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Assignment 5 Solutions (Fall 2008)

# Assignment 5 Solutions (Fall 2008) - P207 Fall 2008...

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P207 - Fall 2008 Solutions Assignment 5 1 (a) F friction W F N ? 6 b (b) The sum of the forces in the horizontal direction F x = ma x . The acceleration is a x = - 1 2 v 2 d , (1) where v is the initial velocity and d the distance over which the velocity slows to zero. The horizontal force is due to the friction of tires on the road, F x = F friction = - μ k F N . F N = mg is the normal force. Combining the above equations we can write F x = - mgμ k = - 1 2 mv 2 d (2) Then solve for μ k = 1 2 v 2 gd = 1 2 (30 m / s) 2 (10 m / s 2 )(40 m) = 1.125 (c) We rearrange Equation 2 to solve for d = v 2 2 k (3) 1

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and we see that the stopping distance d is independent of the mass of the car . It takes more force to stop (decelerate) a car in proportion to its mass. But the force of kinetic friction also scales as the mass of the car. (d) According to Equation 3 the stopping distance is inversely proportional to μ k . (i) If μ k = 0 . 25, d = (1 . 125 / 0 . 25)(40 m) = 180 m (ii) If μ k = 0 . 1, d = (1 . 125 / 0 . 1)(40 m) = 450 m (e) F friction W F N ? y b with F friction F N . (f) Choose the x-axis parallel to the slope of the hill. Then X F x = W sin θ - F friction = ma x . (4) The frictional force F friction = μ k F N . In order to determine the normal force we note that the acceleration in the y direction is zero. Therefore X F y = - W cos θ + F N = 0 and F N = W cos θ = mg cos θ (5) Solve Equation 4 for a x and use Equation 5 to get a x = 1 m ( mg sin θ - μ k mg cos θ ) = g (sin θ - μ k cos θ ) 2
Note: if μ k cos θ > sin θ , a x is negative, as expected. Then using Equation 1 we write stopping distance d in terms of velocity v and acceleration a d = - 1 2 v 2 a x = - 1 2 v 2 g (sin θ - μ k cos θ ) For a 12% grade, θ = tan - 1 (0 . 12) = 6 . 8 . Then d = - 1 2 (30 m / s) 2 (10 m / s 2 )(sin 6 . 8 - 1 . 125 cos 6 . 8 ) = 46 m (g) As long as the tires are not skidding, the frictional force between tires and road is static. When the tires begin to skid, the frictional force is kinetic. Since static friction is typically greater than kinetic friction, braking distance is minimized if the tires do not skid. 2. Chapter 6, Problem 20 (a) Suppose that the x and y axis are parallel and perpendicular to the plane respectively. The plane is inclined at an angle θ = 20 . We are looking for the least force F required to keep the sled from slipping. That implies that the force of friction is directed up the plane, opposite that of gravity. Then F x = F + F friction - W sin θ (6) = F + μ s F N - W sin θ (7) F y = F N - W cos θ (8) where F is the force applied in the positive x-direction, - W sin θ is the component of the force of gravity in the x-direction. F friction is the force of static friction opposing the force of gravity. To minimize

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Assignment 5 Solutions (Fall 2008) - P207 Fall 2008...

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