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 `h/2`, where `h` is the distance between the hinges.

Options

A. `312N`
B. `280N`
C. `412N`
D. `480N`

Answer

C

Explanation

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.
`sumF_(x)=0`, or `F_(2)-H=0`
`sum F_(y)=0` or `V-400 N=0`
We find from these that
`H=100 N` and `V=400 N`
To find the resultant force `vec(R)` on hinge at `A`, we have `R=sqrt(((400)^(2)+(100)^(2))=412 N)`