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Chapter 04 - 4.1 a For the magnitude of the sum to be the...

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4.1: 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 ( 29 , cos 2 2 2 2 2 2 θ F F F F - = + from which cos 0 θ = , 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 . 4.2: 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 N 32 323 cos ) N 50 ( 233 cos ) N 120 ( - = ° + ° = x R . N 124 323 sin ) N 50 ( 233 sin ) N 120 ( ) N 250 ( = ° + ° + = y R b) , N 128 2 2 = + = y x R R R ( 29 ° = - 104 arctan 32 124 . The results have the same magnitude, and the angle has been changed by the amount ) 37 ( ° that the coordinates have been rotated. 4.3: The horizontal component of the force is N 1 . 7 45 cos ) N 10 ( = ° to the right and the vertical component is N 1 . 7 45 sin ) N 10 ( = ° down. 4.4: a) , cos θ F F x = where θ is the angle that the rope makes with the ramp ( ° = 30 θ in this problem), so . N 3 . 69 30 cos N 0 . 60 cos = = = = ° θ x F F F b) N. 6 . 34 tan sin = = = θ F θ F F x y
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4.5: Of the many ways to do this problem, two are presented here. Geometric: From the law of cosines, the magnitude of the resultant is . N 494 60 cos ) N 300 )( N 270 ( 2 ) N 300 ( ) N 270 ( 2 2 = ° + + = R 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 ( 29 . 7 . 31 N 494 ) N 300 ( 120 sin arcsin ° = ° Components: Taking the x + -direction to be along dog A ’s rope, the components of the resultant are N 420 60 cos ) N 300 ( ) N 270 ( = ° + = x R , N 8 . 259 60 sin ) N 300 ( = ° = y R so ( 29 . 7 . 31 arctan , N 494 ) N 8 . 259 ( ) N 420 ( 420 8 . 259 2 2 ° = = = + = θ R 4.6: a) N 10 . 8 ) 9 . 126 ( cos ) N 00 . 6 ( 120 cos ) N 00 . 9 ( 2 1 - = ° - + ° = + x x F F N. 00 . 3 ) 9 . 126 ( sin ) N 00 . 6 ( 120 sin ) N 00 . 9 ( 2 1 + = ° - + ° = + y y F F b) N. 8.64 N) (3.00 N) 10 . 8 ( 2 2 2 2 = + = + = y x R R R 4.7: 2 s / m 2.2 kg) (60 N) 132 ( / = = = / m F a (to two places). 4.8: . N 189 ) m/s kg)(1.40 135 ( 2 = = = ma F 4.9: kg. 16.00 ) m/s N)/(3.00 0 . 48 ( / 2 = = = a F m 4.10: a) The acceleration is 2 s) (5.00 ) m 0 . 11 ( 2 2 s / m 88 . 0 2 2 = = = t x a . The mass is then kg. 9 . 90 2 m/s 0.88 N 0 . 80 = = = a F m b) The speed at the end of the first 5.00 seconds is m/s 4 . 4 = at , and the block on the frictionless surface will continue to move at this speed, so it will move another m 0 . 22 = vt in the next 5.00 s.
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4.11: a) During the first 2.00 s, the acceleration of the puck is 2 m/s 563 . 1 / = m F (keeping an extra figure). At s 00 . 2 = t , the speed is m/s 13 . 3 = at and the position is m 13 . 3 2 / 2 / 2 = = vt at . b) The acceleration during this period is also 2 m/s 563 . 1 , and the speed at 7.00 s is m/s 6.26 s) 00 . 2 )( m/s (1.563 m/s 13 . 3 2 = + . The position at s 00 . 5 = t is m 125 s) 2.00 s m/s)(5.00 (3.13 m 13 . 3 = - + = x , and at s 00 . 7 = t is m, 21.89 s) )(2.00 m/s 3 (1/2)(1.56 s) m/s)(2.00 (3.13 m 12.5 2 2 = + + or 21.9 m to three places. 4.12: a) . m/s 31 . 4 kg 5 . 32 / N 140 / 2 = = = m F a x b) With m 215 , 0 2 2 1 0 = = = at x v x .
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