Approach

We consider the equilibrium of the pulley. The pulley is supported by the tension in the string passing underneath it. The total downward force acting on the pulley’s axle is the weight of the pulley itself plus the weight of the block suspended from it. This total load is balanced by the vertical components of the tension from the main string. Once the tension is found, we can determine the normal reaction per unit length on the pulley surface using the standard relationship for a string under tension wrapped around a curved surface.

1. Geometry of the System

Let $l$ be the total length of the string and $x$ be the horizontal distance between the nails. The pulley rests at the midpoint due to symmetry. The string forms two hypotenuses of length $l/2$ each.

From the right-angled triangle formed by one half of the string, the horizontal span is $x/2$. If $\theta$ is the angle the string makes with the horizontal, we have: $$ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{x/2}{l/2} = \frac{x}{l} $$

Substituting the given values ($x = 10\sqrt{2}$ m, $l = 20$ m): $$ \cos \theta = \frac{10\sqrt{2}}{20} = \frac{\sqrt{2}}{2} \implies \theta = 45^\circ $$

2. Force Equilibrium

The downward force acting on the pulley is the sum of the pulley’s weight $W_p$ and the suspended block’s weight $W_b$. $$ F_{\text{down}} = W_p + W_b = 10\sqrt{2} + 10\sqrt{2} = 20\sqrt{2} \text{ N} $$

This downward force is balanced by the vertical components of the tension $T$ from both sides of the string: $$ 2T \sin \theta = F_{\text{down}} $$ $$ 2T \sin(45^\circ) = 20\sqrt{2} $$ $$ 2T \left( \frac{1}{\sqrt{2}} \right) = 20\sqrt{2} $$ $$ T\sqrt{2} = 20\sqrt{2} \implies T = 20 \text{ N} $$

3. Normal Reaction per Unit Length

For a string under tension $T$ wrapped around a cylinder of radius $r$, the normal force exerted by the string on the surface per unit length, often denoted as $dF/dl$, is given by: $$ \frac{dN}{dl} = \frac{T}{r} $$

Substituting $T = 20$ N and $r = 10 \text{ cm} = 0.1$ m: $$ \frac{dN}{dl} = \frac{20}{0.1} = 200 \text{ N/m} $$