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(C) On a banked road with friction,the forces acting on the vehicle are the normal reaction $R$,the frictional force $f = \mu R$,the weight $mg$,and t

Vedclass · Accepted Answer

$	an 	heta=\frac{v^2-\mu r g}{r g+\mu v^2}$

Vedclass · Answer

(C) On a banked road with friction,the forces acting on the vehicle are the normal reaction $R$,the frictional force $f = \mu R$,the weight $mg$,and the centripetal force $\frac{mv^2}{r}$.Resolving forces in the horizontal and vertical directions:Horizontal: $R \sin 	heta + f \cos 	heta = \frac{mv^2}{r}$Vertical: $R \cos 	heta - f \sin 	heta = mg$Dividing the horizontal equation by the vertical equation:$\frac{R \sin 	heta + \mu R \cos 	heta}{R \cos 	heta - \mu R \sin 	heta} = \frac{mv^2/r}{mg}$$\frac{\sin 	heta + \mu \cos 	heta}{\cos 	heta - \mu \sin 	heta} = \frac{v^2}{rg}$Dividing numerator and denominator by $\cos 	heta$:$\frac{	an 	heta + \mu}{1 - \mu 	an 	heta} = \frac{v^2}{rg}$Cross-multiplying:$rg(	an 	heta + \mu) = v^2(1 - \mu 	an 	heta)$$rg 	an 	heta + \mu rg = v^2 - \mu v^2 	an 	heta$$	an 	heta(rg + \mu v^2) = v^2 - \mu rg$$	an 	heta = \frac{v^2 - \mu rg}{rg + \mu v^2}$

$A$ vehicle is moving with speed $v$ on a curved road of radius $r$. The coefficient of friction between the vehicle and the road is $\mu$. The angle $\theta$ of banking needed is given by

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