कोई कार त्रिज्या R की वक्रित सड़क पर गतिमान ह | vedclass

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Question

$\begin{array}{l}
For\,vertical\,equilibrium\,on\,the\,road,\
N\cos 	heta  = mg + f\sin 	heta \
mg = N\cos 	heta  - f\sin 	heta \,\

Vedclass · Accepted Answer

$\sqrt {gR\frac{{{\mu _s} + tan	heta }}{{1 - {\mu _s}tan	heta }}} \;\;\;\;$

Vedclass · Answer

$\begin{array}{l}
For\,vertical\,equilibrium\,on\,the\,road,\
N\cos 	heta  = mg + f\sin 	heta \
mg = N\cos 	heta  - f\sin 	heta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)\
Centripetal\,force\,for\,safe\,turning\,,\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N\sin 	heta  + f\cos 	heta  = \frac{{m{v^2}}}{R}\,\,\,\,\,\,...\left( {ii} ight)\
From\,eqns.\,\left( i ight)\,and\,\left( {ii} ight),\,we\,get\
\,\,\,\,\,\,\,\,\,\,\,\frac{{{v^2}}}{{Rg}} = \frac{{N\sin 	heta  + f\cos 	heta }}{{N\cos 	heta  - f\sin 	heta }}
\end{array}$

$\begin{array}{l}
 \Rightarrow \,\frac{{{v^2}_{\max }}}{{Rg}} = \frac{{N\sin 	heta  + {\mu _s}N\cos 	heta }}{{N\cos 	heta  - {\mu _s}N\sin 	heta }}\
\,\,\,\,\,\,{v_{\max }} = \sqrt {Rg\left( {\frac{{{\mu _s} + 	an 	heta }}{{1 - {\mu _s}	an 	heta }}} ight)} 
\end{array}$

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