# Physics: If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the...?

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

Update:

(a) Calculate the ideal speed to take a 120-m radius curve banked at 13.0°.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

Relevance

banked curve data :

r = 120 m

Θ = 13°

aequilibrium in ideal conditions requires that centripetal force Fc = m*ac = m*V^2/r must equate inner sliding force Fi = m*g*tang Θ....so :

m*V^2/r = m*g*tang Θ

m cross

V = √9.8*0.231*120 = 16.482 m/sec (59.34 km/h ; 36.87 mph)

if real speed Vr differs from the ideal speed V, then aequilibrium is maintaines by tires transversal friction μk such that :

m*Vr^2/r±m*g*μk = m*g*tang Θ

in case Vr<V , then m*g*μk is a positive quantity and must sum up to centripetal force and vice-versa for Vr>V

m cross

μk = (g*tanΘ-Vr^2/r)/g = (9.8*0.231-20^2/(3.6^2*120))/9.8 = 0.205

• (a) "Ideal speed" means there is no friction, and the downslope component of gravity is equal to the upslope component of centripetal acceleration.

gsinΘ = (v²/r)cosΘ

v = √(grtanΘ) = √(9.8m/s² * 120m * tan13º) = 16.5 m/s = 59.3 km/h

(b) At speeds less than the ideal speed, the friction points upslope.

normal acceleration a_n = gcosΘ + (v²/r)sinΘ

upslope friction acceleration a_f = µ*a+n = µgcosΘ + µ(v²/r)sinΘ

upslope centrip accel a_c = (v²/r)cosΘ

downslope gravity a_g = gsinΘ

Then gsinΘ = µgcosΘ + µ(v²/r)sinΘ + (v²/r)cosΘ

gsinΘ - (v²/r)cosΘ = µ(gcosΘ + (v²/r)sinΘ

µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ)

v² = (20km/h * 1m/s / 3.6km/h)² = 30.9 m²/s², so

µ = (9.8sin13 - (30.9/120)cos13) / (9.8cos13 + (30.9/120)sin13)

µ = 0.203

Hope this helps!