(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door’s weight (Fig. 9–69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.

The mathematical expression for the translational equilibrium is given as:

\(\sum {F_{{\rm{net}}}} = 0\)

Here,\({F_{{\rm{net}}}}\)is the total force working on a system.

The mathematical expression for the rotational equilibrium is given as:

\(\sum {\tau _{{\rm{net}}}} = 0\)

Here,\({\tau _{{\rm{net}}}}\)is the net torque working on the system.

Given data:

The height of the door is\(h = 2.30\;{\rm{m}}\).

The width of the door is \(w = 1.30\;{\rm{m}}\).

The distance of the hinge point from the top and bottom is \(d = 0.40\;{\rm{m}}\).

The mass of the door is \(m = 13.0\;{\rm{kg}}\).

Draw a free body diagram of the door.

Here, \({F_{{\rm{Ay}}}}\) is the vertical force at the top hinge, \({F_{{\rm{By}}}}\) is the vertical force at the bottom hinge, \({F_{{\rm{Ax}}}}\) is the horizontal force at the top hinge, and \({F_{{\rm{Bx}}}}\) is the horizontal force at the bottom hinge.

Since the hinge from the top and that from the bottom support half the door’s weight each, therefore you can write:

\({F_{{\rm{Ay}}}} = {F_{{\rm{By}}}} = \frac{1}{2}mg\)

Take torques about the bottom hinge, with the counterclockwise torques, as positive.

\(\begin{array}{c}\sum \tau = 0\\mg\frac{w}{2} - {F_{{\rm{Ax}}}}\left( {h - 2d} \right) = 0\\{F_{{\rm{Ax}}}} = \frac{{mgw}}{{2\left( {h - 2d} \right)}}\\{F_{{\rm{Ax}}}} = \frac{{\left( {13\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {1.3\;{\rm{m}}} \right)}}{{2\left[ {\left( {2.30\;{\rm{m}}} \right) - \left( {2 \times 0.4\;{\rm{m}}} \right)} \right]}}\\{F_{{\rm{Ax}}}} = 55.2\;{\rm{N}}\end{array}\)

Apply the fore equilibrium condition along the horizontal direction.

\(\begin{array}{c}\sum {F_{\rm{x}}} = 0\\{F_{{\rm{Ax}}}} - {F_{{\rm{Bx}}}} = 0\\{F_{{\rm{Bx}}}} = {F_{{\rm{Ax}}}}\\{F_{{\rm{Bx}}}} = 55.2\;{\rm{N}}\end{array}\)

Apply the force equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{{\rm{Ay}}}} + {F_{{\rm{By}}}} - mg = 0\\{F_{{\rm{Ay}}}} + {F_{{\rm{By}}}} = mg\end{array}\)

As each hinge supports half the weight, therefore,

\(\begin{array}{c}{F_{{\rm{Ay}}}} = {F_{{\rm{By}}}} = \frac{1}{2}mg\\{F_{{\rm{Ay}}}} = {F_{{\rm{By}}}} = \frac{1}{2}\left( {13\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\\{F_{{\rm{Ay}}}} = {F_{{\rm{By}}}} = 63.7\;{\rm{N}}\end{array}\)

Thus, the horizontal and vertical force components exerted by the top hinge are \(55.2\;{\rm{N}}\) and \(63.7\;{\rm{N}}\), respectively, and the horizontal and vertical force components exerted by the bottom hinge are \(55.2\;{\rm{N}}\) and \(63.7\;{\rm{N}}\), respectively.