Problem_Set_6

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Mechanical Engineering

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Jan 9, 2024

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Anthony Putrino Problem Set #6- Problems 2, 10, 18, 26, 48 2. Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops by applying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than, less than, or equal to the stopping distance of driver 2? (b) Choose the best explanation from among the following: I. Locking up the brakes gives the greatest possible braking force. II. The same tires on the same road result in the same force of friction. III. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction. Given: Two drivers are moving at the same speed and try to stop by applying the brake when they see a deer ahead of them. *First driver stops by locking up his brakes and second driver stops by applying her brakes just to verge of locking. The first driver applied a locking brake so there is no relative velocity in wheels because in rolling the point of contact on the ground has zero velocity therefore static rolling friction acts on them. On the flip side, the second driver truck's wheels have some relative velocity therefore kinetic friction acts on them. The static rolling friction force has less value than the kinetic friction force. f k >f r Hence the stopping distance of first driver is greater than stopping distance for second driver. * Locked-up brakes lead to sliding friction, which is less than rolling friction. Final Answer: The stopping distance of the driver 1 has greater than the stopping distance of the driver 2.

10. A person places a cup of coffee on the roof of his car while he dashes back into the house for a forgotten item. When he returns to the car, he hops in and takes off with the coffee cup still on the roof. (a) If the coefficient of static friction between the coffee cup and the roof of the car is 0.24, what is the maximum acceleration the car can have without causing the cup to slide? Ignore the effects of air resistance. (b) What is the smallest amount of time in which the person can accelerate the car from rest to 15 m/s and still keep the coffee cup on the roof? Solution: a) Solve using Newton’s Second Law & solve for a. SigmaF x = f s = µ s mg = ma a = µ s g = (.24)(9.81m/s 2 ) = 2.35m/s 2 (b) Use the kinematics equation to solve for the minimum time. t = (v-v 0 )/a = (15m/s - 0)/2.35m/s 2 = 6.4sec 18. A backpack full of books weighing 52.0 N rests on a table in a physics laboratory classroom. A spring with a force constant of 150 N/m is attached to the backpack and pulled horizontally. (a) If the spring is pulled until it stretches 2.00 cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? (b) Does your answer to part (a) change if the mass of the backpack is doubled? Explain. Solution: (a) Solve using Newton’s Second Law. Then use Hooke’s Law to solve for force of friction. SigmaF x = f s – F = ma x = 0 f s = F = -kx = -(150N/m)(-.0200m) = 3.0N

(b) If the mass of the backpack was doubled, it would not change the answer towards the force of friction exerted on the backpack by the table. This is because the mass of the backpack is independent from the spring’s force. 26. After a skiing accident, your leg is in a cast and supported in a traction device as shown in Figure -43. Find the magnitude of the force exerted by the leg on the small pulley. (By Newton’s third law, the small pulley exerts an equal and opposite force on the leg.) Let the mass m be 2.27. Solution: (a) Calculate the rope tension and solve it with Newton’s Second Law. T1 - T2 = mg SigmaF x = -F + T 1 (cos30degrees) + T 2 (cos30degrees) = 0 (b) Calculate the magnitude using F. F = 2mg(cos30degrees) = 2*2.27kg*(9.81m/s 2 )*(cos30degrees) = 38.6N or 8.67lb 48. To test the effects of high acceleration on the human body, the National Aeronautics and Space Administration (NASA) constructed a large centrifuge at the Manned Spacecraft Center in Houston. In this device, astronauts were placed in a capsule that moved in a circular path with a radius of 15 m. If the astronauts in this centrifuge experienced a centripetal acceleration 9.0 times that of gravity, what was the linear speed of the capsule? Solution: Use centripetal equation and solve for V to calculate linear speed. a cp = (v 2 )/r → v = sqrt(ra cp ) = sqrt(15m*9*(9.81m/s 2 )) = 37m/s

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