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We can solve this problem by using static equilibrium equations. Let's break it down step by step.

### Given Information:
- The beam is subjected to two forces:
  1. A 12 kN force at an angle of 60° applied at the center (distance of 10 m from point A).
  2. A vertical 6 kN force applied at a distance of 5 m from point B (right end).
- The beam is supported at point A (roller) and point B (pin).

### Static Equilibrium Equations:
1. **Sum of vertical forces = 0** (since the beam is in static equilibrium):
   \[
   \sum F_y = 0
   \]
2. **Sum of horizontal forces = 0**:
   \[
   \sum F_x = 0
   \]
3. **Sum of moments about any point = 0**. For convenience, we will take the moment about point A (to eliminate the reaction at A):
   \[
   \sum M_A = 0
   \]

### Step 1: Resolve the 12 kN Force
The 12 kN force is inclined at 60°, so it has both horizontal and vertical components:
- Vertical component: \( 12 \, \text{kN} \cdot \sin(60^\circ) \)
- Horizontal component: \( 12 \, \text{kN} \cdot \cos(60^\circ) \)

\[
12 \cdot \sin(60^\circ) = 12 \cdot \frac{\sqrt{3}}{2} = 10.39 \, \text{kN}
\]
\[
12 \cdot \cos(60^\circ) = 12 \cdot \frac{1}{2} = 6 \, \text{kN}
\]

### Step 2: Sum of Vertical Forces
Let \( R_A \) be the reaction at point A, and \( R_B \) be the vertical reaction at point B. The total vertical forces must sum to zero:
\[
R_A + R_B = 6 \, \text{kN} + 10.39 \, \text{kN}
\]
\[
R_A + R_B = 16.39 \, \text{kN} \quad \text{(Equation 1)}
\]

### Step 3: Sum of Moments about Point A
Taking moments about point A to eliminate \( R_A \), we use the following:
- Moment arm of the vertical component of the 12 kN force (at 10 m): \( 10.39 \, \text{kN} \cdot 10 \, \text{m} \)
- Moment arm of the 6 kN force (at 25 m): \( 6 \, \text{kN} \cdot 25 \, \text{m} \)
- Moment arm of \( R_B \) (at 20 m): \( R_B \cdot 20 \, \text{m} \)

Setting the sum of moments equal to zero:
\[
10.39 \cdot 10 + 6 \cdot 25 - R_B \cdot 20 = 0
\]
\[
103.9 + 150 = R_B \cdot 20
\]
\[
253.9 = R_B \cdot 20
\]
\[
R_B = \frac{253.9}{20} = 12.695 \, \text{kN}
\]

### Step 4: Solve for \( R_A \)
Now substitute \( R_B = 12.695 \, \text{kN} \) into Equation 1:
\[
R_A + 12.695 = 16.39
\]
\[
R_A = 16.39 - 12.695 = 3.695 \, \text{kN}
\]

### Final Answer:
The vertical reaction at point B is approximately **12.70 kN**.

Would you like further details or clarifications?

Here are some related questions:
1. What would happen if the angle of the inclined force changes to 45°?
2. How would the reaction forces change if the length of the beam increases?
3. Can you calculate the horizontal reaction force at point A?
4. How does the support type (roller vs. pin) affect the reaction forces?
5. What would the moments look like if we considered point B instead of point A?

**Tip:** When solving for reactions, always start with the moment equation at a point where you can eliminate unknowns, simplifying the calculation.

Based on the structural system below, the vertical component of the reaction force B is equal to ____ kN.

This problem involves calculating the vertical reaction force at point B in a beam supported at two points and subjected to a 12 kN inclined force and a 6 kN vertical force. Using static equilibrium equations, the solution demonstrates key concepts like moments, equilibrium of forces, and vector components. Suitable for university-level engineering students.

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