A uniform wood door has mass `m`, height `h`, and width `w`. It is hanging from two hinges attached to one side , the hinges are located `h//3` and `2h//3` form the bottom of the door. Suppose that `m= 20 kg, h = 2.20 m`, and the bottom smooth hinge is not screwed into the door frame. Find the forces acting on the door.
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Answer

`F_(d//x) = (3mgw)/(2h)` and `F_(duy) = mg`

Explanation

For balance of weight

`mg = N_(1) + N_(2)`
Torque about `A`
`R_(1) xx (h)/(3) = mg xx(w)/(2) rArr R_(1) = (3 mgw)/(2h)`
`F_(x) = (3mgw)/(2h)` and `F_(y) = mg`.