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The problem shown in the image involves calculating the forces exerted on two supports, \( CA \) and \( CB \), by a system under a vertical load \( F = 4 \, \text{kN} \). The load is placed between the supports at a distance of \( 5 \, \text{m} \) from \( CA \) and \( 3 \, \text{m} \) from \( CB \).

### Approach:
This is a classic static equilibrium problem where the goal is to decompose the total force acting on the beam into the forces acting on the supports \( CA \) and \( CB \). The system is in equilibrium, meaning the sum of forces and the sum of moments (torques) about any point must be zero.

#### Step 1: Equilibrium of Forces
The total vertical force acting on the beam must equal the sum of the forces exerted by the supports:
\[
F = F_{CA} + F_{CB}
\]
where \( F_{CA} \) and \( F_{CB} \) are the reaction forces at supports \( CA \) and \( CB \), respectively.

#### Step 2: Equilibrium of Moments
To calculate the forces at the supports, we'll apply the moment equilibrium equation around one of the supports. Taking moments around \( CA \) (to eliminate \( F_{CA} \) from the equation), we have:
\[
\sum M_{CA} = 0
\]
The moment caused by the vertical force \( F = 4 \, \text{kN} \) acts at a distance of \( 5 \, \text{m} + 3 \, \text{m} = 8 \, \text{m} \). The moment at \( CB \) acts at a distance of \( 5 \, \text{m} \) from \( CA \). This gives the equation:
\[
F \times 5 \, \text{m} = F_{CB} \times (5 \, \text{m} + 3 \, \text{m})
\]
Substituting the values:
\[
4 \, \text{kN} \times 5 \, \text{m} = F_{CB} \times 8 \, \text{m}
\]
\[
20 \, \text{kN} \cdot \text{m} = 8 \, F_{CB}
\]
\[
F_{CB} = \frac{20 \, \text{kN} \cdot \text{m}}{8 \, \text{m}} = 2.5 \, \text{kN}
\]

#### Step 3: Solve for \( F_{CA} \)
Now, using the sum of vertical forces:
\[
F_{CA} + F_{CB} = 4 \, \text{kN}
\]
\[
F_{CA} + 2.5 \, \text{kN} = 4 \, \text{kN}
\]
\[
F_{CA} = 4 \, \text{kN} - 2.5 \, \text{kN} = 1.5 \, \text{kN}
\]

### Final Answers:
- \( F_{CA} = 1.5 \, \text{kN} \)
- \( F_{CB} = 2.5 \, \text{kN} \)

These forces can now be placed on the diagram.

Would you like any further details or clarifications?

Here are 5 related questions to deepen your understanding:
1. How does changing the position of the load affect the forces on the supports?
2. What would the reaction forces be if the load were centered between the supports?
3. How would you calculate the reaction forces if the beam were not horizontal but inclined?
4. How can the moment of a force be calculated if the load is applied at an angle?
5. What would happen to the reaction forces if the beam were longer but the load remained at the same position?

**Tip**: Always check units carefully when performing calculations involving forces and moments to avoid scaling errors.

Hallar las cargas que recibe cada una de las columnas CA y CB según el esquema dado, donde una carga F=4KN se aplica en una barra soportada por dos columnas con distancias de 5.0m a CA y 3.0m a CB.

Solve the static equilibrium problem of a beam with a 4KN load applied, supported by two columns at different distances. Learn the step-by-step approach to calculate the forces acting on the supports CA and CB using force and moment equilibrium principles. Suitable for university-level engineering students.

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