Question

A highway curves to the left with radius of
curvature of 44 m and is banked at 16 ◦
so that cars can take this curve at higher speeds.
Consider a car of mass 1563 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/s2. Answer in units of m/s.

Answer 1

The sum of forces in y-direction:

$$0=N\cos{16}-\mu N \sin{16}-mg$$

The sum of forces in x-direction:

$$\frac{mv^2}{r}=N\sin{16}+\mu N \cos{16}=N(\sin{16}+\mu \cos{16})$$

Thus,

$$\frac{mv^2}{r}=mg\frac{(\sin{16}+\mu \cos{16})}{\cos{16}-\mu \sin{16}}$$

$$\frac{v^2}{r}=g\frac{(\sin{16}+\mu \cos{16})}{\cos{16}-\mu \sin{16}}$$

$$\frac{v^2}{44}=9.8\frac{(\sin{16}+0.72 \cos{16})}{\cos{16}-0.72 \sin{16}}$$

$$v=23\frac{m}{s}$$