91Ó°ÊÓ

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

Net force, $\underset{\mathbf{}}{\mathbf{∑}}\mathbf{F}\mathbf{=}\mathbf{ma}$

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.

$\underset{}{∑}\mathrm{τ}=0\mathrm{implies}$

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

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.