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Unformatted text preview: homework 11 YOO, HEE Due: Feb 12 2008, 4:00 am 1 Question 1, chap 5, sect 2. part 1 of 2 10 points A book is at rest on an incline as shown above. A hand, in contact with the top of the book, produces a constant force F hand vertically downward. F hand B o o k The following figures show several attempts at drawing free-body diagrams for the book. Which figure has the correct directions for each force? Note: The magnitude of the forces are not necessarily drawn to scale. 1. weight force friction normal correct 2. weight friction force normal 3. normal force friction weight 4. normal friction weight force 5. weight force normal friction 6. weight friction normal force 7. weight force normal friction 8. weight normal friction force Explanation: The normal force points perpendicular to the surface of the inclined plane. The weight force points down. The F hand also points down. The friction force keeps the book from sliding and consequently points up the incline. Question 2, chap 5, sect 2. part 2 of 2 10 points For the normal force exerted on the book by the wedge in the diagram, which force(s) homework 11 YOO, HEE Due: Feb 12 2008, 4:00 am 2 complete(s) the force pair for Newtons third law (action-reaction)? 1. the component of F hand pointing perpen- dicular to the surface of the incline 2. the pull of the book on the earth 3. the component of gravity pointing per- pendicular to the surface of the incline 4. the normal force exerted on the wedge by the book correct 5. the pull of the earth on the book 6. the component of gravity pointing parallel to the surface of the incline 7. the sum of the component of gravity per- pendicular to the surface of the incline and the component of F hand perpendicular to the surface of the incline Explanation: The force that completes the third law pair with the normal force of the wedge on the book is the normal force of the book on the wedge. Question 3, chap 5, sect 6. part 1 of 1 10 points The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 3 kg 7 kg 6 kg F F r = 0 If F = 21 N and F r = 4 N, what is the magnitude of the force exerted on the block with mass 7 kg by the block with mass 6 kg? Correct answer: 10 . 375 N (tolerance 1 %). Explanation: m 1 m 2 m 3 F F r = 0 Given : vector F = +21 N , vector F r = 4 N , = 0 , m 1 = 3 kg , m 2 = 7 kg , m 3 = 6 kg , and g = 9 . 8 m / s 2 . Note: F is acting on the combined mass of the three blocks, resulting in a common acceleration after accounting for friction. m 1 F 21 F m 2 F 32 F 12 m 3 F r F 23 Let F , F r , F 32 represent the force exerted on the system from the right, from the left, and the force exerted on m 2 by m 3 , respec- tively....
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