An almirah door is 6 ft high and 1.5 ft wide, weighing 8 kg . It is...

Let the acceleration due to gravity be $g = 10 \text{ m/s}^2$.
Weight of the door, $W = mg = 8 \times 10 = 80 \text{ N}$.

Let the horizontal and vertical components of the reaction force at the upper hinge be $H_1$ and $V_1$, and at the lower hinge be $H_2$ and $V_2$.

For translational equilibrium in the horizontal and vertical directions:
$\Sigma F_x = 0 \Rightarrow H_1 + H_2 = 0 \Rightarrow H_1 = -H_2$
$\Sigma F_y = 0 \Rightarrow V_1 + V_2 = W = 80 \text{ N}$

It is given that the magnitudes of the forces exerted by both hinges are identical:
$\sqrt{H_1^2 + V_1^2} = \sqrt{H_2^2 + V_2^2}$

Since $H_1^2 = H_2^2$, we get $V_1^2 = V_2^2$.
As $V_1 + V_2 = 80 \text{ N}$, we must have $V_1 = V_2 = 40 \text{ N}$.

The hinges are placed $1 \text{ ft}$ from each end of the $6 \text{ ft}$ high door.
Distance between the hinges, $d = 6 - 1 - 1 = 4 \text{ ft}$.
The center of mass of the door is at its geometric center, so its horizontal distance from the axis of the hinges is $x = \dfrac{1.5}{2} = 0.75 \text{ ft}$.

For rotational equilibrium, taking torque about the lower hinge:
$H_1 \times d + W \times x = 0$
$H_1 \times 4 + 80 \times 0.75 = 0$
$4 H_1 = -60 \Rightarrow H_1 = -15 \text{ N}$

The magnitude of the force exerted by each hinge is:
$F = \sqrt{H_1^2 + V_1^2} = \sqrt{(-15)^2 + 40^2}$
$F = \sqrt{225 + 1600} = \sqrt{1825} \approx 42.72 \text{ N}$

Rounding to the nearest integer, the magnitude is $43 \text{ N}$.

Answer: $43 \text{ N}$