OLDHW13 - oldhomewk 13 PAPAGEORGE MATT Due 10:00 pm...

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oldhomewk 13 – PAPAGEORGE, MATT – Due: Feb 17 2008, 10:00 pm 1 Question 1, chap 6, sect 3. part 1 of 4 10 points A curve in a road is banked. There is a car on the curve. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 5 Mg μ = 0 . 13 31 What is the component of its weight paral- lel to the incline? Correct answer: 7571 . 06 N (tolerance ± 1 %). Explanation: Let : M = 1500 kg , v c = 31 . 732 m / s , Part2 r = 171 m , θ = 31 , and μ = 0 . 13 . Consider the free body diagram for the car M g sin θ N = M g cos θ μ N |W bardbl - M a bardbl | m g Solution: The car is on an incline that makes an angle θ with the horizontal direction so the component of its weight parallel to the incline is W bardbl = M g sin θ = (1500 kg)(9 . 8 m / s 2 ) sin 31 = 7571 . 06 N . Question 2, chap 6, sect 3. part 2 of 4 10 points If the radius of curvature is 171 m, what is the ideal speed of the car such that it doesn’t rely on friction to keep from sliding sideways? Correct answer: 31 . 732 m / s (tolerance ± 1 %). Explanation: Since there is no frictional force keeping the car from sliding sideways, W bardbl = M a bardbl M g sin θ = M v 2 c r cos θ v 2 c = g r sin θ cos θ v c = radicalbig g r tan θ = radicalBig (9 . 8 m / s 2 )(171 m) tan(31 ) = 31 . 732 m / s . Question 3, chap 6, sect 3. part 3 of 4 10 points The next curve that the car approaches also has a radius of curvature 171 m. It is banked at an angle of 30 . The ideal speed for this curve is v c (banked so that the car experiences no frictional force). The speed of the car v s as it rounds this curve is v s = 0 . 543 v c . If the mass of the car is 1500 kg, what is the magnitude of the frictional force needed to keep it from sliding sideways? Correct answer: 5182 . 86 N (tolerance ± 1 %). Explanation: For this curve there is a new v c since r is different and θ is different. The frictional force acts either up or down the incline W bardbl + f = M a bardbl = M v 2 r cos θ = (0 . 543) 2 M v 2 c r cos θ .
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oldhomewk 13 – PAPAGEORGE, MATT – Due: Feb 17 2008, 10:00 pm 2 But we know that at the ideal speed, f critical = 0 and M v 2 c r cos θ = W bardbl . So f = (0 . 543) 2 M v 2 c r cos θ - W bardbl = (0 . 543) 2 W bardbl - W bardbl = [(0 . 543) 2 - 1] W bardbl = [(0 . 543) 2 - 1] M g sin θ = [ - 0 . 705151] (1500 kg) (9 . 8 m / s 2 ) sin 30 | f | = 5182 . 86 N .
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