04_InstSolManual_PDF

# 04_InstSolManual_PDF - NEWTONS LAWS OF MOTION 4 11 A D 12 A...

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4-1 N EWTON S L AWS OF M OTION 4 Answers to Multiple-Choice Problems 1. C 2. B 3. B 4. A 5. D 6. C 7. B 8. D 9. A 10. C 11. A, D 12. A 13. B 14. E 15. B 16. C Solutions to Problems 4.1. Set Up and Solve: The horizontal component of the force is to the right, and the vertical component is downward. 4.2. Set Up: Take to be along the direction of the pull of A . The coordinates and forces are shown in Figure 4.2. and Figure 4.2 Solve: makes an angle of with the rope of dog A . 31.8° R S t a n u5 R y R x 5 260 N 420 N . R 5 " R x 2 1 R y 2 5 494 N. R y 5 F Ay 1 F By 5 260 N. 270 N 1 150 N 5 420 N. R x 5 F Ax 1 F Bx 5 F By 5 F B sin 60.0 ° 5 260 N. F Bx 5 F B cos 60.0 ° 5 150 N. F Ay 5 0. F Ax 5 F A 5 270 N. 60.0 ° y x F B F A R y 5 g F y . R x 5 g F x 1 x 1 10 N 2 sin 45 ° 5 7.1 N, 1 10 N 2 cos 45 ° 5 7.1 N,

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4.3. Set Up: The force and its horizontal and vertical components are shown in Figure 4.3. The force is directed at an angle of above the horizontal. Figure 4.3 Solve: Refect: We could also f nd the components of in the directions parallel and perpendicular to the incline but the problem asks for horizontal and vertical components. 4.4. Set Up: Take to be upward, so The strap on each side of the jaw exerts a force F directed at an angle of above the horizontal, as shown in Figure 4.4. Figure 4.4 Solve: so 4.5. Set Up: Let and The angles that each force makes with the are and Solve: (a) (b) so u5 78.1 ° tan R y R x R 5 " R x 2 1 R y 2 5 886 N; R y 5 F 1 y 1 F 2 y 1 F 3 y 5 847 N R x 5 F 1 x 1 F 2 x 1 F 3 x 5 179 N; F 3 y 5 F 3 sin u 3 52 328 N F 3 x 5 F 3 cos u 3 247 N; F 2 y 5 F 2 sin u 2 5 668 N F 2 x 5 F 2 cos u 2 418 N; F 1 y 5 F 1 sin u 1 5 507 N F 1 x 5 F 1 cos u 1 5 844 N; u 3 5 233 ° . u 2 5 122 ° , u 1 5 31 ° , 1 x -axis u F 3 5 411 N. F 2 5 788 N, F 1 5 985 N, F 5 3.15 N. g F y 5 2 F sin 52.5 ° 5 5.00 N, O F F x y 52.5 ° 52.5 ° 52.5 ° g F y 5 5.00 N. 1 y F S F v 5 F sin 50.0 ° 5 287 N. F h 5 F cos 50.0 ° 5 241 N; x y 50.0 ° F v F F h 20.0 ° 1 30.0 ° 5 50.0 ° 4-2 Chapter 4
and its components are shown in Figure 4.5. Figure 4.5 Refect: Adding the forces as vectors gives a very different result from adding their magnitudes. 4.6. Set Up: Let be the direction of the force and acceleration. Solve: gives 4.7. Set Up: Let be the direction of the force. Use a constant acceleration equation to calculate the displacement of the probe. Solve: (a) (b) 4.8. Set Up: Take to be the direction in which the skater is moving initially. (comes to rest). Solve: so The only horizontal force on the skater is the friction force, so The force is 46.7 N, directed opposite to the motion of the skater. 4.9. Set Up: Take to be in the direction in which the cheetah moves. Solve: (a) so (b) The force is exerted on the cheetah by the ground. Refect: The net force on the cheetah is in the same direction as the acceleration of the cheetah. 4.10. Set Up: Let be the direction of the force. Use to calculate the acceleration of the puck. Then use constant acceleration equations to f nd the speed and displacement of the puck for Solve: v x 5 v 0 x 1 a x t 5 1 1.56 m / s 2 21 2.00 s 2 5 3.12 m / s x 2 x 0 5 v 0 x t 1 1 2 a x t 2 5 1 2 1 1.56 m / s 2 2.00 s 2 2 5 3.12 m v 0 x 5 0. a x 5 g F x m 5 0.250 N 0.160 kg 5 1.56 m / s 2 .

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04_InstSolManual_PDF - NEWTONS LAWS OF MOTION 4 11 A D 12 A...

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