Solution
We analyze the equilibrium of the rope by considering the forces acting at the peg B.
Total length of rope $L = 13$ m.
Length hanging between pegs A and B is $L_{curve} = 8$ m.
Length of the vertical tail is $L_{tail} = 13 – 8 = 5$ m.
Let $\lambda$ be the mass per unit length.
The tension at peg B due to the vertical tail is equal to the weight of the tail:
$$ T_{tail} = \lambda \cdot 5 \cdot g = 5\lambda g $$
Since the peg B is frictionless, the magnitude of the tension is continuous. Thus, the tension in the curved part at point B is also $T = 5\lambda g$.
Consider the equilibrium of the right half of the curved rope (from the lowest point to B).
The vertical component of tension at B supports the weight of half the curved rope ($8/2 = 4$ m).
$$ T_y = \text{Weight of half curve} = \lambda \cdot 4 \cdot g = 4\lambda g $$
Let $\alpha$ be the angle the rope makes with the vertical (wall) at peg A. By symmetry, this is the same as the angle the rope makes with the vertical at peg B.
Resolving the tension $T$ at B:
$$ T \cos \alpha = T_y $$ $$ (5\lambda g) \cos \alpha = 4\lambda g $$ $$ \cos \alpha = \frac{4}{5} $$ $$ \alpha = 37^\circ $$(Since $\cos 37^\circ \approx 0.8$)