. Physics, 6th Edition *6-63. (Cont.) s1 = 0.0271 vo2 ; vB = - 60 ft/s; t = 0.03125 vo vo soB = 300 ft 300 s2 = soB + vBt = 300 ft + (-60 ft/s)t s2 = 300 – 60 (0.03125 vo) = 300 – 1.875 vo 0.0271 vo2 = 300 – 1.875 vo The quadratic solution gives: s=0 vo = 76.2 ft/s 64 s = 300 ft Now, set s1 = s2 and solve for vo vo2 + 69.2 vo – 11,070 = 0 or s1 = s2
  65. 65. Physics, 6th Edition Chapter 7. Newton’s Second Law Newton’s Second Law 7-1. A 4-kg mass is acted on by a resultant force of (a) 4 N, (b) 8 N, and (c) 12 N. What are the resulting accelerations? (a) a  4N  1 m/s2 4 kg (b) a  8N  2 m/s2 4 kg (c) a  12N  3 m/s2 4 kg 7-2. A constant force of 20 N acts on a mass of (a) 2 kg, (b) 4 kg, and (c) 6 kg. What are the resulting accelerations? (a) a  20N  10 m/s2 2 kg (b) a  20N  5 m/s2 4 kg (c) a  20N  3.33 m/s2 6 kg 7-3. A constant force of 60 lb acts on each of three objects, producing accelerations of 4, 8, and 12 N. What are the masses? m 60 lb  15 slugs 4 ft / s2 m 60 lb  7.5 slugs 8 ft / s2 m 60 lb  5 slugs 12 ft / s2 7-4. What resultant force is necessary to give a 4-kg hammer an acceleration of 6 m/s2 ? F = ma = (4 kg)(6 m/s2); F = 24 N 7-5. It is determined that a resultant force of 60 N will give a wagon an acceleration of 10 m/s 2. What force is required to give the wagon an acceleration of only 2 m/s2 ? m 60 N  6 slugs ; 10 m / s2 F = ma = (6 slugs)(2 m/s2); F = 12 N 7-6. A 1000-kg car moving north at 100 km/h brakes to a stop in 50 m. What are the magnitude and direction of the force? 2 2as  v 2  vo ; f a Convert to SI units: 100 km/h = 27.8 m/s 2 v 2  vo f 2s  (0) 2  (27.8 m / s) 2 ; 2(50 m) F = ma = (1000 kg)(7.72 m/s2); 65 a  7.72 m / s2 F = 772 N, South.
  66. 66. Physics, 6th Edition The Relationship Between Weight and Mass 7-7. What is the weight of a 4.8 kg mailbox? What is the mass of a 40-N tank? W = (4.8 kg)(9.8 m/s2) = 47.0 N ; m 40 N = 4.08 kg 9.8 m / s2 7-8. What is the mass of a 60-lb child? What is the weight of a 7-slug man? m 60 lb = 1.88 slugs ; 32 ft / s2 W = (7 slugs)(32 ft/s2) = 224 lb 7-9. A woman weighs 180 lb on earth. When she walks on the moon, she weighs only 30 lb. What is the acceleration due to gravity on the moon and what is her mass on the moon? On the Earth? Her mass is the same on the moon as it is on the earth, so we first find the constant mass: me  180 lb  5.625 slugs; 32 ft / s2 Wm = mmgm gm  mm = me = 5.62 slugs ; 30 lb ; 5.625 slugs gm = 5.33 ft/s2 7-10. What is the weight of a 70-kg astronaut on the surface of the earth. Compare the resultant force required to give him or her an acceleration of 4 m/s2 on the earth with the resultant force required to give the same acceleration in space where gravity is negligible? On earth: W = (70 kg)(9.8 m/s2) = 686 N ; FR = (70 kg)(4 m/s2) = 280 N Anywhere: FR = 280 N The mass doesn’t change. 7-11. Find the mass and the weight of a body if a resultant force of 16 N will give it an acceleration of 5 m/s2. m 16 N = 3.20 kg ; 5.0 m / s2 W = (3.20 kg)(9.8 m/s2) = 31.4 N 66
  67. 67. Physics, 6th Edition 7-12. Find the mass and weight of a body if a resultant force of 200 lb causes its speed to increase from 20 ft/s to 60 ft/s in a time of 5 s. a 60 ft / s - 20 ft / s  8 ft / s2 ; 5s m W = mg = (25.0 slugs)(32 ft/s2); 200 lb = 25.0 slugs 8 ft / s2 W = 800 lb 7-13. Find the mass and weight of a body if a resultant force of 400 N causes it to decrease its velocity by 4 m/s in 3 s. a v 4 m / s  ; a  133 m / s2 ; . t 3s m W = mg = (300 kg)(9.8 m/s2); 400 N ; 133 m / s2 . m = 300 kg W = 2940 N Applications for Single-Body Problems: 7-14. What horizontal pull is