05_InstSolManual_PDF

# 05_InstSolManual_PDF - APPLICATIONS OF NEWTONS LAWS 5 11 C...

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5-1 A PPLICATIONS OF N EWTON S L AWS 5 Answers to Multiple-Choice Problems 1. A 2. D 3. B 4. D 5. A 6. A 7. A 8. B 9. A 10. C 11. C 12. B 13. D Solutions to Problems 5.1. Set Up: Constant speed means Use coordinates where is upward. The breaking strength for the rope in each part is (a) 50 N, (b) 4000 N, and (c) The free-body diagram for the bucket plus its contents is given in Figure 5.1. Figure 5.1 Solve: gives so (a) so 35 N of cement. The mass of cement is 3.6 kg. (b) so 3985 N of cement. The mass of cement is 410 kg. (c) so of cement. The mass of cement is Refect: Since the upward pull of the rope equals the total weight of the object. 5.2. Set Up: Use coordinates where is upward. for each object. Call the tension in the wires (top), (middle), and (bottom). The free-body diagram for each object is given in Figure 5.2. Figure 5.2 T b a 5 0 x y W T m T b a 5 0 x y W T t T m a 5 0 x y W T b T m T t a y 5 0 1 y a 5 0, 6 3 10 3 kg. 6 3 10 4 N w 5 6 3 10 4 N, w 5 4000 N, w 5 50 N, T 5 w . T 2 w 5 0 g F y 5 ma y T a 5 0 x y w 6 3 10 4 N. 1 y a 5 0.

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Solve: (a) and (b) for each object gives and and The top wire has the greatest tension. If then and and 5.3. Set Up: for each object. Apply to each weight and to the pulley. Take upward. The pulley has negligible mass. Let be the tension in the rope and let be the tension in the chain. Solve: (a) The free-body diagram for each weight is the same and is given in Figure 5.3a. gives Figure 5.3 (b) The free-body diagram for the pulley is given in Figure 5.3b. Refect: The tension is the same at all points along the rope. 5.4. Set Up: For each object use coordinates where is upward. Each object has Call the objects 1 and 2, with and Solve: (a) The free-body diagrams for each object are shown in Figure 5.4. and are replaced by their x and y components. Figure 5.4 (b) For object 2, gives and (c) For object 2, gives so gives T C 5 128 N T B 5 T A 1 w 1 2 sin 60° 5 150 N 1 72 N 2 sin 60° 5 128 N; T B sin 60 ° 1 T C sin 60 ° 2 T A 2 w 1 5 0. g F y 5 ma y T B 5 T C T C cos 60 ° 2 T B cos 60 ° 5 0 g F x 5 ma x T A 5 150 N. T A 2 w 2 5 0 g F y 5 ma y T B sin60 ° T A sin60 ° T A a 5 0 x y w 2 5 150N T B T A a 5 0 x w 1 5 72N T B cos60 ° T A cos60 ° object 1 object 2 y 60 ° 60 ° T C T B w 2 5 150 N. w 1 5 72 N a 5 0. 1 y T c 5 2 T r 5 50.0 N. T r a 5 0 x y w 5 25.0N ( a ) T c T r a 5 0 x y T r ( b ) T r 5 w 5 25.0 N. g F y 5 ma y T c T r 1 y g F y 5 ma y a 5 0 T b 5 W 5 25.0 N. T m 5 2 W 5 50.0 N W 5 T t / 3 5 25.0 N T t 5 75.0 N T t 5 T m 1 W 5 3 W . T m 5 T b 1 W 5 2 W , T b 5 W , T t 2 T m 2 W 5 0. T m 2 W 2 T b 5 0, T b 2 W 5 0, g F y 5 ma y 5-2 Chapter 5
5.5. Set Up: Take upward. Apply to the person. Solve: (a) The free-body diagram for the person is given in Figure 5.5. The two sides of the rope each exert a force with vertical component gives Figure 5.5 (b) Set and solve for and Re f ect: Only the vertical components of the tension on each side of the person act to hold him up. The tension in the rope is much greater than his weight.

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## This note was uploaded on 06/10/2010 for the course PHY 185363 taught by Professor R during the Summer '09 term at Florida College.

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05_InstSolManual_PDF - APPLICATIONS OF NEWTONS LAWS 5 11 C...

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