Ahighway curves to the left with radius of

curvature of 39 m and is banked at 30 ◦

so that cars can take this curve at higher speeds.

consider a car of mass 919 kg whose tires

have a static friction coefficient 0.72 against

the pavement.

how fast can the car take this curve without

skidding to the outside of the curve? the

acceleration of gravity is 9.8 m/s^2.

answer in units of m/s.

(use image as needed)

29.131 m/s

Explanation:

Draw a free body diagram.  There are three forces: weight force mg pulling down, normal force N pushing perpendicular to the road, and friction force Nμ pushing parallel to the road downwards.

Sum of forces in the centripetal direction:

∑F = ma

N sin θ + Nμ cos θ = m v² / r

N (sin θ + μ cos θ) = m v² / r

Sum of forces in the vertical direction:

∑F = ma

N cos θ − Nμ sin θ − mg = 0

N (cos θ − μ sin θ) = mg

N = mg / (cos θ − μ sin θ)

Substituting:

mg (sin θ + μ cos θ) / (cos θ − μ sin θ) = m v² / r

g (sin θ + μ cos θ) / (cos θ − μ sin θ) = v² / r

v² = gr (sin θ + μ cos θ) / (cos θ − μ sin θ)

Plugging in values:

v² = (9.8 m/s²) (39 m) (sin 30° + 0.72 cos 30°) / (cos 30° − 0.72 sin 30°)

v = 29.131 m/s

true tan speed = ang vel x radius