Mark asked in Science & MathematicsPhysics · 6 years ago

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?

2 Answers

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  • oubaas
    Lv 7
    6 years ago
    Favorite Answer

    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

  • NCS
    Lv 7
    6 years ago

    (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!

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