There is a uniform door as shown in the figure. Its width is h/2 and weight 500N and is fixed with the help of hinges A and B. If the lower hinge supports the entire weight of the door, the magnitude of the force exerted on the door at the lower hinge is

The FBD given shows all the forces acting on the door. The given Condition is R1y = 0 (since the lower hinge only supports the entire weight of the door)

For equilibrium,

${\mathrm{\Sigma F}}_{\mathrm{x}}={\mathrm{R}}_{1\mathrm{x}}+{\mathrm{R}}_{2\mathrm{x}}=0$

$\text{ i.e., }{\mathrm{R}}_{1\mathrm{x}}=-{\mathrm{R}}_{2\mathrm{x}}$

${\mathrm{\Sigma F}}_y={\mathrm{R}}_{2y}-500=0$

$\therefore R_{2\mathrm{y}}=500\mathrm{N}$

Considering the torque about lower hinge,

$\mathrm{\Sigma \tau }={\mathrm{R}}_{1\mathrm{x}}\mathrm{h}+500\times \frac{\mathrm{h}}{4}=0$

$\therefore R_{1x}h=-125h$

$\Rightarrow R_{1x}=-125N\text{ and }$

$R_{2x}=+125N$

Thus R₁ pulls the door to the left. Magnitude of R₂ is

$\left|{\overline{R}}_2\right|=\sqrt{R_{2x}^2+R_{2y}^2}$

$=\sqrt{{500}^2+{125}^2}$

= 515. 4N