Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 04 | a For the magnitude of the sum to be the sum of the magnitudes the forces must be parallel and the angle between them is zero. b The forces form the sides of a right isosceles triangle and the angle between them is 90 . Alternatively the law of cosines may be used as F2 F2 2F 2 - 2F2 cosO from which cosO 0 and the forces are perpendicular. c For the sum to have 0 magnitude the forces must be antiparallel and the angle between them is 180 . In the new coordinates the 120-N force acts at an angle of 53 from the - x -axis or 233 from the x -axis and the 50-N force acts at an angle of 323 from the x -axis. a The components of the net force are Rx 120 N cos 233 50 N cos 323 -32 N Ry 250 N 120 N sin 233 50 N sin 323 124 N. b R A r2 R2 128 N arctan 104 . The results have the same magnitude x y 32 and the angle has been changed by the amount 37 that the coordinates have been rotated. The horizontal component of the force is 10 N cos 45 N to the right and the vertical component is 10N sin 45 down. a Fx Fcos 0 where 0 is the angle that the rope makes with the ramp Q 30 in this problem so F 1 1 -FF 60 N N. z cosO cos 30 b Fy FsinQ Fx tanQ . Of the many ways to do this problem two are presented here. Geometric From the law of cosines the magnitude of the resultant is R J 270 N 2 300 N 2 2 270 N 300 N cos 60 494 N. The angle between the resultant and dog A s rope the angle opposite the side corresponding to the 250-N force in a vector diagram is then . sin 120 300N arcsin I---- --- r----- I . 494 N Components Taking the x -direction to be along dog A s rope the components of the resultant are Rx 270 N 300 N cos 60 420 N Ry 300N sin 60 N so R 420 N 2 259 8N 2 494 N 3 arctan 317 . 42 . . a F1X F2x N cos 120 N cos F1y F2y sin 120 sin . b R J R2 R2 7 2 N 2 N. a F m 132 N 60 kg m s2 to two places . F ma 135 kg s2 189N. m F a s2 .
