Question

A curved stretch of an express way of radius of curvature 300 m is banked at an angle of 5 degree to the horizontal.&
If the limiting friction between the road surface and the tyres of a car is 4000N, what is the maximum speed of the car sothat it will not skip off the road?

Answer 1

Second Newton's law :

$$\left\{ \begin{array}{c} O X: \frac {m v ^ {2}}{R} \cos (\alpha) + m g \sin (\alpha) = F _ {f r} = \mu N \\ O Y: m g \cos (\alpha) = N + \frac {m v ^ {2}}{R} \sin (\alpha) \end{array} \right.$$ $$F _ {f r} = 4 0 0 0 N - \text {friction force}$$ $$\mu - \text {coefficient of friction}$$ $$v - \text {maximum speed}$$ $$R = 3 0 0 m - \text {radius of curvature}$$ $$\alpha = 5 \text { degrees}$$ $$m - \text {mass of the car}$$ $$N - \text {force of normal reaction of the surface}$$ $$\frac {v ^ {2}}{R} - \text {centripetal acceleration}$$ $$g - \text {free fall acceleration}$$

The input data given is not enough to solve this problem. Let we assume that the mass of the car $m$ is given.

Then, from equation (1)

$$v = \sqrt {\frac {F _ {f r} - m g \sin (\alpha)}{m \cos (\alpha)} R}$$

The average mass of the car is about $m = 1500 \, kg$. Then

$$v = 2 3, 3 \frac {m}{s}$$

Answer:

$$v = \sqrt {\frac {F _ {f r} - m g \sin (\alpha)}{m \cos (\alpha)} R}$$

For $m = 1500 \, kg$ $= >$ $v = 23,3 \, \frac{m}{s}$