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The problem in the image involves a rod supported by smooth journal bearings at points \( A \), \( B \), and \( C \), and subjected to two forces (one of 400 N acting vertically at point C). We are tasked with determining the reaction forces at these three supports.

### Step-by-Step Process:

1. **Assumptions**:
   - The bearings allow rotation but prevent translation in both the \( x \) and \( y \) directions, hence producing reaction forces.
   - At point \( A \), the reactions are \( A_x \) and \( A_z \) (due to smooth bearings).
   - At point \( B \), the reactions are \( B_x \) and \( B_z \).
   - At point \( C \), the reaction is vertical, denoted by \( C_y \), as there is a vertical load of 400 N acting at this point.

2. **Equilibrium Equations**:
   The system is in static equilibrium, so we apply the following equilibrium conditions:
   - \( \Sigma F_x = 0 \): The sum of the forces in the \( x \)-direction is zero.
   - \( \Sigma F_z = 0 \): The sum of the forces in the \( z \)-direction is zero.
   - \( \Sigma M = 0 \): The sum of the moments about any point must be zero.

### Equilibrium Conditions:

1. **Forces in the \( x \)-direction**:
   \[
   A_x + B_x = 0 \quad \text{(since there are no external forces in the x-direction)}
   \]

2. **Forces in the \( z \)-direction**:
   \[
   A_z + B_z = 400 \, \text{N} \quad \text{(because the 400 N vertical force at C is counteracted by vertical reactions)}
   \]

3. **Moments about Point A**:
   Taking moments about point \( A \) (counterclockwise as positive):
   - The 400 N force at \( C \) creates a moment, and the perpendicular distance is 1.2 m.
   - The vertical force \( B_z \) also creates a moment, with a distance of 0.6 m.

   The moment equation about \( A \) becomes:
   \[
   400 \, \text{N} \times 1.2 \, \text{m} - B_z \times 0.6 \, \text{m} = 0
   \]

   Solving for \( B_z \):
   \[
   B_z = \frac{400 \times 1.2}{0.6} = 800 \, \text{N}
   \]

4. **Substitute \( B_z \) into the vertical equilibrium equation**:
   \[
   A_z + 800 = 400
   \]
   Therefore, \( A_z = -400 \, \text{N} \) (which means it is acting downward).

5. **Conclusion**:
   - \( A_x = 0 \)
   - \( B_x = 0 \)
   - \( A_z = -400 \, \text{N} \)
   - \( B_z = 800 \, \text{N} \)
   - \( C_y = 400 \, \text{N} \)

These are the reaction forces at points \( A \), \( B \), and \( C \).

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Do you have any questions or would you like further details on the calculations?

### Related Questions:
1. What would happen if the 400 N force at point C were increased to 600 N?
2. How would the reactions change if the rod was supported by only two bearings?
3. What if the force acted at an angle instead of vertically at point C?
4. How can we calculate the reaction forces if there were an additional force acting along the rod?
5. How would the moment equilibrium equation change if the distances between the points were altered?

**Tip**: Always resolve forces and moments in terms of known dimensions and directions for easier calculation.

Determine the reactions at the supports A, B, and C for the rod subjected to two forces and supported by smooth journal bearings.

Calculate the reactions at supports A, B, and C for a rod subjected to a 400 N force using statics principles. The problem involves applying equilibrium conditions for forces and moments, ideal for high school or early college students.

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