Question

In the figure shown, for minimum value of $\mu$, mass $M$ does not move. If $\frac{1}{\mu }=(\alpha {)}^{3/2}$, then find the value of $\alpha$.
(Here $m_1=1kg,m_2=2kg,M=3/2kg$ and take $g=10m{s}^{-2}$ ]

Answer 1

Answer: 3

Net horizontal force on a bigger block,
$\left(N_2-N_1\right)\sin \theta =\mu N_{\text{Net }}$
Now,
$N_{\text{Net }}=\left(N_1+N_2\right)\cos \theta +Mg$

$N_1=m_{1}g\cos \theta$
$N_2=m_{2}g\cos \theta$
$\Rightarrow \mu =\frac{\left(m_2-m_1\right)g\cos \theta \sin \theta }{\left(m_1+m_2\right)g{\cos }^2\theta +Mg}$
$\mu =\frac{(1)\left(\frac{1}{2}\right)\frac{\sqrt{3}}{2}}{(3)\frac{1}{4}+\frac{3}{2}}=\frac{\sqrt{3}}{9}=(3{)}^{-3/2}$
$\Rightarrow \frac{1}{\mu }=(\alpha {)}^{3/2}$
$\Rightarrow \alpha =3$