Answer to Question #86963 in Mechanics | Relativity for Maria

Question

A curve of radius 75 m is banked for a design speed of 75 km/h.
If the coefficient of static friction is 0.37 (wet pavement), at what range of speeds can a car safely make the curve? [Hint: Consider the direction of the friction force when the car goes too slow or too fast.]
Express your answers using two significant figures separated by a comma.

Answer 1

A banked curve means a car should do the turn at the design speed without participation of acceleration. This information allows us to find out the angle of incline of the road:

$\alpha = \arctan\frac{v_0^2}{gR}; \\$

For smaller speeds (e.g. car sliding downwards):

$F_n\sin\alpha - F_f\cos\alpha = m\frac{v^2}{R}; \\ F_n\cos\alpha + F_f\sin\alpha - mg = 0; \\$

Remembering that $F_f = {\mu}F_n; \\$ and dividing the equations we get

$v_{min} = \sqrt\frac{gR(\tan{(\alpha)}-\mu)}{1+{\mu}\tan\alpha} = \sqrt\frac{gR(v_0^2-{\mu}gR)}{gR+{\mu}v_0^2} \approx 41\frac{km}{h}. \\$

Similarly, for greater speeds (car sliding upwards):

$F_n\sin\alpha + F_f\cos\alpha = m\frac{v^2}{R}; \\ F_n\cos\alpha - F_f\sin\alpha - mg = 0; \\$

$v_{max} = \sqrt\frac{gR(\tan{(\alpha)}+\mu)}{1-{\mu}\tan\alpha} = \sqrt\frac{gR(v_0^2+{\mu}gR)}{gR-{\mu}v_0^2} \approx 108\frac{km}{h}. \\$

When doing the calculation please note that km/h should be converted to m/s and back likewise.

Answer: 41, 108.