hw4solutions - Niu (qn269) HW04 Tsoi (58020) This print-out...

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Niu (qn269) – HW04 – Tsoi – (58020) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 5) 10.0 points A car travels at a speed oF 22 m / s around a curve oF radius 31 m. The acceleration oF gravity is 9 . 8 m / s 2 . 2 . 9 Mg μ 30 What is the net centripetal Force needed to keep the car From skidding sideways? Correct answer: 45277 . 4 N. Explanation: Let : m = 2900 kg , v = 22 m / s , r = 31 m , θ = 30 , and μ = 0 . 529117 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 = 45277 . 4 N . 002 (part 2 oF 5) 10.0 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: 16408 . 3 N. Explanation: In the absence oF Friction, there are only two Forces acting on the car, namely its weight mVg and the normal Force V 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 = 16408 . 3 N . 003 (part 3 oF 5) 10.0 points Now suppose the Friction Force is su±cient 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 frst question beFore proceeding with this and the Following questions. Correct answer: 47251 . 2 N. Explanation: Consider the Free body diagram For the car mg sin θ N = cos μN | W b ma mg W b = mg sin θ a b = v r 2 cos θ Now in addition to the car’s weight mVg and the normal Force V N there is also the Friction Force V F f whose direction is across the road but along the road’s surFace. In other words, V F f lies in the same vertical plane as the net centripetal Force V F c that makes the car Follow
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Niu (qn269) – HW04 – Tsoi – (58020) 2 the curving road and at the angle θ below the horizontal direction in that plane. The net force v F net = mvg + v N + v F f must be equal to the centripetal force v 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) 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
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hw4solutions - Niu (qn269) HW04 Tsoi (58020) This print-out...

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