{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chap04 - 4.1 a For the magnitude of the sum to be the sum...

This preview shows pages 1–4. Sign up to view the full content.

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 , cos 2 2 2 2 2 2 F F F F from which cos 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 . 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 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 F F x where 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 x F F F b) N. 6 . 34 tan sin θ F θ F F x y

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 . 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 . 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.
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 20

chap04 - 4.1 a For the magnitude of the sum to be the sum...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online