4.5k views
0 votes
Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. Neglect the effect of air drug and rolling friction. If the coefficient of static friction between and the road and the tires is .30 what is the safe range of speed?

User Jlliagre
by
5.0k points

1 Answer

4 votes

Answer:


v = 15.56 m/s


v = 56 km/h

Step-by-step explanation:

When coefficient of friction is approximately zero then we have


F_ncos\theta = mg


F_n sin\theta = (mv^2)/(R)


tan\theta = (v^2)/(Rg)

here we know that


v = 40 km/h = 11.11 m/s

R = 30 m


tan\theta = (11.11^2)/(30* 9.81)


\theta = 22.75 degree

now when friction coefficient is 0.30 then we have


F_n cos\theta = mg + F_f sin\theta


F_f cos\theta + F_n sin\theta = (mv^2)/(R)

now we have


v = \sqrt{Rg((\mu + tan\theta)/(1 - \mu tan\theta))}


v = \sqrt{30(9.81)((0.30 + tan22.75)/(1 - (0.30) tan22.75))}


v = 15.56 m/s


v = 56 km/h

User Austin Wagner
by
5.1k points