Steps to Solve

1. Identify Forces and Moments

In the system, we have three external forces acting downwards:

• A force of $3 , \text{kN}$ at point C,
• A force of $1.5 , \text{kN}$ also at point C,
• A force of $2 , \text{kN}$ at point D.

The reaction forces at points A and D will be denoted as $R_A$ and $R_D$ respectively.

2. Sum of Vertical Forces

According to the equilibrium condition, the sum of the vertical forces must equal zero:

$$ R_A + R_D - 3 , \text{kN} - 1.5 , \text{kN} - 2 , \text{kN} = 0 $$

This simplifies to:

$$ R_A + R_D = 6.5 , \text{kN} \quad \text{(Equation 1)}$$

3. Sum of Moments about Point A

Now, we take moments about point A to find $R_D$. The moments due to the forces $3 , \text{kN}$ and $1.5 , \text{kN}$ are calculated as follows:

$$ \text{Moment from } 3 , \text{kN} = 3 , \text{kN} \cdot 2 , \text{m} = 6 , \text{kN m} $$

$$ \text{Moment from } 1.5 , \text{kN} = 1.5 , \text{kN} \cdot 4 , \text{m} = 6 , \text{kN m} $$

$$ \text{Moment from } 2 , \text{kN} = 2 , \text{kN} \cdot 6 , \text{m} = 12 , \text{kN m} $$

Setting the sum of moments around point A equal to zero gives us:

$$ -6 , \text{kN m} - 6 , \text{kN m} + R_D \cdot 6 , \text{m} = 0 $$

This simplifies to:

$$ R_D \cdot 6 = 12 \quad \Rightarrow \quad R_D = 2 , \text{kN} $$

4. Substitute $R_D$ back into Equation 1

Substituting $R_D = 2 , \text{kN}$ into Equation 1:

$$ R_A + 2 , \text{kN} = 6.5 , \text{kN} $$

This gives:

$$ R_A = 6.5 , \text{kN} - 2 , \text{kN} = 4.5 , \text{kN} $$