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Homework Assignment 08: ..........40 points total --------------------------------------------------------------------------------------------------------------------- 4.16 Find the tension in the two wires that support the 100 N light fixture in Figure P4.16 Solution: From Fx 0 , T1 cos40. T2 cos40. 0 0 0 or T1 T2 Then, Fy 0 gives 2 T1 si 40. 100 N 0 n 0 yielding T1 T2 77. N 8 ----------------------------------------------------------------------------------------------------------- 4.20: Two people are pulling a boat through the water as in Figure P4.20. Each exerts a force of 600 N directed at a 30 degree angle relative to the forward motion of the boat. If the boat moves with constant velocity, find the resistive force F exerted by the water on the boat. Solution: The resultant force exerted on the boat by the people is 2 600 N cos30. 1. 103 N in the forward direction. If the boat moves 0 04 with constant velocity, the total force acting on it must be zero. Hence, the resistive force exerted on the boat by the water must be f 1. 103 N i t r w ar di ecton 04 n he ear d r i ----------------------------------------------------------------------------------------------------------- 4.32 a) an elevator of mass m moving upward has two forces acting on it, the upward force of tension in the cable and the downward force due to gravity. When the elevator is accelerating upward, which is greater, T or W? b) When the elevator is moving at a constant velocity upward, which is greater, T or W? c) When the elevator is moving upward, but the acceleration is downward, which is greater, T or W? d) Let the elevator have a mass of 1500 kg and an upward acceleration of 2.5 m/s2. Find T. Is your answer consistent with the answer you gave in part a). e) The elevator of part d) is moving with constant upward velocity of 10 m/s. Find T. Is your answer consistent with your answer to part b)? f) Having initially moved upward with a constant velocity, the elevator begins to accelerate downward at 1.5 m/s2 . Find T. Is your answer consistent with the answer to part c)? Solution: (a) When the acceleration is upward, the total upward force T must exceed the total downward force w m g 1500 kg 9. m s2 1. 104 N 80 47 (b) When the velocity is constant, the acceleration is zero. The total upward force T and the total downward force w must be equali m agniude . n t (c) If the acceleration is directed downward, the total downward force w must exceed the total upward force T. (d) Fy m ay T m g m ay 1500 kg 9. m s2 2. m s2 1. 104 N 80 50 85 Yes , T w . (e) Fy m ay T m g m ay 1500 kg 9. m s2 0 1. 104 N 80 47 Yes , T w . (f) Fy m ay T m g m ay 1500 kg 9. m s 1. m s 1. 104 N 80 2 50 2 25 Yes , T w . 4.62 Three objects are connected by light strings as shown in Fig. P4.62. The string connecting the 4 kg object and the 5 kg object passes over a light frictionless pulley. Determine a) the acceleration of each object and the b) tension in the two strings. Solution: Let m 1 5. kg,m 2 4. kg,and m 3 3. kg . Let T1 be the tension in the 00 00 00 string between m 1 and m 2 , and T2 the tension in the string between m 2 and m 3 . (a) We may apply Newton’s second law to each of the masses. for m1: m 1a T1 m 1g (1) for m 2 : m 2a T2 m 2 g T1 (2) for m 3 : m 3a m 3 g T2 (3) Adding these equations yields m 1 m 2 m 3 a m 1 m 2 m 3 g , so m 1 m 2 m 3 2. kg 00 a m1 m2 m3 0 80 g 12. kg 9. m s 1. m s 2 63 2 (b) From Equation (1), T1 m 1 a g 5. kg 11. m s2 57. N , and 00 4 2 17 from Equation (3), T2 m 3 g a 3. kg 8. m s2 24. N 00 5 Homework Assignment: 09..................... 40 points total Conceptual Question 17: 4.17 Suppose you are driving a car at a high speed. Why should you avoid slamming on your brakes when you want to stop in the shortest possible distance? Answer : The reason for antilock brakes is that based on comparison of the static and kinetic coefficients of friction. For most materials, the static coefficient is slightly larger than the kinetic coefficient of friction, which means that a larger friction force will be developed is two bodies are not sliding over each other. If you hit the brakes, you start sliding and the friction force is smaller than if your tires roll, but do not slide. Since the friction force is larger for rolling tires than sliding tires, you should be able to stop in a shorter distance. Realistically, as long as your tires don’t slide you also can steer the car, which you can’t do if you start sliding, so antilock brakes provide you with a way to stop the car faster and maintain control of the car as you stop. -------------------------------------------------------------------------------------------------------------------- Problems: 4.35 A dock worker loading crates on a ship finds that a 20 kg crate initially at rest on a horizontal surface requires a 75 N horizontal force to set it in motion. After the crate is in motion, a horizontal force of 60 N is required to keep it moving at constant speed. Find the coefficients of static and kinetic friction between the crate and the floor. Solution: When the block is on the verge of moving, the static friction force has a magnitude f f m ax sn . s s Since equilibrium still exists and the applied force is 75 N, we have Fx 75 N f 0 or f m ax 75 N s s In this case, the normal force is just the weight of the crate, or n m g . Thus, the coefficient of static friction is s f s m ax f s m ax 75 N 0. 