The free body diagram of the door is shown in figure `B`. The forces exerted on theinges are equal in magnitude. The door is in equilibrium in horizontal as well as in vertical direction, i.e., the horizontal and vertial components of the forces will be equal. For the vertical equilibrium of door , we have
`Sigma F_(v)=0, 2F_(v)=500impliesF_(v)=250N`
For rotational equilibrium of the door, we have `Sigmatau=0`
Taking moment of all forces acting on the door about lower hinge (may be upper higne or any other point), we get
`500xx1.0-F_(u)xx2=0` which gives `F_(H)=250N`
The force on the door exerted by either hinge, in magnitude is
`F=sqrt(F_(H)^(2)+F_(V)^(2))=sqrt((250)^(2)+(250)^(2))=250sqrt(2)N`