A car is negotiating a curved road of radius | vedclass

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Question

(A) For vertical equilibrium on the road:$N \cos	heta = mg + f \sin	heta$$mg = N \cos	heta - f \sin	heta$ ... $(i)$For safe turning,the centripeta

Vedclass · Accepted Answer

$\sqrt{gR\frac{\mu_s + 	an	heta}{1 - \mu_s	an	heta}}$

Vedclass · Answer

(A) For vertical equilibrium on the road:$N \cos	heta = mg + f \sin	heta$$mg = N \cos	heta - f \sin	heta$ ... $(i)$For safe turning,the centripetal force is provided by the horizontal components of the normal force and friction:$N \sin	heta + f \cos	heta = \frac{mv^2}{R}$ ... $(ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{v^2}{Rg} = \frac{N \sin	heta + f \cos	heta}{N \cos	heta - f \sin	heta}$At maximum velocity,the friction force $f$ reaches its limiting value,$f = \mu_s N$. Substituting this into the equation:$\frac{v_{\max}^2}{Rg} = \frac{N \sin	heta + \mu_s N \cos	heta}{N \cos	heta - \mu_s N \sin	heta}$Dividing the numerator and denominator by $N \cos	heta$:$\frac{v_{\max}^2}{Rg} = \frac{	an	heta + \mu_s}{1 - \mu_s 	an	heta}$Therefore,the maximum safe velocity is:$v_{\max} = \sqrt{gR \left( \frac{\mu_s + 	an	heta}{1 - \mu_s 	an	heta} ight)}$

$A$ car is negotiating a curved road of radius $R$. The road is banked at an angle $\theta$. The coefficient of friction between the tyres of the car and the road is $\mu_s$. The maximum safe velocity on this road is

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