Step 1

We are given that the force applied at point A is ${F_A} = 20000\;{\rm{lb}}$, the force applied at point B is ${F_B} = 8000\;{\rm{lb}}$.

We are asked to calculate the uniform distribution load at points A and B.

We have the distance between end point and point at which force applied at A is ${d_1} = 0.25\;{\rm{ft}}$.

We have the distance between end point and point at which force applied at B is ${d_2} = 1.5\;{\rm{ft}}$.

We have the distance at which uniform load at A is distributed is ${d_A} = 2\;{\rm{ft}}$.

We have the distance at which uniform load at B is distributed is ${d_B} = 3\;{\rm{ft}}$.

We have the distance between A and B $d = 8\;{\rm{ft}}$.

 
Step 2

The free body diagram of the system is shown as:

 
Step 3

Applying the moment of force about point A:

\[ - {F_B}\left( {\frac{{{d_A}}}{2} + d + \frac{{{d_B}}}{2}} \right) + {w_B}\left( {{d_B}} \right)\left( {\frac{{{d_A}}}{2} + d + \frac{{{d_B}}}{2}} \right) + {F_A}\left( {\frac{{{d_A}}}{2} - {d_1}} \right) = 0\]
 
Step 4

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} - \left( {8000\;{\rm{lb}}} \right)\left( {\frac{{2\;{\rm{ft}}}}{2} + 8\;{\rm{ft}} + \frac{{3\;{\rm{ft}}}}{2}} \right) + \\ {w_B}\left( {3\;{\rm{ft}}} \right)\left( {\frac{{2\;{\rm{ft}}}}{2} + 8\;{\rm{ft}} + \frac{{3\;{\rm{ft}}}}{2}} \right) + \\ \left( {20000\;{\rm{lb}}} \right)\left( {\frac{{2\;{\rm{ft}}}}{2} - 0.25\;{\rm{ft}}} \right) \end{array} \right\} = 0\\ {w_B}\left( {3\;{\rm{ft}}} \right)\left( {10.5\;{\rm{ft}}} \right) = \left( {8000\;{\rm{lb}}} \right)\left( {10.5\;{\rm{ft}}} \right) - \left( {20000\;{\rm{lb}}} \right)\left( {0.75\;{\rm{ft}}} \right)\\ {w_B} = \frac{{84000\;{\rm{lb}} \cdot {\rm{ft}} - 15000\;{\rm{lb}} \cdot {\rm{ft}}}}{{31.5\;{\rm{f}}{{\rm{t}}^2}}}\\ = 2190.5\;{\rm{lb/ft}} \end{array}\]
 
Step 5

Applying equation of force equilibrium along y-axis:

\[{w_B}{d_B} - {F_A} - {F_B} + {w_A}{d_A} = 0\]
 
Step 6

Substitute the known values in the equation:

\[\begin{array}{c} \left\{ \begin{array}{l} \left( {2190.5\;{\rm{lb/ft}}} \right)\left( {3\;{\rm{ft}}} \right) - 20000\;{\rm{lb}}\\ - 8000\;{\rm{lb}} + {w_A}\left( {2\;{\rm{ft}}} \right) \end{array} \right\} = 0\\ {w_A}\left( {2\;{\rm{ft}}} \right) = - 6571.5\;{\rm{lb}} + 28000\;{\rm{lb}}\\ {w_A} = \frac{{21428.5\;{\rm{lb}}}}{{2\;{\rm{ft}}}}\\ = 10714.25\;{\rm{lb/ft}} \end{array}\]