Step 1

We are given that the applied force to the beam is ${F_a} = 26\;{\rm{kN}}$ and the force applied at point B of the beam is $F = 40\;{\rm{kN}}$.

We are asked to obtain the reaction at point A and tension in cord BC.

We are given that the distance between point A and applied force is ${d_1} = 2\;{\rm{m}}$.

We are given that the distance between point of applied force and B is ${d_2} = 4\;{\rm{m}}$.

We are given that the right-angled triangle of applied force has the value of base ${b_1} = 5$ , the value of perpendicular is ${p_1} = 12$ and the value of hypotenuse is ${h_1} = 13$.

We are given that the right-angled triangle of force applied at point B has the value of base ${b_2} = 4$ , the value of perpendicular is ${p_2} = 3$ and the value of hypotenuse is ${h_2} = 5$.

 
Step 2

The free body diagram of the beam is shown as,

 
Step 3

Applying the moment of force about x-axis:

\[ - {F_a}\left( {\frac{{{p_1}}}{{{h_1}}}} \right){d_1} - F\left( {{d_1} + {d_2}} \right) + \left( {\frac{{{p_2}}}{{{h_2}}}} \right){T_{BC}}\left( {{d_1} + {d_2}} \right) = 0\]
 
Step 4

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} - \left( {26\;{\rm{kN}}} \right)\left( {\frac{{12}}{{13}}} \right)\left( {2\;{\rm{m}}} \right) - \\ \left( {40\;{\rm{kN}}} \right)\left( {2\;{\rm{m}} + 4\;{\rm{m}}} \right) + \\ \left( {\frac{3}{5}} \right){T_{BC}}\left( {2\;{\rm{m}} + 4\;{\rm{m}}} \right) \end{array} \right\} = 0\\ \left( {\frac{3}{5}} \right){T_{BC}}\left( {6\;{\rm{m}}} \right) = 48\;{\rm{kN}} \cdot {\rm{m}} + 240\;{\rm{kN}} \cdot {\rm{m}}\\ {T_{BC}} = \frac{{288\;{\rm{kN}} \cdot {\rm{m}}}}{{3.6\;{\rm{m}}}}\\ = 80\;{\rm{kN}} \end{array}\]
 
Step 5

Applying the equilibrium of force equation along x-axis:

\[\begin{array}{c} \sum {F_x} = 0\\ {T_{BC}}\left( {\frac{{{b_2}}}{{{h_2}}}} \right) - {A_x} - {F_a}\left( {\frac{{{b_1}}}{{{h_1}}}} \right) = 0 \end{array}\]
 
Step 6

Substitute the known values in the equation:

\[\begin{array}{c} \left( {80\;{\rm{kN}}} \right)\left( {\frac{4}{5}} \right) - {A_x} - \left( {26\;{\rm{kN}}} \right)\left( {\frac{5}{{13}}} \right) = 0\\ {A_x} = 64\;{\rm{kN}} - 10\;{\rm{kN}}\\ = 54\;{\rm{kN}} \end{array}\]
 
Step 7

Applying the equilibrium of force equation along y-axis:

\[\begin{array}{c} \sum {F_y} = 0\\ {A_y} - {F_a}\left( {\frac{{{p_1}}}{{{h_1}}}} \right) - F + {T_{BC}}\left( {\frac{{{p_2}}}{{{h_2}}}} \right) = 0 \end{array}\]
 
Step 8

Substitute the known values in the equation:

\[\begin{array}{c} {A_y} - \left( {26\;{\rm{kN}}} \right)\left( {\frac{{12}}{{13}}} \right) - 40\;{\rm{kN}} + \left( {80\;{\rm{kN}}} \right)\left( {\frac{3}{5}} \right) = 0\\ {A_y} = 24\;{\rm{kN}} + 40\;{\rm{kN}} - 48\;{\rm{kN}}\\ = 16\;{\rm{kN}} \end{array}\]