Given that

Mass of an object m = 6 kg

Length of the rope = 2 m

Horizontal force applied at mid point F = 50 N

When a given weight is in equilibrium with gravitational force acting on it, the result is T_2 = 6 * 10 = 60   N

Consider the point P's equilibrium under the influence of three forces: the tensions T_1 \text{ and } T_2 , as well as the horizontal force 50 N. The resulting force's horizontal and vertical components must all vanish separately:

T_1 \cos \theta = T_2 = 60N

T_1 \sin \theta = 50N

Direction of the vector \tan \theta= \frac{50}{60}

\theta= \tan^{-1} (\frac{5}{6})

\theta = 40\degree

It is important to note that the answer is independent of both the length of the rope (assumed massless) and the point at which the horizontal force is applied.