A door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two hinges, one 0.50 m from the top and the other 0.50 m from the bottom. Each hinge supports half the total weight of the door. Assuming that the door’s center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

Given that a door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two hinges, one 0.50 m from the top and the other 0.50 m from the bottom.

Each hinge supports half the total weight of the door.

Weight of the door, w = 330 N

Let $\overrightarrow{{\mathrm{H}}_1}$and $\overrightarrow{{\mathrm{H}}_2}$be the forces exerted by the upper and the lower hinges respectively.

Net force, $\mathbf{\sum }_{\mathbf{}}\mathbf{F}\mathbf{=}\mathbf{ma}$

Where m is the mass of the object and a is the acceleration

Let the bottom hinge be the origin.

Since each hinge supports half the total weight of the door, so

${\mathbf{H}}_{\mathbf{1}\mathbf{v}}\mathbf{=}{\mathbf{H}}_{\mathbf{2}\mathbf{v}}\mathbf{=}\frac{\mathbf{w}}{\mathbf{2}}\mathbf{=}\mathbf{165}\mathbf{}\mathbf{N}$

The horizontal components of the hinge force are equal and opposite in direction.

So, ${\mathrm{H}}_{1\mathrm{h}}={\mathrm{H}}_{2\mathrm{h}}$

Since ${\mathrm{H}}_{1\mathrm{v}},{\mathrm{H}}_{2\mathrm{v}}$and${\mathrm{H}}_{2\mathrm{h}}$ all have zero moment arm and hence have zero torque about the origin.

Therefore, the total torque is zero.

$\sum _{}\mathrm{\tau }=0\mathrm{implies}$

${\mathrm{H}}_{1\mathrm{h}}\left(1\mathrm{m}\right)-\mathrm{w}\left(0.5\mathrm{m}\right)=0$

Thus

$\begin{array}{r}{\mathrm{H}}_{1\mathrm{h}}=\mathrm{w}\left(\frac{0.5\mathrm{m}}{1\mathrm{m}}\right)\\ =\frac{1}{2}\left(330\mathrm{N}\right)\\ =165\mathrm{N}\end{array}$

Hence, the horizontal component of each hinge force is 165 N.