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HW8.sol - homework 08 RAHMAN TARIQUE Due 3:00 am Question 1...

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homework 08 – RAHMAN, TARIQUE – Due: Feb 20 2008, 3:00 am 1 Question 1, chap 6, sect 1. part 1 of 2 10 points A block of mass m is accelerated across a rough surface by a force of magnitude F that is exerted at an angle φ with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude f . f F φ m What is the magnitude of the acceleration vectora of the block? 1. | vectora | = F m 2. | vectora | = F cos φ f m correct 3. | vectora | = F f m 4. | vectora | = F cos φ m 5. | vectora | = F sin φ m g m Explanation: From Newton’s second law of motion, the acceleration is the total force in the horizontal direction divided by the mass. There are two forces in the horizontal direction: one is the friction force; the other is the horizontal com- ponent of the dragging force F , but they are in the opposite directions, so the acceleration of the block is | vectora | = F cos φ f m . Question 2, chap 6, sect 1. part 2 of 2 10 points Which of the following expressions for the coefficient of friction is correct? 1. μ = m g f 2. μ = f m g F cos φ 3. μ = f m g F sin φ correct 4. μ = f m g 5. μ = m g F cos φ f Explanation: By definition, the coefficient of kinetic fric- tion is the ratio of the friction force and the normal force in the vertical direction. And it is easy to see that the normal force in the vertical direction is just N = m g F sin φ , So the coefficient of friction is μ = f m g F sin φ . Question 3, chap 6, sect 1. part 1 of 3 10 points A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are μ s = 0 . 87 and μ k = 0 . 74, respectively. The acceleration of gravity is 9 . 8 m / s 2 . 27 kg μ 37 What is the frictional force acting on the 27 kg mass? Correct answer: 159 . 24 N (tolerance ± 1 %). Explanation:
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homework 08 – RAHMAN, TARIQUE – Due: Feb 20 2008, 3:00 am 2 F f N m g 37 The forces acting on the block are shown in the figure. Since the block is at rest, the magnitude of the friction force should be equal to the component of the weight on the plane of the incline F f = M g sin θ = (27 kg) (9 . 8 m / s 2 ) sin 37 = 159 . 24 N . Question 4, chap 6, sect 1. part 2 of 3 10 points What is the largest angle which the incline can have so that the mass does not slide down the incline? Correct answer: 41 . 0233 (tolerance ± 1 %). Explanation: The largest possible value the static friction force can have is F f,max = μ s N , where the normal force is N = Mg cos θ . Thus, since F f = M g sin θ , M g sin θ m = μ s M g cos θ m tan θ m = μ s θ m = tan 1 ( μ s ) = tan 1 (0 . 87) = 41 . 0233 .
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