Question

A racing car of 1000kg moves round a banked track at a constant speed of 108kmh^-1. Assuming the total reaction at the wheels is normal to the track, and the horizontal radius of the track is 100m, calculate the angle of inclination of the track to the horizontal and the reaction of the wheels.

Answer 1

Answer on Question #73904, Physics / Mechanics | Relativity

Question. A racing car of $1000 \, kg$ moves round a banked track at a constant speed of $108 \, km/h$. Assuming the total reaction at the wheels is normal to the track, and the horizontal radius of the track is $100 \, m$, calculate the angle of inclination of the track to the horizontal and the reaction of the wheels.

Solution.

Given. $m = 1000 \, kg$; $v = 108 \, km/h = 30 \, m/s$; $r = 100 \, m$.

Find. $\alpha, R - ?$

Solution.

Applying Newton's second law, we write

$$+ \uparrow \sum F_y = 0: R \cos \alpha - m g = 0 \rightarrow R = \frac{m g}{\cos \alpha}.$$ $$\stackrel{+}{\rightarrow} \sum F_x = m a_n: R \sin \alpha = m \frac{v^2}{r} \rightarrow \frac{m g}{\cos \alpha} \sin \alpha = m \frac{v^2}{r} \rightarrow \tan \alpha = \frac{v^2}{g r}.$$ $$\alpha = \arctg \alpha = \arctg \frac{v^2}{g r} = \arctg \frac{30^2}{9.81 \cdot 100} = 42.53{}^\circ.$$ $$R = \frac{m g}{\cos \alpha} = \frac{1000 \cdot 9.81}{\cos 42.53{}^\circ} = 13312 \, N.$$

Answer. $\alpha = 42.53{}^\circ$; $R = 13312 \, N$.

Answer provided by https://www.AssignmentExpert.com