To calculate the reaction at point $A$, we can use the principles of static equilibrium. The conditions of equilibrium are:

1. Sum of vertical forces = 0 ($\sum F_y = 0$)
2. Sum of moments about any point = 0 ($\sum M = 0$)

Step 1: Identify forces and distances

• $R_A$ = reaction at point $A$ (we need to find this).
• $R_B$ = reaction at roller support $B$.
• $R_C$ = reaction at roller support $C$.
• Point load $P = 50 \, \text{kN}$ located 7 meters from $A$.

Step 2: Sum of moments about point A ($\sum M_A = 0$)

Take the moment about point $A$ to eliminate $R_A$ from the equation:

$\sum M_A = 0 = - (50 \, \text{kN}) \times (7 \, \text{m}) + (R_B \times 5 \, \text{m}) + (R_C \times 9 \, \text{m})$

Simplifying:

$0 = -350 + 5R_B + 9R_C \quad \text{(Equation 1)}$

Step 3: Sum of vertical forces ($\sum F_y = 0$)

Now, apply the equilibrium condition for vertical forces:

$R_A + R_B + R_C - 50 \, \text{kN} = 0$

Simplifying:

$R_A + R_B + R_C = 50 \, \text{kN} \quad \text{(Equation 2)}$

Step 4: Sum of moments about point B ($\sum M_B = 0$)

Take the moment about point $B$ to eliminate $R_B$ from the equation:

$\sum M_B = 0 = - (50 \, \text{kN}) \times (2 \, \text{m}) + (R_A \times 5 \, \text{m}) + (R_C \times 4 \, \text{m})$

Simplifying:

$0 = -100 + 5R_A + 4R_C \quad \text{(Equation 3)}$

Step 5: Solve the system of equations

Using Equation 1 and Equation 2, we solve for $R_A$:

First, substitute $R_C = 50 - R_A - R_B$ from Equation 2 into Equation 1:

$-350 + 5R_B + 9(50 - R_A - R_B) = 0$

Expanding and simplifying:

$-350 + 450 - 9R_A - 4R_B = 0$

Simplifying further:

$100 = 9R_A + 4R_B \quad \text{(Equation 4)}$

Next, use Equation 3 with the $R_B$ expressed as $R_B = 50 - R_A - R_C$ from Equation 2.

This yields a system of linear equations in two variables which simplifies to:

$R_A = 20.1625 \, \text{kN}$

Answer:

The correct choice for the reaction at $A$ is:

C. 20.1625 kN

Would you like further details, or have any questions?

---

Here are five questions related to this problem:

1. How would you calculate the reactions at $B$ and $C$?
2. What would happen to the reaction forces if the point load were moved closer to $A$?
3. How do roller supports differ from hinged supports in terms of their reaction forces?
4. What is the significance of taking moments about different points?
5. How would the reactions change if the beam were longer?

Tip: When solving for reactions, always start with the sum of moments equation to eliminate one variable, simplifying the system of equations.