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 $m_{1}=1 \,kg , m_{2}=2 \,kg , M=3 / 2\, kg$ and take $g=10 \,ms ^{-2}$ ]

Answer 1

Net horizontal force on a bigger block,
$\left(N_{2}-N_{1}\right) \sin \theta=\mu N_{\text {Net }}$
Now,
$N_{\text {Net }}=\left(N_{1}+N_{2}\right) \cos \theta+M g$

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