05-203-08hwk-Ch6-7

# 05-203-08hwk-Ch6-7 - Chapter 6: Applications of Newton's...

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Chapter 6: Applications of Newton’s Laws Solutions to Problems 4. Picture the Problem : The book slides in a straight line across the top of the tabletop. Strategy: The minimum force required to get the book moving is related to the maximum coefficient of static friction, and the force required to keep the book sliding at constant speed is equal to the magnitude of the kinetic friction force, from which µ k can be determined. Solution: 1. When the book begins sliding, the applied force equals the maximum static friction force: ( ) ( ) app s s app s 2 2.25 N 0.127 1.80 kg 9.81 m/s F f mg F mg μ = = = = = 2. When the book is sliding at constant speed, the applied force equals the kinetic friction force: ( ) ( ) app k k app k 2 1.50 N 0.0849 1.80 kg 9.81 m/s F f mg F mg = = = = = Insight: The coefficient of kinetic friction is usually smaller than the coefficient of static friction. This is the basic idea behind antilock brakes, which seek to keep the tire of a car rolling so that the friction between the tire and the road remains in the static regime, where there is a greater force to stop the car and improved handling during the stop. ------------------------------------------------------------------------------------------------------------------------------ 43. Picture the Problem : Your car travels along a circular path at constant speed. Strategy: Static friction between your tires and the road provides the centripetal force required to make the car travel along a circular path. Set the static friction force equal to the centripetal force and calculate its value. Solution: Set the static friction force equal to the centripetal force: ( )( ) 2 2 s cp cp 1200 kg 15 m/s 4.7 kN 57 m mv f F ma r = = = = = Insight: The maximum static friction force is ( )( )( ) 2 s 0.88 1200 kg 9.81 m/s 10.4 kN mg = = which corresponds to a maximum cornering speed (without skidding) of 22 m/s. ------------------------------------------------------------------------------------------------------------------------------ 44. Picture the Problem : The test tube travels along a circular path at constant speed. Strategy: Solve equation 6-15 for the speed required to attain the desired acceleration. Solution: Solve equation 6-15 for the speed: ( ) ( )( )( ) 2 cp 52,000 0.075 m 52,000 9.81 m/s 200 m/s 0.20 km/s v ra r g = = = = = Insight: This speed corresponds to 25,000 revolutions per minute for the centrifuge, or 415 revolutions per second.

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46. Picture the Problem : The forces acting on the car are depicted at right. Strategy: Write Newton’s Second Law in the horizontal and vertical directions and combine them to obtain the radius of the curve. The procedure is similar to Example 6-9 in the text. Solution: 1. Write Newton’s Second Law in the x direction and solve for r : 2 cp 2 sin sin x mv F N ma r mv r N θ = = = = 2. Write Newton’s Second Law in the y direction and solve for N : cos 0 cos y F N mg mg N = - = = 3. Substitute for N in the equation from step 1: ( ) ( ) 2 2 2 2 22.7 m/s cos 85.7 m sin tan 9.81 m/s tan31.5 mv v r mg g = = = = ° Insight: Note that the car follows the curved path without any help from friction at all. This is because the inward component of the normal force is sufficient to provide the necessary centripetal force.
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## This note was uploaded on 04/14/2008 for the course PHYSICS 203 taught by Professor Croft during the Spring '08 term at Rutgers.

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05-203-08hwk-Ch6-7 - Chapter 6: Applications of Newton's...

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