Rope between inclines

Let the rope have total length $l$ , the hanging portion have length $x$ , the coefficient of static friction be $\mu$ . We wish to maximize $x/l$ .

There are 5 forces that act on the rope: 2 normal forces $N$ , two static friction forces $f$ , and the gravitational force. The forces balance in horizontal direction due to symmetry. In the vertical direction, $2Ncos\theta + 2fsin\theta = mg$ $2N(cos\theta + \mu sin\theta)\ge mg$ .

The normal force comes solely from the portion resting on the platform. The length of rope on either platform is $\frac{l-x}{2}$ , so $N = \frac{l-x}{2}\frac{m}{l}gcos\theta$ .

Substituting into inequality, we have $x/l \le 1- (cos\theta(cos\theta + \mu sin\theta))^{-1}$ , which has minimum $\boxed{0.172}$ at $\theta = \pi/8$ .

I haven't solved the problem myself. I've, in fact, got the answer from here: https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol89.pdf