Step 1

We are given the door of weight $W = 100\;{\rm{lb}}$.

We are asked to determine the components of reaction at hinges A and B if hinge B resists only forces in the x and y directions and A resists forces in the x, y, z directions.

 
Step 2

First draw the free body diagram for the given system.

Figure-(1)

We have distance between point $A$ and floor along $z$-axis as $a = 18\;{\rm{in}}{\rm{.}}$

We have distance between point $A$ and $G$ along $z$-axis as $b = 24\,{\rm{in}}{\rm{.}}$

We have distance between force $B$ and $G$ along $z$-axis ad $c = 24\,{\rm{in}}{\rm{.}}$

We have distance between point $B$ and $G$ along $y$-axis as $d = 18\,{\rm{in}}{\rm{.}}$

We have angle of door with wall as $\theta = 30^\circ $.

 
Step 3

To find ${A_y}$ by taking moment along $x$-axis under equilibrium condition use the following relation.

\[{A_y}\left( a \right) + {B_y}\left( {a + b} \right) - W\left( d \right) = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{l} {A_y}\left( {18\,{\rm{in}}{\rm{.}}} \right) + {B_y}\left( {18\;{\rm{in}}{\rm{. + 24}}\;{\rm{in}}{\rm{.}} + 24\,{\rm{in}}{\rm{.}}} \right)\\ - \left( {100\;{\rm{lb}}} \right)\left( {18\;{\rm{in}}{\rm{.}}} \right) \end{array} \right] = 0\\ {A_y} = \frac{{1800\;{\rm{lb}} \cdot {\rm{in}}{\rm{.}} - {B_y}\left( {66\,{\rm{in}}{\rm{.}}} \right)}}{{18\,{\rm{in}}{\rm{.}}}} \end{array}\]
 
Step 4

Find ${B_y}$ by adding all the forces along $y$-axis using the following relation.

\[{A_y} + {B_y} = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} \left( {\frac{{1800\;{\rm{lb}} \cdot {\rm{in}}{\rm{.}} - {B_y}\left( {66\,{\rm{in}}{\rm{.}}} \right)}}{{18\,{\rm{in}}{\rm{.}}}}} \right) + {B_y} = 0\\ 1800\;{\rm{lb}} \cdot {\rm{in}}{\rm{.}} - {B_y}\left( {66\,{\rm{in}}{\rm{.}}} \right) + {B_y}\left( {18\,{\rm{in}}{\rm{.}}} \right) = 0\\ {B_y} = \frac{{1800\;{\rm{lb}} \cdot {\rm{in}}{\rm{.}}}}{{48\,{\rm{in}}{\rm{.}}}}\\ {B_y} = 37.5\,{\rm{lb}} \end{array}\]
 
Step 5

To find ${A_y}$ use the following relation.

\[{A_y} + {B_y} = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {A_y} + 37.5\;{\rm{lb}} = 0\\ {A_y} = - 37.5\;{\rm{lb}} \end{array}\]
 
Step 6

Find ${A_z}$ by adding all the forces along $z$-axis using the following relation.

\[{A_z} - W = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {A_z} - 100\;{\rm{lb}} = 0\\ {A_z} = 100\;{\rm{lb}} \end{array}\]
 
Step 7

To find ${A_x}$ by taking moment along $y$-axis under equilibrium condition use the following relation.

\[{A_x}\left( a \right) + {B_x}\left( {a + b + c} \right) = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {A_x}\left( {18\,{\rm{in}}{\rm{.}}} \right) + {B_x}\left( {18\;{\rm{in}}{\rm{.}} + 2{\rm{4}}\,{\rm{in}}{\rm{.}} + 24\,{\rm{in}}{\rm{.}}} \right) = 0\\ {A_x} = \frac{{ - {B_x}\left( {66\,{\rm{in}}{\rm{.}}} \right)}}{{18\,{\rm{in}}{\rm{.}}}} \end{array}\]
 
Step 8

Find ${B_x}$ by adding all the forces along $x$-axis using the following relation.

\[{A_x} + {B_x} = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} \frac{{ - {B_x}\left( {66\,{\rm{in}}{\rm{.}}} \right)}}{{18\,{\rm{in}}{\rm{.}}}} + {B_x} = 0\\ {B_x} = 0 \end{array}\]
 
Step 9

Find reaction ${A_x}$ using the following relation.

\[{A_x} + {B_x} = 0\]

Substitute the known values in the above expression.

\[\begin{array}{c} {A_x} + 0 = 0\\ {A_x} = 0 \end{array}\]