Practice 05 - practice 05 – GADHIA TEJAS – Due 2:00 pm 1 Question 1 part 1 of 5 10 points A car travels at a speed of 20 m s around a curve of

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Unformatted text preview: practice 05 – GADHIA, TEJAS – Due: Feb 15 2007, 2:00 pm 1 Question 1 part 1 of 5 10 points A car travels at a speed of 20 m / s around a curve of radius 40 m. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 9 M g μ 21 ◦ What is the net centripetal force needed to keep the car from skidding sideways? Correct answer: 29000 N (tolerance ± 1 %). Explanation: Let : m = 2900 kg , v = 20 m / s , r = 40 m , θ = 21 ◦ , and μ = 0 . 457386 Part5 . The centripetal acceleration of the car rounding a curve is a c = v 2 R and the net centripetal force needed to provide such ac- celeration is F c = ma c = mv 2 R = 29000 N . Question 2 part 2 of 5 10 points Were there no friction between the car’s tires and the road, what centripetal force could be provided just by the banking of the road? Correct answer: 10909 . 4 N (tolerance ± 1 %). Explanation: In the absence of friction, there are only two forces acting on the car, namely its weight mvectorg and the normal force vector N provided by the road. The weight is directed vertically down while the normal force is directed perpendicular to the banked road surface and thus at the angle θ from the vertical. The centripetal force due to banking comes from the horizontal component of the normal force, F banking c = N sin θ . The free body diagram in the vertical di- rection gives N cos θ = mg or N = mg cos θ and horizontally gives F banking c = N sin θ = mg cos θ sin θ = mg tan θ = 10909 . 4 N . Question 3 part 3 of 5 10 points Now suppose the friction force is sufficient to keep the car from skidding. Calculate the magnitude of the normal force exerted on the car by the road’s surface. Hint: Check the correctness of your answer to the first question before proceeding with this and the following questions. Correct answer: 36925 N (tolerance ± 1 %). Explanation: Consider the free body diagram for the car m g s i n θ N = m g c o s θ μ N | W bardbl- m a bardbl | mg W bardbl = mg sin θ a bardbl = v r 2 cos θ Now in addition to the car’s weight mvectorg and the normal force vector N there is also the friction force vector F f whose direction is across the road but along the road’s surface. In other words, vector F f lies in the same vertical plane as the net centripetal force vector F c that makes the car follow practice 05 – GADHIA, TEJAS – Due: Feb 15 2007, 2:00 pm 2 the curving road and at the angle θ below the horizontal direction in that plane. The net force vector F net = mvectorg + vector N + vector F f must be equal to the centripetal force vector F c since otherwise the car would skid. Writing this vector equation in terms of the horizontal and the vertical components, we have F c = N sin θ + F f cos θ (1) 0 = N cos θ- F f sin θ- mg . (2) Solve Eqs. 1 & 2 equations for N , by isolating F f we obtain F f sin θ = N cos θ- mg F f cos θ = F c- N sin θ , dividing the two equation we have sin θ cos θ = N cos θ- mg F c- N sin θ , by cross multiplication we have N cos 2...
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Practice 05 - practice 05 – GADHIA TEJAS – Due 2:00 pm 1 Question 1 part 1 of 5 10 points A car travels at a speed of 20 m s around a curve of

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