A man has fallen into a ditch of width d , and two of his friends...

Let $T$ be the force applied by each friend. Since the man is pulled up slowly, he is in equilibrium.

The horizontal components of the tension forces balance each other, which implies the man is exactly in the middle of the ditch. Thus, the horizontal distance from the man to each pulley is $\dfrac{d}{2}$.

Let $\theta$ be the angle each rope makes with the vertical. From the geometry of the system, the length of the rope from the pulley to the man is given by:

$L = \sqrt{h^2 + \left(\dfrac{d}{2}\right)^2} = \dfrac{\sqrt{4h^2 + d^2}}{2}$

The cosine of the angle $\theta$ is:

$\cos\theta = \dfrac{h}{L} = \dfrac{2h}{\sqrt{d^2 + 4h^2}}$

Balancing the forces in the vertical direction, the upward components of the tension from both ropes must equal the weight of the man:

$2T \cos\theta = mg$

Substituting the value of $\cos\theta$ into the equation:

$2T \left( \dfrac{2h}{\sqrt{d^2 + 4h^2}} \right) = mg$

$\dfrac{4Th}{\sqrt{d^2 + 4h^2}} = mg$

Solving for $T$, we get the force applied by each friend:

$T = \dfrac{mg\sqrt{d^2 + 4h^2}}{4h}$