Step 1

We are given the forces acting over the different columns are ${F_1} = 60\;{\rm{kN}}$, ${F_2} = 80\;{\rm{kN}}$, and ${F_3} = 50\;{\rm{kN}}$.

We are asked the intensities ${w_1}$ and ${w_2}$ of distributed load to support the column loadings.

 
Step 2

The following is the diagram of the system:

We have the length of distributed load is $l = 8\;{\rm{m}}$.

We have the distance of force 1 from point A is ${d_1} = 1\;{\rm{m}}$.

We have the distance of force 2 from point A is ${d_2} = 3.5\;{\rm{m}}$.

We have the distance of force 3 from point A is ${d_3} = 7\;{\rm{m}}$.

 
Step 3

The moment of all the forces acting over the system about point B is,

\[\begin{array}{c} \Sigma M = 0\\ \left[ \begin{array}{c} {w_2}l\left( {\frac{l}{2} - \frac{1}{3} \times l} \right) + {F_1}\left( {\frac{1}{3} \times l - {d_1}} \right)\\ - {F_2}\left( {{d_2} - \frac{1}{3} \times l} \right)\\ - {F_3}\left( {{d_3} - \frac{1}{3} \times l} \right) \end{array} \right] = 0 \end{array}\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{c} {w_2}\left( {8\;{\rm{m}}} \right)\left( {\frac{{8\;{\rm{m}}}}{2} - \frac{1}{3} \times 8\;{\rm{m}}} \right) + \left( {60\;{\rm{kN}}} \right)\left( {\frac{1}{3} \times 8\;{\rm{m}} - 1\;{\rm{m}}} \right)\\ - \left( {80\;{\rm{kN}}} \right)\left( {3.5\;{\rm{m}} - \frac{1}{3} \times 8\;{\rm{m}}} \right)\\ - \left( {50\;{\rm{kN}}} \right)\left( {7\;{\rm{m}} - \frac{1}{3} \times 8\;{\rm{m}}} \right) \end{array} \right] = 0\\ {w_2}\left( {8\;{\rm{m}}} \right)\left( {1.33\;{\rm{m}}} \right) + \left( {60\;{\rm{kN}}} \right)\left( {1.66\;{\rm{m}}} \right) = \left( {80\;{\rm{kN}}} \right)\left( {0.83\;{\rm{m}}} \right) + \left( {50\;{\rm{kN}}} \right)\left( {4.33\;{\rm{m}}} \right)\\ {w_2} = 17.22\;{\rm{kN/m}} \end{array}\]
 
Step 5

The force equation of equilibrium along vertical axis is,

\[\begin{array}{c} \Sigma F = 0\\ \left[ \begin{array}{c} \frac{1}{2}\left( {{w_1} - {w_2}} \right)\left( l \right) + {w_2}\left( l \right)\\ - {F_1} - {F_2} - {F_3} \end{array} \right] = 0\\ \frac{1}{2}\left( {{w_1} - {w_2}} \right)\left( l \right) + {w_2}\left( l \right) = {F_1} + {F_2} + F \end{array}\]
 
Step 6

Substitute the values in the above expression.

\begin{array}{c} \frac{1}{2}\left( {{w_1} - 17.22\;{\rm{kN/m}}} \right)\left( {8\;{\rm{m}}} \right) + \left( {17.22\;{\rm{kN/m}}} \right)\left( {8\;{\rm{m}}} \right) = 60\;{\rm{kN}} + 80\;{\rm{kN}} + 50\;{\rm{kN}}\\ \frac{1}{2}\left( {{w_1} - 17.22\;{\rm{kN/m}}} \right)\left( {8\;{\rm{m}}} \right) = 52.25\;{\rm{kN}}\\ {w_1} = 30.28\;{\rm{kN/m}} \end{array}