Physics - NLM Question with Solution | TestHub

case (i)



$F_1=Mg sin\theta +f_r$

$=Mg \left(sin\theta +\mu cos \theta \right)$... (i)

Case (ii)



$F_2+f_{\mathrm{r}}=Mg sin\theta$

$F_2=Mg cos\theta -\mu Mg\cos \theta$

$=Mg \left(sin\theta -\mu cos\theta \right)$... (ii)

$F_1=2F_2 \left(Given\right)$

From Eqs. (i) and (ii)

$Mg \left(sin\theta +\mu cos\theta \right)=2Mg \left(sin\theta -\mu cos\theta \right)$

$⟹sin\theta +\frac{1}{4}\cos \theta =2\left(sin\theta -\frac{1}{4}cos\theta \right)$

$⟹cos\theta \left(\frac{1}{4}+\frac{1}{2}\right)=sin\theta$

$⟹\frac{3}{4} cos \theta =sin \theta$

$⟹ tan \theta =\frac{3}{4}$

$⟹tan\theta =\frac{3}{4}$

$⟹\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$