As shown in figure, the hinges $A$ and $B$ hold a uniform $400N$ door in place. the upper hinge supports the entire weight of the door. find the resultant force exerted on the door at the hinges . the width of the door is $\frac{h}{2}$, where $h$ is the distance between the hinges?

 

 

Only a horizontal force acts at hinge $B$, because hinge $A$ is assumed to support the door,$s$ weight.Let us take torques about $A$ as axis.

$\sum _{ }^{ }{F}_x=0$   or   $F_2-H=0$

$\sum _{ }^{ }{F}_y=0$   or   $V-400N=0$

We find from these that

$H=100N$   and   $V=400N$

To find the resultant force $\overrightarrow{R}$ on the hinge at $A$,

we have $R=\sqrt{({400}^2}+{\left(100\right)}^2)=412N$