required to drag a 6-kg sled with an acceleration of 4 m/s2 if a 6 kg friction force of 20 N opposes the motion? P – 20 N = (6 kg)(4 m/s2); 20 N P P = 44.0 N 7-15. A 2500-lb automobile is speeding at 55 mi/h. What resultant force is required to stop the car in 200 ft on a level road. What must be the coefficient of kinetic friction? We first find the mass and then the acceleration. (55 mi/h = 80.7 m/s) m a 2500 lb  78.1 slugs; 32 ft / s2 2 v 2  v0 f 2s 2 Now recall that: 2as  v 2  v0 f (0)  (80.7 ft / s) 2  ; 2(200 ft) and F = ma = (78.1 slugs)(-16.3 ft/s2); Fk  k N ; k  1270 lb ; 2500 lb 67 a  - 16.3 m / s2 F = -1270 lb k = 0.508
  68. 68. Physics, 6th Edition 7-16. A 10-kg mass is lifted upward by a light cable. What is the tension in the cable if the + acceleration is (a) zero, (b) 6 m/s2 upward, and (c) 6 m/s2 downward? T Note that up is positive and that W = (10 kg)(9.8 m/s2) = 98 N. 10 kg W = mg (a) T – 98 N = (10 kg)(0 m/sand T = 98 N (b) T – 98 N = (10 kg)(6 m/sand T = 60 N + 98 N (c) T – 98 N = (10 kg)(-6 m/sand T = - 60 N + 98 N or or T = 158 N T = 38.0 N 7-17. A 64-lb load hangs at the end of a rope. Find the acceleration of the load if the tension in + the cable is (a) 64 lb, (b) 40 lb, and (c) 96 lb. (a) T  W  T W  64 lb  a; 64 lb  64 lb =  a; 2  g  32 ft/s  m = W/g a=0 W (b) T  W  W  64 lb  a; 40 lb  64 lb =  a; 2  g  32 ft/s  a = -12.0 ft/s2 (b) T  W  W  64 lb  a; 96 lb  64 lb =  a; 2  g  32 ft/s  a = 16.0 ft/s2 7-18. An 800-kg elevator is lifted vertically by a strong rope. Find the acceleration of the elevator if the rope tension is (a) 9000 N, (b) 7840 N, and (c) 2000 N. + T Newton’s law for the problem is: T – mg = ma (up is positive) (a) 9000 N – (800 kg)(9.8 m/s2) = (800 kg)a ; (a) 7840 N – (800 kg)(9.8 m/s2) = (800 kg)a ; (a) 2000 N – (800 kg)(9.8 m/s2) = (800 kg)a ; m a = 1.45 m/s2 mg a=0 a = -7.30 m/s2 7-19. A horizontal force of 100 N pulls an 8-kg cabinet across a level floor. Find the acceleration of the cabinet if k = 0.2. F = kN = k mg 100 N – F = ma; F F = 0.2(8 kg)(9.8 m/s 100 N – 15.7 N = (8 kg) a; 68 a = 10.5 m/s2 N mg 100 N
  69. 69. Physics, 6th Edition 7-20. In Fig. 7-10, an unknown mass slides down the 300 inclined plane. N What is the acceleration in the absence of friction? 300 Fx = max; mg sin 300 = ma ; a = (9.8 m/s2) sin 300 = a = g sin 300 300 4.90 m/s2, down the plane mg 7-21. Assume that k = 0.2 in Fig 7-10. What is the acceleration? N F Why did you not need to know the mass of the block? Fx = max; mg sin 300 - kN = ma ; 300 N = mg cos 300 300 mg mg sin 30 - k mg cos 30 = ma ; a = g sin 30 - k g cos 30 0 0 0 a = (9.8 m/s2)(0.5) – 0.2(9.8 m/s2)(0.866); 0 a = 3.20 m/s2, down the plane. *7-22. Assume that m = I 0 kg and k = 0. 3 in Fig. 7- 10. What push P directed up and along the incline in Fig.7-10 will produce an acceleration of 4 m/s2 also up the incline? N F = kN = kmg cos 300; F = 0.3(10 kg)(9.8 m/s2)cos 300 = 25.5 N Fx = ma; P – F – mg sin 300 = ma P – 25.5 N – (10 kg)(9.8 m/s2)(0.5) = (10 kg)(4 m/s2) P – 25.5 N – 49.0 N = 40 N; F + 300 P 300 mg P = 114 N *7-23. What force P down the incline in Fig. 7-10 is required to cause the acceleration DOWN the plane to be 4 m/s2 ? Assume that in = IO kg and k = 0. 3. See Prob. 7-22: F is up the plane now. P is down plane (+). Fx = ma; P - F + mg sin 300 = ma ; Still, F = 25.5 N P - 25.5 N + (10 kg)(9.8 m/s2)(0.5) = (10 kg)(4 m/s2) P - 25.5 N + 49.0 N = 40 N; P = 16.5 N 69 N + P 300 F 300 mg