38 n mg 20 kg 9. m s2 80 After motion exists, the friction force is that of kinetic friction, f kn k Since the crate moves with constant velocity when the applied force is 60 N, we find that Fx 60 N f 0 or f 60 N . Therefore, the coefficient of kinetic k k friction is f f 60 N k k k 0. 31 n m g 20 kg 9. m s2 80 ---------------------------------------------------------------------------------------------------------------- 4.44 A student decides to move a box of books into her dormitory room by pulling a rope attached to the box. She pulls with a force of 80 N at an angle of 25 degrees above the horizontal. The box has a mass of 25 kg and the coef. of kinetic friction between the floor and the box is 0.3. a) find the acceleration of the box. B) the student now starts moving the box up a 10 degree incline, keeping her 80 N force directed at 25 degrees (a) above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box? Solution: Find the normal force n on the 25.0 kg box: Fy n 80. N si 25. 245 N 0 0 n 0 or n 211 N Now find the friction force, f, as f kn 0. 211 N 63. N 300 4 From the second law, we have Fx m a, or 80. N cos25. 63. N = 25. kg a 0 0 4 0 which yields a 0. 366 m s2 (b) When the box is on the incline, Fy n 80. N si 25. 245 N cos10. 0 0 n 0 0 giving n 207 N The friction force is f kn 0. 207 N 62. N 300 2 The net force parallel to the incline is Fx 80. N cos25. 245 N s n10. 62. N =- 3 N 0 0 i 0 2 32. Fx 32. N 3 Thus, a 1. m s2 , or 1. m s2 dow n t i i 29 29 he nclne m 0 25. kg 4.53 In figure P4.53, the coefficient of kinetic friction between the two blocks is 0.3. The surface of the table and the pulleys are frictionless. A) Draw the FBD for each block. B) Determine the acceleration of each block c) find the tension in the strings. Solution: (a) (b) For the 10-kg object: Fy m ay T2 98 N 10 kg a or T2 98 N 10 kg a (1) For the 2.0-kg object: Fy m ay n1 20 N 0 or n1 20 N so f kn1 0. 20 N 6. N k 30 0 Also, Fx m ax 6. N T1 2. kg a 0 0 or T1 6. N 2. kg a 0 0 (2) Finally, for the 3.0-kg object: Fx m ax T2 T1 6. N 3. kg a 0 0 or T2 T1 6. N 3. kg a 0 0 (3) Substituting Equations (1) and (2) into Equation (3) yields: 98 N 10 kg a 6. N 2. kg a 6. N 3. kg a or 0 0 0 0 86 N a 5. m s2 7 15 kg (c) Substituting the computed value for the magnitude of the acceleration into 7 Equations (1) and (2) gives: T1 6. N 2. kg 5. m s2 17 N 0 0 and T2 98 N 10 kg 5. m s2 41 N 7 Homework Assignment: 10 .................30 points total 3.54 A boy and girl are tossing an apple back and forth between them. Figure P3.54 shows one path the apple follows when watched by an observer looking on from the side. The apple is moving from left to right. Five points mark the path. Ignore air resistance. A) Make a copy of this figure. At each of the marked points draw an arrow that shows the magnitude and direction of the apple’s velocity when it passes through that point. b) make another copy of this figure. This time at each marked point place an arrow indicating the magnitude and direction of any acceleration the apple exhibits at that point. solution: (a) Velocity vector at several points (b) Acceleration vector at several points -------------------------------------------------------------------------------------------------------------------- 4.41 The coefficient of static friction between the 3 kg crate and the 35 degree incline it rests on is 0.300 What minimum force F directed perpendicular to the inclined plane is needed to prevent the crate from sliding down the incline? Solution: The normal force acting on the crate is given by n F m gcos . The net force tending to move the crate down the incline is m g si f , where f is the force of static n s s friction between the crate and the incline. If the crate is in equilibrium, then m gsi f 0 , so that f Fg si n s s n But, we also know f sn s F m g cos s Therefore, we may write m g si s F m g cos , or n si n35. si F n s cos m g 0. 0 300 cos35. 3. kg 9. m s2 32. N 0 00 80 1 --------------------------------------------------------------------------------------------------------- 4.68 A 3 kg object hangs at one end of a rope that is attached to a support on a railroad car. When the car accelerates to the right, the rope makes an angle of 4 degrees with the vertical. Find the acceleration of the car. Solution: In the vertical direction, we have mg Fy T cos4. m g 0 , or 0 T cos4. 0 In the horizontal direction, the second law becomes: Fx T s n 4. m a, so i 0 T s n 4. i 0 a gt 4. 0. m s2 an 0 69 m

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