Block on Incline (2)

The normal force $F_N$ on the block is given by: $F_N=Mg(sin\theta+cos\theta)$ Therefore, the frictional force: $F=0.5Mg(sin\theta+cos\theta)$ The resultant force on the block is given by: $F_R=Mg(cos\theta-sin\theta)$ The block is at rest if: $F_N \le |F_R|$ $\Rightarrow 0.5Mg(sin\theta+cos\theta) \le |Mg(sin\theta+cos\theta)|$ $\Rightarrow 0.5(sin\theta +cos\theta) \ge \begin{cases} cos\theta -sin\theta \\ \\ sin\theta -cos\theta \end{cases}$ $\div cos\theta \quad \Rightarrow \quad 0.5(tan\theta +1)\ge \begin{cases} 1-tan\theta \\ \\ tan\theta -1 \end{cases}$ $\Rightarrow \quad \begin{cases} 1.5tan\theta \ge 0.5 \\ \\ 1.5\ge 0.5tan\theta \end{cases}\Rightarrow \quad \frac { 1 }{ 3 } \le tan\theta \le 3$ $\Rightarrow ab=\frac{1}{3} \times 3 = \boxed{1}$

Decompose the forces into normal and parallel axes (to the inclined plane). We find that the forces along the plane is $|Mg(\sin\theta-\cos\theta)|$ and that pressing against the wedge is $Mg(\sin\theta+\cos\theta)$ Identify the limiting cases where the whole friction is either acting downwards along the plane or upwards along. We will get the equations as $\frac{\sin\theta - \cos\theta}{\sin\theta+\cos\theta} = \pm \frac { 1 }{ 2 }$ Thus we get $\tan\theta = 3, \frac{1}{3}$ Hence the answer $\boxed{1}$