Engineering a highway curve . If a car goes through acurve too fast, the car ten

Question

Engineering a highway curve. If a car goes through acurve too fast, the car tends to slide out of the curve. For abanked curve with friction, a frictional force acts on a fast carto oppose the tendency to slide out of the curve; the force isdirected down the bank (in the direction water would drain).Consider a circular curve of radius R = 220 m and bankangle , where the coefficient of static friction betweentires and pavement is s. A car (without negativelift) is driven around the curve as shown in Figure 6-13. Find anexpression for the car speed vmax that puts thecar on the verge of sliding out, in terms of R, ,and s. In kilometers per hour, evaluatevmax for a bank angle of = 11° andfor (a) s = 0.53 (dry pavement)and (b) s = 0.047 (wet or icypavement). (Now you can see why accidents occur in highway curveswhen icy conditions are not obvious to drivers, who tend to driveat normal speeds.)

Explanation / Answer

From free body diagram we can write                    -FNr Sin =m(-v2/R)         ........1                      FNy Cos =Fg                   ........2 Where FNr = sFNy         ........3            Fg = mg From Eq 1,2 and 3 we get                     - sFg Sin / Cos = m(-v2/R)                        sFg Tan =mv2/R Therefore       v = (R sFgTan / m)          = (Rs g Tan )        Here we have                       R = 220 m                       s = 0.53 (dry pavement)                            = 0.047 (wet or icy pavement)                        g = 9.8 m/s2                         = 11.0o a)   vmax = (220 *0.53*9.8*tan11)                = 14.90 m/s b) vmax = (220 *0.047*9.8*tan11)                = 4.438 m/s

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