We are dealing with elastic cords cut from the same long cord. This implies they share the same material properties ($Y$) and cross-sectional area ($A$). The spring constant (stiffness) of a cord is inversely proportional to its natural length:

$$ k = \frac{YA}{L} \propto \frac{1}{L} $$

From the geometry (assuming small extension where angles $\theta$ are preserved):

• Natural length of middle cord: $L_m = h$.
• Natural length of outer cords: $L_s = \frac{h}{\cos \theta}$.

Therefore, the relationship between their stiffnesses is:

$$ k_m \propto \frac{1}{h} \quad , \quad k_s \propto \frac{\cos \theta}{h} = k_m \cos \theta $$

Let the knot P move down by a small distance $\delta$.

• Extension in middle cord: $\Delta_m = \delta$.
• Extension in outer cords: $\Delta_s = \delta \cos \theta$ (component of displacement along the cord).

Calculate the Tensions ($T = k \Delta$):

$$ T_{mid} = k_m \delta $$ $$ T_{side} = k_s \Delta_s = (k_m \cos \theta)(\delta \cos \theta) = k_m \delta \cos^2 \theta $$

Thus, $T_{side} = T_{mid} \cos^2 \theta$.

Consider the vertical equilibrium of the knot P:

$$ T_{mid} + 2 T_{side} \cos \theta = mg $$

Substitute $T_{side}$ in terms of $T_{mid}$:

$$ T_{mid} + 2 (T_{mid} \cos^2 \theta) \cos \theta = mg $$ $$ T_{mid} (1 + 2 \cos^3 \theta) = mg $$

Solving for the tension in the middle cord ($T = T_{mid}$):

$$ T = \frac{mg}{1 + 2 \cos^3 \theta} $$