Block A of mass M is placed on an inclined plane, connected to a string, passing over a pulley as shown in the fig. The other end of the string also carries a block B of mass M. The system is held in the position shown such that triangle APQ lies in a vertical plane with horizontal line AQ in the plane of the inclined surface.

 

Find the minimum coefficient of friction between the inclined surface and block A such that the system remains at rest after it is released. Take  $\theta =\alpha ={45}^o$

$N=\frac{mg}{\sqrt{2}}-\frac{T}{2}=Mg\left[\frac{\sqrt{2}-1}{2}\right]$Fig (a) shows component of tension (acting A) along horizontal and vertical. Fig. (b) Shows forces along the line of greatest slope $(XX\text{'})$ and perpendicular to the incline. In equilibrium $T=mg$ [considering block B] -------- (1)
 

$N=\frac{mg}{\sqrt{2}}-\frac{T}{2}=Mg\left[\frac{\sqrt{2}-1}{2}\right]$ ……………(2)
Forgot about the friction for a moment 
 

Unbalanced force along  $XX\text{'}=$
$N=\frac{mg}{\sqrt{2}}-\frac{T}{2}=Mg\left[\frac{\sqrt{2}-1}{2}\right]$ ………..(3)
And there is a force along  $AQ=\frac{T}{\sqrt{2}}=\frac{mg}{\sqrt{2}}$---------- (4)
The friction will balance the resultant of the force given by (3) and (4)
The resultant has been shown in fig. (C)
 

$$\begin{gathered} \therefore f=Mg\sqrt{{\left(\frac{\sqrt{2}-1}{2}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2} \\ =\frac{\sqrt{5-2\sqrt{2}}}{2}Mg \\ f\le \mu N \\ \begin{array}{l}=\frac{\sqrt{5-2\sqrt{2}}}{2}Mg\le \mu \left(\frac{\sqrt{2}-1}{2}\right)mg\\ \Rightarrow \frac{\sqrt{5-2\sqrt{2}}}{\sqrt{2}-1}\le \mu \end{array} \end{gathered}$$