  70. 70. Physics, 6th Edition Applications for Multi-Body Problems 7-24. Assume zero friction in Fig. 7-11. What is the acceleration of the system? What is the tension T in the connecting cord? 2 kg Resultant force = total mass x acceleration 80 N = (2 kg + 6 kg)a; 6 kg 80 N T a = 10 m/s2 To find T, apply F = ma to 6-kg block only: 80 N – T = (6 kg)(10 m/s 10 kg 7-25. What force does block A exert on block B in Fig, 7-12? F = mTa; 45 N = (15 kg) a;  45 N A a = 3 m/s2 Force ON B = mB a = (5 kg)(3 m/s2); 5 kg B F = 15 N *7-26. What are the acceleration of the system and the tension in the N T connecting cord for the arrangement shown in Fig. 7-13? +a Assume zero friction and draw free-body diagrams. T m1 g (m1g is balanced by N) For total system: m2g = (m1 + m2)a a m2 g 2 m2 g (6 kg)(9.8 m / s ) ;  m1  m2 4 kg + 6 kg  F = m1 a a = 5.88 m/s2 Now, to find T, consider only m1 T = m1a = (4 kg)(5.88 m/s *7-27. If the coefficient of kinetic friction between the table and the 4 kg block is 0.2 in Fig. 7-13, what is the acceleration of the system. What is the tension in the cord? Fy = 0; N = m1g; F = kN = km1g Fk N T +a For total system: m2g - km1g = (m1 + m2)a a m2 g   k m1 g (6 kg)(9.8 m / s )  (0.2)(4 kg)(9.8 m / s )  m1  m2 4 kg + 6 kg 2 2 70 m1 g T m2 g
  71. 71. Physics, 6th Edition *7-27. (Cont.) a 58.8 N  7.84 N 10 kg a = 5.10 m/s2 or N Fk To find T, consider only m2 and make down positive: T +a T m1 g Fy = m2a ; m2g – T = m2a; T = m2g – m2a (6 kg)(9.8 m/s2) – (6 kg)(5.10 m/s2); m2 g T = 28.2 N *7-28. Assume that the masses m1 = 2 kg and m2 = 8 kg are connected by a cord +a that passes over a light frictionless pulley as in Fig. 7-14. What is the acceleration of the system and the tension in the cord? T T Resultant force = total mass of system x acceleration m2g – m1g = (m1 + m2)a a a m2 g  m1 g m1  m2 (8 kg)(9.8 m / s2 ) - (2 kg)(9.8 m / s2 ) 2 kg + 8 kg m2g m1g a = 5.88 m/s Now look at m1 alone: T - m1g = m1 a; T = m1(g + a) = (2 kg)(9.8 m/s2 – 5.88 m/s2); T = 31.4 N +a *7-29. The system described in Fig. 7-15 starts from rest. What is the N acceleration assuming zero friction? (assume motion down plane) Fx = mT a; T m1g sin 320 – m2g = (m1 + m2) a 320 (10 kg)(9.8 m/s2)sin 320 – (2 kg)(9.8 m/s2) = (10 kg + 2 kg)a a T 519 N - 19.6 N . 12 kg 320 m1g a = 2.69 m/s2 *7-30. What is the acceleration in Fig. 7-15 as the 10-kg block moves down the plane against friction (k = 0.2). Add friction force F up plane in figure for previous problem. m1g sin 320 – m2g – F = (m1 + m2) a ; 71 Fy = 0 ; N = m1g cos 320 m2g
  72. 72. Physics, 6th Edition *7-30. (Cont.) m1g sin 320 – m2g – F = (m1 + m2) a ; F = kN = k m1g cos 320 m1g sin 320 – m2g – k m1g cos 320 = (m1 + m2) a ; a = 1.31 m/s2 *7-31 What is the tension in the cord for Problem 7-30? Apply F = ma to mass m2 only: T = m2(g + a) = (2 kg)(9.8 m/s2 + 1.31 m/s2); T – m2g = m2 a; T = 22.2 N Challenge Problems 7-32. A 2000-lb elevator is lifted vertically with an acceleration of 8 ft/s 2. T +a Find the minimum breaking strength of the cable pulling the elevator? m Fy = ma; T – mg = ma; W 2000 lb  ; m  62.5 slugs g 32 ft / s2 T = m(g + a); T = (62.5 slugs)(32 ft/s2 + 8 ft/s2); 2000 lb T = 2500 lb 7-33. A 200-lb worker stands on weighing scales in the elevator of Problem 7-32. What is the reading of the scales as he is lifted at 8 m/s? +a The scale reading will be equal to the normal force N on worker. N – mg = ma; N = m(g + a); T = (62.5 slugs)(32 ft/s2 + 8 ft/s2); N 200 lb T = 2500 lb 7-34. A 8-kg load is accelerated upward with a cord whose breaking strength is 200 N. What is the maximum acceleration? Tmax – mg = ma Tmax  mg 200 N - (8 kg)(9.8 m / s2 ) a  m 8 kg a = 15.2 m/s2 72 T = 200 N +a 8 kg mg

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