Step 1

We are given the following data:

The maximum magnitude of the uniformly varying load is $W = 3\;{\rm{kN/m}}$.

The length of the first uniformly varying load is ${l_1} = 3\;{\rm{m}}$.

The length of the second uniformly varying load is ${l_2} = 1.5\;{\rm{m}}$.

We are asked to replace the distributed loading with an equivalent resultant force and specify its location on the beam measured from point $O$.

 
Step 2

The formula to calculate the effective point force of the first uniformly varying load is given by,

\[{F_1} = \left( {\frac{1}{2} \times W \times {l_1}} \right)\]

Here, ${F_1}$ represent the effective point force of the first uniformly varying load.

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_1} = \frac{1}{2} \times \left( {3\;{\rm{kN/m}}} \right) \times \left( {3\;{\rm{m}}} \right)\\ = \frac{{\left( {9\;{\rm{kN}}} \right)}}{2}\\ = 4.5\;{\rm{kN}} \end{array}\]
 
Step 3

The formula to calculate the effective point force of the second uniformly varying load is given by,

\[{F_2} = \left( {\frac{1}{2} \times W \times {l_2}} \right)\]

Here, ${F_2}$ represent the effective point force of the second uniformly varying load.

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_2} = \frac{1}{2} \times \left( {3\;{\rm{kN/m}}} \right) \times \left( {1.5\;{\rm{m}}} \right)\\ = \frac{{\left( {4.5\;{\rm{kN}}} \right)}}{2}\\ = 2.25\;{\rm{kN}} \end{array}\]
 
Step 4

The formula to calculate the distance of the first effective point force from the point $O$ is given by,

\[{x_1} = \frac{2}{3}{l_1}\]

Here, ${x_1}$ represent the distance of the first effective point force from the point $O$.

Substitute all the known values in the above formula.

\[\begin{array}{c} {x_1} = \frac{2}{3} \times \left( {3\;{\rm{m}}} \right)\\ = \frac{{\left( {6\;{\rm{m}}} \right)}}{3}\\ = 2\;{\rm{m}} \end{array}\]
 
Step 5

The formula to calculate the distance of the second effective point force from the point $O$ is given by,

\[{x_2} = \left( {{l_1} + \frac{1}{3}{l_2}} \right)\]

Here, ${x_2}$ represent the distance of the second effective point force from the point $O$.

Substitute all the known values in the above formula.

\[\begin{array}{c} {x_2} = \left( {3\;{\rm{m}}} \right) + \left( {\frac{1}{3} \times \left( {1.5\;{\rm{m}}} \right)} \right)\\ = \left( {3\;{\rm{m}}} \right) + \left( {0.5\;{\rm{m}}} \right)\\ = 3.5\;{\rm{m}} \end{array}\]
 
Step 6

The free-body diagram of the horizontal beam is given below.

The formula to calculate the equivalent resultant force is given by,

\[{F_{\rm{R}}} = {F_1} + {F_2}\]

Here, ${F_{\rm{R}}}$ represent the equivalent resultant force.

Substitute all the known values in the above formula.

\[\begin{array}{c} {F_{\rm{R}}} = \left( {4.5\;{\rm{kN}}} \right) + \left( {2.25\;{\rm{kN}}} \right)\\ = 6.75\;{\rm{kN}} \end{array}\]
 
Step 7

The formula to calculate the magnitude of the equivalent moment about point $O$ is given by,

\[\left( { - {F_{\rm{R}}} \times \bar x} \right) = \left( { - {F_1}{x_1}} \right) + \left( { - {F_2}{x_2}} \right)\]

Here, $\bar x$ represent the location of the equivalent force measured from the point $O$.

Substitute all the known values in the above formula.

\[\begin{array}{c} \left( { - 6.75\;{\rm{kN}}} \right) \times \left( {\bar x} \right) = \left[ {\left( { - 4.5\;{\rm{kN}}} \right) \times \left( {2\;{\rm{m}}} \right)} \right] + \left[ {\left( { - 2.25\;{\rm{kN}}} \right) \times \left( {3.5\;{\rm{m}}} \right)} \right]\\ \left( { - 6.75\;{\rm{kN}}} \right) \times \left( {\bar x} \right) = \left( { - 16.875\;{\rm{kN}} \cdot {\rm{m}}} \right)\\ \bar x = 2.5\;{\rm{m}} \end{array}\]