A 1.03 tonne car travels on a circular section of flat track with instantaneous

Question

A 1.03 tonne car travels on a circular section of flat track with instantaneous radius of curvature of rho = 176m. The speed of the car is v = 65 km/h. g = 9.8m/s^2: The car is increasing speed at a rate of a_t = 4.9m/s^2. (a) What is the sideways static friction force f_N exerted by the road on the car? This is the force that turns the car! (b) What is the minimum coefficient of static friction mu_s between car and road so that the car does not slide? (c) Neglecting aerodynamic drag (Unlikely to be reasonable with a car at this speed!), What is the net static friction force f exerted by the road on the car?

Explanation / Answer

sideways static frictional force is fs = centripetal force = m*v^2/r =

given that m = 1.03 tonne = 1030 kg

v = 65*5/18 = 18.05 m/sec

then

fs = m*v^2/r = (1030*18.05^2)/176

fs = 1906.7 N

b) fs = mu_s*N

N is the normal force = m*g

then

fs = mu_s*m*g

1906.7 = mu_s*1030*9.8

mu_s = 0.188

C) Net static firctional force is Fnet = m*a = 1030*4.9 = 5047 N

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