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The problem you provided is about a simply supported beam under two forces, \( F_1 \) and \( F_2 \). Let's break down the steps to solve it.

### Given:
- \( F_1 = 6 \, \text{kN} \), inclined at 60° to the horizontal axis.
- \( F_2 = 6 \, \text{kN} \), applied vertically downward.
- Lengths of beam segments: \( 2.40 \, \text{m} \), \( 1.70 \, \text{m} \), and \( 4.40 \, \text{m} \).
- Total length of the beam = \( 2.40 + 1.70 + 4.40 = 8.50 \, \text{m} \).
- Supports at points A and B.

### Tasks:
1. **Vertical force sum (\( F_v \)):**
   You need to sum the vertical components of the applied forces.
   
   - The vertical component of \( F_1 \): 
     \[
     F_{1v} = F_1 \sin(60^\circ) = 6 \times \sin(60^\circ) = 6 \times 0.866 = 5.196 \, \text{kN}
     \]
   - The vertical component of \( F_2 \) is already vertical: \( F_{2v} = 6 \, \text{kN} \).

   Therefore, the total vertical force is:
   \[
   F_v = F_{1v} + F_2 = 5.196 \, \text{kN} + 6 \, \text{kN} = 11.196 \, \text{kN}
   \]

2. **Vertical reaction forces at A and B:**
   - Let \( R_A \) and \( R_B \) be the reaction forces at points A and B, respectively.
   - To solve for these reaction forces, we apply the equilibrium conditions:
     
     **Sum of vertical forces:**
     \[
     R_A + R_B = F_v = 11.196 \, \text{kN}
     \]
     
     **Sum of moments about A:**
     \[
     \text{Moment at A} = 0 = R_B \times 8.50 - F_1 \sin(60^\circ) \times 2.40 - F_2 \times 4.10
     \]
     Substituting values:
     \[
     R_B \times 8.50 - 5.196 \times 2.40 - 6 \times 4.10 = 0
     \]
     Simplifying:
     \[
     8.50 R_B = 12.47 + 24.6
     \]
     \[
     8.50 R_B = 37.07 \implies R_B = \frac{37.07}{8.50} = 4.36 \, \text{kN}
     \]

     Now substitute \( R_B \) into the equation \( R_A + R_B = 11.196 \):
     \[
     R_A = 11.196 - 4.36 = 6.84 \, \text{kN}
     \]

3. **Horizontal reaction force at B:**
   - Only the horizontal component of \( F_1 \) needs to be considered since \( F_2 \) is vertical.
   - The horizontal component of \( F_1 \) is:
     \[
     F_{1h} = F_1 \cos(60^\circ) = 6 \times \cos(60^\circ) = 6 \times 0.5 = 3 \, \text{kN}
     \]
     Therefore, the horizontal reaction force at B must balance this force:
     \[
     R_{Bh} = F_{1h} = 3 \, \text{kN}
     \]

### Summary:
- \( F_v = 11.196 \, \text{kN} \)
- \( R_A = 6.84 \, \text{kN} \)
- \( R_B = 4.36 \, \text{kN} \)
- Horizontal force at \( B = 3 \, \text{kN} \)

Would you like any further details or clarifications?

Here are 5 related questions to further explore the topic:
1. How would the solution change if the angle of \( F_1 \) were different?
2. What would happen if the beam were supported at only one point?
3. How do the moments of inertia affect beam deflection in this scenario?
4. What would be the impact of a third force applied horizontally?
5. How does the beam's material affect the reactions at supports?

**Tip**: When solving for reaction forces, always ensure to sum forces and moments to zero for a static system.

A simply supported beam is loaded with a vertical point load F2 and an inclined point load F1 at 60° to the beam's axis. Calculate: a) the total vertical force Fv, b) the vertical reaction forces at points A and B, and c) the horizontal reaction force at point B.

Solve a statically determinate beam problem subjected to vertical and inclined point loads. Includes the calculation of vertical and horizontal reaction forces using trigonometry and moment equilibrium. Suitable for undergraduate engineering students.

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