4-2.Friction
hard

$W$ વજનવાળો બ્લોક સમક્ષિતિજ સપાટી પર સ્થિત ઘર્ષણાંક $\mu$ થી સ્થિર છે. બ્લોક પર ન્યુનત્તમ મૂલ્યનું બળ લગાવીને તેને ગતિ કરાવવામાં આવે છે. સમક્ષિતિજથી એવો ખૂણો $\theta $ કે જ્યાથી બળ લગાવવામાં આવે અને બળનું મૂલ્ય અનુક્રમે શું થાય?

A

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

B

$\theta \, = {\tan ^{ - 1}}\,\left( {\frac{1}{\mu }} \right),\,\,F = \frac{{\mu W}}{{\sqrt {1 + {\mu ^2}} }}$

C

$\theta \, = 0,\,\,F = \mu W$

D

$\theta \, = {\tan ^{ - 1}}\,\left( {\frac{\mu }{{1 + \mu }}} \right),\,\,F = \frac{{\mu W}}{{1 + \mu }}$

(AIEEE-2012)

Solution

$\begin{array}{l}
Let\,the\,force\,F\,is\,applied\,at\,an\,angle\,\theta \\
with\,the\,horizontal.\\
For\,horizontal\,equilibrium,\,\\
F\cos \theta  = \mu R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
For\,vertical\,equilibrium,\\
R + F\sin \theta  = mg\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
or,\,R = mg – F\sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,….\left( {ii} \right)\\
Substituting\,this\,value\,of\,R\,in\,eq.\left( i \right),
\end{array}$

$\begin{array}{l}
we\,get\\
F\cos  = \mu \left( {mg – F\sin \theta } \right)\\
 = \mu \,mg – \,\mu F\sin \theta \\
or,\,F\,\left( {\cos \theta  + \mu \sin \theta } \right) = \mu mg\\
or,\,\,F = \frac{{\mu mg}}{{\cos \theta  + \mu \sin \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {iii} \right)\\
For\,F\,to\,be\,m{\rm{inimum,}}\,{\rm{the}}\,{\rm{denominator}}\\
\left( {\cos \theta  + \mu \sin \theta } \right)\,should\,be\,{\rm{maximum}}.
\end{array}$

$\begin{array}{l}
\therefore \,\frac{d}{{d\theta }}\left( {\cos \theta  + \mu \sin \theta } \right) = 0\\
or,\, – \sin \theta  + \mu \cos \theta  = 0\\
or,\,\tan \theta  = \mu \\
or,\theta  = {\tan ^{ – 1}}\left( \mu  \right)\\
Then,\,\sin \theta \, = \frac{\mu }{{\sqrt {1 + {\mu ^2}} }}\,and
\end{array}$

$\begin{array}{l}
\cos \theta  = \frac{1}{{\sqrt {1 + {\mu ^2}} }}\\
Hence,\,{F_{\min }}\\
 = \frac{{\mu w}}{{\frac{1}{{\sqrt {1 + {\mu ^2}} }} + \frac{{{\mu ^2}}}{{\sqrt {1 + {\mu ^2}} }}}} = \frac{{\mu w}}{{\sqrt {1 + {\mu ^2}} }}
\end{array}$

Standard 11
Physics

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