Let the rod be tilted by a small angle $\alpha$. The extensions in the springs are $x_1$ and $x_2$. The vertical distance from the ceiling to the rod ends is $L_{spring}$. $$ L_{spring1} = D – \frac{L}{2}\sin\alpha $$ $$ L_{spring2} = D + \frac{L}{2}\sin\alpha $$ Extension $x = L_{spring} – l_0$. Taking torque about the central hinge: $$ \tau_{net} = 0 $$ $$ (k_1 x_1) \frac{L}{2} \cos\alpha = (k_2 x_2) \frac{L}{2} \cos\alpha $$ $$ k_1 x_1 = k_2 x_2 $$ Let the mean extension at the horizontal position be $x_0 = D – l_0 = 1.0 – 0.4 = 0.6 \text{ m}$. Let $\delta$ be the vertical displacement of the ends due to tilt ($\delta = \frac{L}{2}\sin\alpha$). Then $x_1 = x_0 – \delta$ and $x_2 = x_0 + \delta$. $$ k_1 (x_0 – \delta) = k_2 (x_0 + \delta) $$ $$ 25(0.6 – \delta) = 15(0.6 + \delta) $$ $$ 15 – 25\delta = 9 + 15\delta $$ $$ 6 = 40\delta \implies \delta = \frac{6}{40} = 0.15 \text{ m} $$

Now we calculate the individual spring forces (upward pull on the rod): $$ F_1 = k_1(0.6 – 0.15) = 25(0.45) = 11.25 \text{ N} $$ $$ F_2 = k_2(0.6 + 0.15) = 15(0.75) = 11.25 \text{ N} $$ Note: The forces are equal, which is consistent with the torque balance about the midpoint. Total upward force from springs: $$ F_s = F_1 + F_2 = 11.25 + 11.25 = 22.5 \text{ N} $$

Equilibrium of the Rod: Let $N_{hinge}$ be the normal force exerted by the pedestal on the rod (upward). $$ F_s + N_{hinge} = Mg $$ $$ 22.5 + N_{hinge} = 10 \times 10 = 100 $$ $$ N_{hinge} = 100 – 22.5 = 77.5 \text{ N} $$ Equilibrium of the Pedestal: The rod exerts a downward force of $N_{hinge} = 77.5 \text{ N}$ on the pedestal. Let $N_{floor}$ be the reaction from the floor. $$ N_{floor} = N_{hinge} + mg $$ $$ N_{floor} = 77.5 + 0.25 \times 10 $$ $$ N_{floor} = 77.5 + 2.5 = 80 \text{ N} $$

Answer: The normal reaction between the pedestal and the floor is 80 N.