CH_5 - Chapter 5 CHAPTER 5 - Further Applications of...

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Chapter 5 Page 1 CHAPTER 5 - Further Applications of Newton’s Laws 1. The friction is kinetic, so F fr = k F N . With constant velocity, the acceleration is zero. Using the force diagram for the crate, we can write F = m a : x -component: F k F N = 0; y -component: F N Mg = 0. Thus F N = Mg , and F = k F N = k Mg = (0.30)(12.0 kg)(9.80 m/s 2 ) = 35 N . If k = 0, there is no force required to maintain constant speed. 2. ( a ) In general, static friction is given by F fr s F N . Immediately before the box starts to move, the static friction force reaches its maximum value: F fr,max = s F N . For the instant before the box starts to move, the acceleration is zero. Using the force diagram for the box, we can write F = m a : x -component: F s F N = 0; y -component: F N Mg = 0. Thus F N = Mg , and F = s F N = s Mg ; 25.0 N = s (6.0 kg)(9.80 m/s 2 ), which gives s = 0.43 . ( b ) When the box accelerates and the friction changes to kinetic, we have F k F N = Ma ; 25.0 N – k (6.0 kg)(9.80 m/s 2 ) = (6.0 kg)(0.50 m/s 2 ), which gives k = 0.37 . 3. ( a ) ( b ) ( c ) In ( a ) the friction is static and opposes the impending motion down the plane. In ( b ) the friction is kinetic and opposes the motion down the plane. In ( c ) the friction is kinetic and opposes the motion up the plane. 4. If we simplify the forces so that there is one normal force, we have the diagram shown. We can write F = m a : x -component: F fr + mg sin = 0; y -component: F N mg cos = 0. When we combine the two equations, we have tan = F fr / F N s . Thus we have tan max = s = 0.8, max = 39° . M g x y F F fr F N M g x y F F fr F N F N m g F fr F N m g F fr F N m g F fr F N m g F fr x y

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Chapter 5 Page 2 5. If we simplify the forces so that there is one normal force, we have the diagram shown. The friction force provides the acceleration. We can write F = m a : x -component: F fr = Ma ; y -component: F N Mg = 0. Thus we have a = F fr / M = s F N / M = s g . The minimum value of s is s,min = a / g = 0.20 . 6. If we simplify the forces so that there is one normal force, we have the diagram shown. The friction force provides the acceleration. We can write F = m a : x -component: F fr = Ma ; y -component: F N Mg = 0. Thus we have a = F fr / M = s F N / M = s g . The maximum value of a is a max = s g = (0.80)(9.80 m/s 2 ) = 7.8 m/s 2 . 7. While the box is sliding down, friction will be up the plane, opposing the motion. From the force diagram for the box, we have F = m a : x -component: mg sin F fr = ma ; y -component: F N mg cos = 0. From the x -equation, we have F fr = mg sin ma = m ( g sin a ) = (15.0 kg)[(9.80 m/s 2 ) sin 30° – (0.30 m/s 2 )] = 69 N . Because the friction is kinetic, we have
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This note was uploaded on 04/08/2008 for the course PHY 53 taught by Professor Mueller during the Fall '08 term at Duke.

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CH_5 - Chapter 5 CHAPTER 5 - Further Applications of...

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