Solution

From the initial moment (left figure) to the moment in question (middle figure), the rope of length $l/6$ moves from the left part to the right part (assuming that the pulley is small). The change in gravitational potential energy is that of the rope of mass $m/6$ moving from the red section to the green section in the middle figure, i.e.

$$\frac{m}{6}\cdot g\cdot\left(-\frac{l}{6}\right)=-\frac{mgl}{36}.$$

If $v$ is the speed of the rope, then conservation of energy yields

$$\frac{1}{2}mv^2-\frac{mgl}{36}=0\qquad\Rightarrow\qquad v^2=\frac{gl}{18}.$$

In a short time interval $\Delta t$ afterward, the rope of length $v\Delta t$ and mass $\Delta m=mv\Delta t/l$ moves from the left part to the right part. The change in gravitational potential energy is that of the rope of mass $\Delta m$ moving from the red section to the green section in the right figure, i.e.

$$\Delta m\cdot g\cdot\left(-\frac{l}{3}-v\Delta t\right)\approx-\frac{mgv\Delta t}{3}.$$

Conservation of energy yields

$$\frac{1}{2}m(v+\Delta v)^2-\frac{1}{2}mv^2-\frac{mgv\Delta t}{3}=0\qquad\Rightarrow\qquad\Delta v\approx\frac{g\Delta t}{3}.$$

Consequently, the change in downward momentum of the right part of the rope is

$$\frac{2m}{3}\cdot\Delta v+\Delta m\cdot v=\frac{5}{18}\cdot mg\Delta t,$$

and the change in upward momentum of the left part of the rope is

$$\frac{m}{3}\cdot\Delta v-\Delta m\cdot v=\frac{1}{18}\cdot mg\Delta t.$$

Thus, the downward impulse on the entire rope is $2mg\Delta t/9$, which means that the total force on the rope is $2mg/9$. Since gravity is $mg$, the upward normal force from the pulley is $7mg/9$, which has the same magnitude as the force that the rope exerts on the pulley.