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Unformatted text preview: homework 05 – GADHIA, TEJAS – Due: Feb 21 2007, 4:00 am 1 Question 1 part 1 of 1 10 points Given: g = 9 . 8 m / s 2 . To test the performance of its tires, a car trav els along a circular track of radius R = 379 m. The track is perfectly flat (no banking). The car increases its speed at uniform rate a t ≡ d  v  dt = 2 . 07 m / s 2 until the tires start skidding. If the tires start skidding when the car reaches the speed v = 30 . 2 m / s, what is the coefficient of static friction between the tires and the road? Correct answer: 0 . 323903 (tolerance ± 1 %). Explanation: First, let us calculate the car’s acceleration just before the tires skidded away. The car travels in a circle rather than along a straight line, hence it has the centripetal accleleration a c = v 2 R = 2 . 40644 m / s 2 . But in addition, the car increases its speed, hence we also have a tangential acceleration a t ≡ d  v  dt = 2 . 07 m / s 2 . The net acclerations of the car is the vector sum vector a = vector a c + vector a t , and since the tangential and the centripetal accelerations have perpendicular directions, the magnitude of the net accelerations is  a  = radicalBig a 2 c + a 2 t = 3 . 17425 m / s 2 . The force providing this acceleration is the static friction force vector F s between the tires and the track — indeed, it is the only horizontal force available on the flat track. The static friction force may have any horizontal direc tion, and any magnitude up the the maximum of F max s = μ s × N = μ s × Mg. Consequenly, the net horizontal acceleration of the car is limited by M  a  ≤ μ s × Mg = ⇒  a  ≤ μ s × g. When the car tries to exceed this acceleration limit, the tires skid. Therefore, we may find the static friction coefficient according to μ s =  a  max g = 0 . 323903 . Question 2 part 1 of 3 10 points A curve of radius r is banked at angle θ so that a car traveling with uniform speed v can round the curve without relying on friction to keep it from slipping to its left or right. The acceleration of gravity is 9 . 8 m / s 2 . m μ ≈ θ Find the component of the net force parallel to the incline summationdisplay vector F bardbl . 1. F = mg tan θ 2. F = mv 2 r cos θ 3. F = mv 2 r sin θ 4. F = mg cot θ 5. F = mg cos θ 6. F = mv 2 r sin θ homework 05 – GADHIA, TEJAS – Due: Feb 21 2007, 4:00 am 2 7. F = mv 2 r tan θ 8. F = mv 2 r tan θ 9. F = mv 2 r cos θ correct 10. F = m radicalbigg g 2 + v 4 r 2 Explanation: Basic Concepts: To keep an object mov ing in a circle requires a force directed toward the center of the circle; the magnitude of the force is F c = ma c = mv 2 r Also remember: vector F = summationdisplay i vector F i Solution: Solution in an Inertial Frame: Watching from the “Point of View of Some one Standing on the Ground”....
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 Spring '08
 Turner
 Physics, Force, Friction, Work, Correct Answer, Wnet

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