# If a curve with a radius of 90 m is properly banked for a car traveling 68 km/h , what must be the coefficient of static friction for a ca..?

If a curve with a radius of 90 m is properly banked for a car traveling 68 km/h , what must be the coefficient of static friction for a car not to skid when traveling at 92 km/h ?

### 1 Answer

- oldschoolLv 74 weeks agoFavourite answer
68km/h = 68,000m/3600s = 170/9 m/s

We know the outward force up the bank is exactly countered by the weight in line with the bank:mV²/r *cosΘ = m*g*sinΘ

V²/(r*g) = tanΘ

Θ = arctan[170²/(9²*90*9.8)] = 22°

92km/h = 230/9 m/s

The outward force has a component up the bank and causes friction with a component normal to the bank. The weight has a component down the bank and a component normal to the bank increasing friction.The force up the bank is exactly countered by the weight down the bank plus friction caused by weight and outward force:

mV²/r *cos22 = mg*sin22 + mV²/r *µsin22 + µmg*cos22 Mass m divides out:

(V²/r *cos22 - g*sin22)/(V²/r *sin22 + g*cos22) = µ

[230²/(9²*90) *cos22 - 9.8*sin22]/(230²/[(9²*90) *sin22 + 9.8*cos22] = 0.259

µ = 0.26 with just 2 s.d.