A trap door in a ceiling is $\mathbf{0}\mathbf{.91}\mathbf{}\mathbf{\text{\hspace{0.17em}m}}$ square, has a mass of$\mathbf{11}\mathbf{\text{\hspace{0.17em}kg}}$ and is hinged along one side, with a catch at the opposite side. If the center of gravity of the door is $\mathbf{10}\mathbf{\text{\hspace{0.17em}cm}}$toward the hinged side from the door’s center, what are the magnitudes of the forces exerted by the door on (a) the catch and (b) the hinge?

A side trap door in the ceiling is$0.91\text{\hspace{0.17em}m}$.

The mass of the trap door is $11\text{\hspace{0.17em}kg}$.

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.

$\mathbf{\sum }{\mathit{\tau }}_{\mathbf{n}\mathbf{e}\mathbf{t}}\mathbf{=}\mathbf{0}$

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

$\frac{\text{0.91\hspace{0.17em}m}}{2}=0.455\text{\hspace{0.17em}m}$

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

$\begin{array}{c}{R}_h=0.455m-0.1m\\ =0.355\text{\hspace{0.17em}m}\end{array}$

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

$\begin{array}{c}{R}_c=0.455m+0.1m\\ =0.555\text{\hspace{0.17em}m}\end{array}$

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

$F_c\times d-mg\times R_h=0$

Solving for force, we get

${\text{F}}_{\text{c}}=42.1\text{\hspace{0.17em}N}$

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

$F_h\times d-mg\times R_c=0$

Solving for force, we get

${\text{F}}_{\text{h}}=65.7\text{\hspace{0.17em}N}$