What Coefficient of Friction Do Cars Need on a Flat Curve? (a) Calculate the centripetal force exerted on a 900.0-kg car that negotiates a 500.0-m radius curve at 25.00 m/s. (b) Assuming an unbanked curve, find the minimum static coefficient of friction between the tires and the road, static friction being the reason that keeps the car from slipping (see figure below). Free-body diagram Ñ C N f W This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway. Strategy 1. We know that Fc = mv2. Thus, F = m.2 (500.0m) (900.0 kg) (25.00 m/s)2 =1125 N. 2. The free-body diagram above shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is pas IV, where JAs is the static coefficient of friction and N is the normal force. The normal force equals the car's weight on level ground, so N = mg. Thus the centripetal force in this situation is F .= f=M,N = 11,mg. Now we have a relationship between centripetal force and the coefficient of friction. Using the equation F. = m .- 22 we obtain m- T = 14smg. We solve this for /45, noting that mass cancels, and obtain rg v2 Substituting the knowns, (25.00 m/s)2 Hs= (500.0m) (9.80 m/s2) =0.13. (Because coefficients of friction are approximate, the answer is given to only two digits.)