A uniform $\mathbf{300}\mathbf{\text{N}}$ trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges

(a) if the upward force is applied at the center and

(b) if the upward force is applied at the center of the edge opposite the hinges.

The object is said to be in an equilibrium state when the net force, as well as net torque acting on it, is equal to zero.

(a)

Given, that the trapdoor is uniform which indicates its center of mass is at the center of its length. The trapdoor weighs 300 N.

Let the whole set up illustrated as a free body diagram for force $\overrightarrow{F}$shown in the figure as:

Considering anticlockwise rotation and upside direction as positive, and applying the condition for rotational equilibrium at the axis of the hinge, we have:

$\begin{array}{rcl}\sum {\tau }_{ext}& =& 0\\ F·\left(\frac{L}{2}\right)-300·\left(\frac{L}{2}\right)& =& 0\\ F& =& 300\text{N}\end{array}$

Now, applying the criteria of translational equilibrium on $x-y$axes, we get:

$\begin{array}{rcl}\sum F_x& =& F_{xh}=0\\ \sum F_y& =& F+F_{yh}-300=0\\ F_{yh}& =& 0\end{array}$

Thus, there is no force on the trapdoor applied by the hinge.

(b)

Given this is the same case as the previous part, except for the fact that the force is acting towards the edge, which is opposite to the hinges.

Let the whole set up illustrated as a free body diagram for force $\overrightarrow{F}$shown in the figure as:

Considering the criteria for rotational equilibrium about the axis of the hinge, we have:

$\begin{array}{rcl}\sum {\tau }_{ext}& =& 0\\ F·L-300·\left(\frac{L}{2}\right)& =& 0\\ F& =& 150\text{N}\end{array}$

Again, applying the criteria of translational equilibrium on $x-y$axes, we get:

$\begin{array}{rcl}\sum F_x& =& F_{xh}=0\\ \sum F_y& =& F+F_{yh}-300=0\\ 150+F_{yh}-300& =& 0\\ F_{yh}& =& 150\text{N}\end{array}$

Thus, there is a force of magnitude 150 N on the trapdoor in the upward direction applied by the hinge.