W વજનવાળો બ્લોક સમક્ષિતિજ સપાટી પર સ્થિત ઘર્ષ | vedclass

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Question

$\begin{array}{l}
Let\,the\,force\,F\,is\,applied\,at\,an\,angle\,	heta \
with\,the\,horizontal.\
For\,horizontal\,equilibrium,\,\
F\cos 	het

Vedclass · Accepted Answer

$	heta \, = {	an ^{ - 1}}\,(\mu ),\,\,F = \frac{{\mu W}}{{\sqrt {1 + {\mu ^2}} }}$

Vedclass · Answer

$\begin{array}{l}
Let\,the\,force\,F\,is\,applied\,at\,an\,angle\,	heta \
with\,the\,horizontal.\
For\,horizontal\,equilibrium,\,\
F\cos 	heta  = \mu R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i ight)\
For\,vertical\,equilibrium,\
R + F\sin 	heta  = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\
or,\,R = mg - F\sin 	heta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {ii} ight)\
Substituting\,this\,value\,of\,R\,in\,eq.\left( i ight),
\end{array}$

$\begin{array}{l}
we\,get\
F\cos  = \mu \left( {mg - F\sin 	heta } ight)\
 = \mu \,mg - \,\mu F\sin 	heta \
or,\,F\,\left( {\cos 	heta  + \mu \sin 	heta } ight) = \mu mg\
or,\,\,F = \frac{{\mu mg}}{{\cos 	heta  + \mu \sin 	heta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} ight)\
For\,F\,to\,be\,m{m{inimum,}}\,{m{the}}\,{m{denominator}}\
\left( {\cos 	heta  + \mu \sin 	heta } ight)\,should\,be\,{m{maximum}}.
\end{array}$

$\begin{array}{l}
	herefore \,\frac{d}{{d	heta }}\left( {\cos 	heta  + \mu \sin 	heta } ight) = 0\
or,\, - \sin 	heta  + \mu \cos 	heta  = 0\
or,\,	an 	heta  = \mu \
or,	heta  = {	an ^{ - 1}}\left( \mu  ight)\
Then,\,\sin 	heta \, = \frac{\mu }{{\sqrt {1 + {\mu ^2}} }}\,and
\end{array}$

$\begin{array}{l}
\cos 	heta  = \frac{1}{{\sqrt {1 + {\mu ^2}} }}\
Hence,\,{F_{\min }}\
 = \frac{{\mu w}}{{\frac{1}{{\sqrt {1 + {\mu ^2}} }} + \frac{{{\mu ^2}}}{{\sqrt {1 + {\mu ^2}} }}}} = \frac{{\mu w}}{{\sqrt {1 + {\mu ^2}} }}
\end{array}$

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