Rope and platforms

Let the total mass of the rope be $m$ , and let a fraction $f$ of it hang in the air. Consider the right half of this section. Its weight, $(f /2)mg$ , must be balanced by the vertical component, $T sin θ$ , of the tension at the point where it joins the part of the rope touching the right platform. The tension at this point therefore equals $T = (f /2)mg/ sin θ$ .

Now consider the part of the rope touching the right platform. This part has mass $(1 - f)m/2$ . The normal force from the platform is $N = (1 - f)(mg/2) cos θ$ , so the maximal friction force equals $(1 - f)(mg/2) cos θ$ , because $µ = 1$ . This friction force must balance the sum of the gravitational force component along the plane, which is $(1 - f)(mg/2) sin θ$ , plus the tension at the lower end, which is $(f /2)mg/ sin θ$ .

Thus, $(1 - f)mg cos θ$ = $(1 - f)mg sin θ$ + $fmg/sin θ$

Let $F(θ) = cos θ sin θ - sin^2 θ$

Then $f = F(θ)/(1 + F(θ))$

The maximum value of $f$ is therefore where $F(θ)$ is the largest. Since $F'(θ)$ = $cos 2θ - sin 2θ$ , $F(θ)$ is max at 22.5 degrees.

Substituting and solving, f has a maximum value of 0.172