No Slip Sliding Away

Suppose that the normal reaction and friction force between the $30$ kg and $40$ kg blocks are $R_1,F_1$ , that the normal reaction and friction forces between the $40$ kg and the $50$ kg blocks are $R_2,F_2$ , and that the normal reaction and friction forces between the $50$ kg block and the slope are $R_3,F_3$ .

Resolving forces on the $30$ kg block perpendicular and parallel to the slope, we see that $R_1 \; = \; 15g\sqrt{3} \hspace{2cm} T + F_1 \; = \; 15g$ Treating the $30$ kg and $40$ kg blocks as a single object, and resolving forces similarly, we see that $R_2 \; = \; 35g\sqrt{3} \hspace{2cm} T + F_2 \; = \; 35g$ Treating the three blocks as a single object, and resolving forces, we see that $R_3 \; = \; 60g\sqrt{3} \hspace{2cm} 2T + F_3 \; = \; 60g$ Since $|F_1| \le \mu R_1, |F_2| \le \mu R_2, |F_3| \le \mu R_3$ , we deduce that $T$ must lie in the intersection of the three intervals $\big[15g - 15g\sqrt{3}\mu , 15g + 15g\sqrt{3}\mu\big] \hspace{1cm} \big[35g - 35g\sqrt{3}\mu , 35g + 35g\sqrt{3}\mu\big] \hspace{1cm} \big[30g - 30g\sqrt{3}\mu , 30g + 30g\sqrt{3}\mu\big]$ If $\mu \ge \frac{1}{\sqrt{3}}$ , then $T=0$ is a possible solution, so the string and pulley are not necessary. Let us suppose that $\mu < \frac{1}{\sqrt{3}}$ so that all these intervals are subsets of $(0,\infty)$ . Then the intersection of these three intervals is $\big[35g - 35g\sqrt{3}\mu , 15g + 15g\sqrt{3}\mu\big]$ (with the largest left-hand end and the smallest right-hand end) provided that $35g - 35g\sqrt{3}\mu \le 15g + 15g\sqrt{3}\mu$ , which happens when $\mu \ge \frac{2}{5\sqrt{3}}$ . Otherwise, the intersection is empty. This makes the smallest possible value of $\mu$ equal to $\boxed{0.230940108}$ .

Note that, at this smallest possible value of $\mu = \tfrac{2}{5\sqrt{3}}$ , friction is limiting between the $30$ kg and $40$ kg blocks, and between the $40$ kg and $50$ kg blocks (but not between the $50$ kg block and the slope). Thus, for a slightly smaller value of $\mu$ , the $40$ kg block will slide out between the $30$ kg and $50$ kg blocks.

There would be 2 values of $\mu$ coming but we have to consider the one for which the 40 kg block does'nt slip ! i'll be adding a force diagram to help u understand the directions of friction(as some people r having trouble considering the frictions) and the other forces hope it helps!/........... force-body diagram !! i have considered the forces keeping in mind u get a minimum co-efficient of friction !

A system is in equilibrium if all its parts have a net force of zero, so we can just consider the force on the middle block.

Let $N_1$ be the normal force on the 40 kg block from the 30 kg block and let $N_2$ be the normal force on the 40 kg block from the 50 kg block.

We know that $f \leq \mu_s (N_1 + N_2)$ where $f$ is the total force of friction (from top and bottom surfaces). Adding the force of gravity in the direction of the slope ( $-200 N$ ) to each side of that equation, we get: $F \leq \mu_s (N_1 + N_2) - 200 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

where $F$ is the total force.

Now let's calculate $N_1$ and $N_2$ :

$N_1 = 30g\sin{30^\circ} = \frac{\sqrt{3}}{2} 30g = 15\sqrt{3}g = 150\sqrt{3}$

and $N_2 = N_1 + 40g\sin(30^\circ) = 150\sqrt{3} + 200\sqrt{3} = 350 \sqrt{3}$

Plugging these into (1) and setting $F = 0$ (because of equilibrium):

$0 \leq \mu_s (150\sqrt{3} + 350\sqrt{3}) - 200 = 500\mu_s\sqrt{3} - 200 \implies \mu_s \geq \frac{2}{5\sqrt{3}}$

This implies the minimum value of $\mu_s$ is $\frac{2}{5\sqrt{3}}$ or 0.23094010767 .