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 = 230 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 = 14° andfor (a) s = 0.75 (dry pavement)and (b) s = 0.040 (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

(A). ?F_vertical = 0 N cos - µ N sin - mg= 0 N(cos - µ sin ) = mg N = mg/[(cos - µ sin )] ?F_horizontal = ma N sin + µ N cos = m (Vmax)²/R N(sin + µ cos ) = m (Vmax)²/R mg/[(cos - µ sin )](sin + µ cos) = m (Vmax)²/R divided both side by m, g(tan + µ)/(1 - µ tan ) =(Vmax)²/R Vmax = [gR(tan + µ)/(1 - µ tan)] (B). V(,µ) = [gR(tan + µ)/(1 - µtan )] So you can plot graph Vmax versus angle and static frictioncoefficient µ with equation above by yourself. Good luck... :)

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.