વિધાન: વિરામકોણ (Angle of repose) એ મર્યાદિત | vedclass

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Question

$\begin{array}{l}
The\,{m{maximum}}\,value\,of\,static\,friction\,up\
to\,which\,body\,does\,not\,move\,is\,called\,\
{m{limiting}}\,friction

Vedclass · Accepted Answer

વિધાન અને કારણ બંને સત્ય છે પણ કારણ એ વિધાનની સાચી સમજૂતી નથી.

Vedclass · Answer

$\begin{array}{l}
The\,{m{maximum}}\,value\,of\,static\,friction\,up\
to\,which\,body\,does\,not\,move\,is\,called\,\
{m{limiting}}\,friction.\
\,\,\,\,\,\,\,\,\,Angle\,of\,repose\,is\,defined\,as\,the\,angle\
\,\,\,\,\,\,\,\,\,\,of\,the\,inclined\,plane\,with\,horizontal\
\,\,\,\,\,\,\,\,\,\,such\,that\,a\,body\,placed\,on\,it\,is\,just\,\
\,\,\,\,\,\,\,\,\,\,begins\,to\,slide.\
\,\,\,\,\,\,\,\,\,In\,{m{limiting}}\,{m{condition,}}\
\,\,\,\,\,\,\,\,\,F = mg\sin \alpha \,and\,R = mg\cos \alpha \
\,\,\,\,\,\,\,\,\,where\,\alpha  - angle\,of\,repose.
\end{array}$

$\begin{array}{l}
\,\,So\,\frac{F}{R} = 	an \alpha \
\,\,\,\,\,\,\,\,\,\,	herefore \frac{F}{R} = {\mu _2} = 	an 	heta  = 	an \alpha \,\,\,\left( {	an 	heta  = {\mu _s}} ight)\
\,\,\,\,\,\,\,\,\,\,or\,	heta  = \alpha \
\,\,\,\,\,\,\,\,\,\,i.e,.\,angle\,of\,friciton\, = \,angle\,of\,repose.
\end{array}$

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