91Ó°ÊÓ

A side trap door in the ceiling is$0.91\text{″¾}$.

The mass of the trap door is $11\text{ k²µ}$.

Using the equation for the equilibrium of torque and properly selecting the pivot point,you can find the force exerted by the door on the hinge and the catch. The equations are given below.

$\mathbf{∑}{\mathit{τ}}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{0}$

The distance of the center of the door from the catch and the hinge will be,

$\frac{\text{0.91″¾}}{2}=0.455\text{″¾}$

So, the distance of the center of gravity from the hinged side is,

$\begin{array}{c}{R}_h=0.455\text{ }m−0.1\text{ }m\\ =0.355\text{″¾}\end{array}$

And the distance of the center of gravity from the catch side is,

$\begin{array}{c}{R}_c=0.455\text{ }m+0.1\text{ }m\\ \text{ }=0.555\text{″¾}\end{array}$

We have, the equation for the equilibrium of torque considering the pivot point at the hinge is,

$F_c×d−mg×R_h=0$

Solving for force, we get

${\text{F}}_{\text{c}}=42.1\text{ N}$

We have, the equation for the equilibrium of torque considering the pivot point at the catch is,

$F_h×d−mg×R_c=0$

Solving for force, we get

${\text{F}}_{\text{h}}=65.7\text{ N}$