Multiple-Concept Example 7 reviews the concepts that play a role in this problem. Car A uses tires for which the coefficient of static friction is 1.1 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 25 m/s. Car B uses tires for which the coefficient of static friction is 0.85 on the same curve. What is the maximum speed at which car B can negotiate the curve?

Concept Example 7

At what maximum speed can a car safely negotiate a horizontal unbanked turn (radius = 51 m) in dry weather (coefficient of static friction = 0.95) and icy weather (coeffi cient of static friction = 0.10)?

Reasoning

The speed υ at which the car of mass m can negotiate a turn of radius r is related to the centripetal force Fc that is available, according to F_c = mυ²/r (Equation 5.3). Static friction provides the centripetal force and can provide a maximum force f_s^MAX given by f_s^MAX = μ_sF_N (Equation 4.7), in which μ_s is the coefficient of static friction and F_N is the normal force. Thus, we will use the fact that F_c = f_s^MAX to obtain the maximum speed from Equation 5.3. It will also be necessary to evaluate the normal force, which we will do by considering that it must balance the weight of the car. Experience indicates that the maximum speed should be greater for the dry road than for the icy road.

Description Symbol Value Comment Radius of turn 51 m Coefficient of static friction 0.95 Dry conditions Coefficient of static friction 0.10 Icy conditions Unknown Variable Speed of car