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### Problem Overview:
You are tasked with determining the tension in cable \( AD \) in the given setup, where a light bar supports a 15 kg block at its midpoint \( C \). Rollers at points \( A \) and \( B \) rest against frictionless surfaces, and a horizontal cable \( AD \) is attached at point \( A \). The block's weight is the only force acting on the system vertically, while the cable exerts a horizontal force to balance the system.

### Given Data:
- Weight of block \( X \): \( 15 \, \text{kg} \)
- Gravitational acceleration \( g \): \( 9.81 \, \text{m/s}^2 \)
- Dimensions of the bar: \( 250 \, \text{mm} \) between points \( C \) and \( B \), and between \( C \) and \( A \)
- Vertical distance between point \( B \) and the horizontal line through \( D \): \( 350 \, \text{mm} \)

### Step-by-Step Explanation:

1. **Calculate the weight of the block:**
   \[
   W = m \cdot g = 15 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 147.15 \, \text{N}
   \]
   This is the downward force acting at point \( C \) due to the block's weight.

2. **Determine forces and moments:**
   Since the system is in equilibrium, we apply the following conditions:
   - The sum of forces in both the horizontal (\( \Sigma F_x = 0 \)) and vertical (\( \Sigma F_y = 0 \)) directions must be zero.
   - The sum of moments about any point (commonly about point \( A \)) must also be zero (\( \Sigma M_A = 0 \)).

3. **Set up the moment equation:**
   Taking moments about point \( A \), the moment caused by the weight \( W \) acts at the midpoint \( C \) (since it is a uniform bar). The distance from \( A \) to \( C \) is \( 250 \, \text{mm} = 0.25 \, \text{m} \), and the moment arm for the force at \( B \) is the vertical distance of \( 350 \, \text{mm} = 0.35 \, \text{m} \).
   \[
   \Sigma M_A = 0 \implies B \cdot 0.35 \, \text{m} - 147.15 \, \text{N} \cdot 0.25 \, \text{m} = 0
   \]
   Solving for the reaction force at \( B \):
   \[
   B = \frac{147.15 \cdot 0.25}{0.35} = 105.1 \, \text{N}
   \]
   This is the force acting in the negative \( x \)-direction at point \( B \) to counterbalance the weight of the block.

4. **Horizontal equilibrium:**
   Now, considering horizontal equilibrium (\( \Sigma F_x = 0 \)), the only horizontal force is the tension \( T_{AD} \) in cable \( AD \), which must balance the force at \( B \):
   \[
   T_{AD} = B = 105.1 \, \text{N}
   \]

### Final Answer:
The tension in the cable \( AD \) is **105.1 N**.

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Would you like further clarification or details on any step? Here are some related questions to deepen your understanding:

1. How would the tension in cable \( AD \) change if the weight of the block increased?
2. What happens to the forces if the bar between \( A \) and \( B \) is not frictionless?
3. How does the horizontal distance between \( A \) and \( B \) affect the tension in the cable?
4. What if the block's weight was applied closer to point \( A \) rather than at midpoint \( C \)?
5. Can you calculate the vertical reaction force at point \( A \)?

**Tip:** When dealing with forces in equilibrium, always break the problem into horizontal and vertical components, and don't forget to apply moment equations when needed.

Learn how to determine the tension in cable AD of a bar system supporting a 15 kg block. This problem covers key concepts in statics, including the equilibrium of forces and moments. Suitable for undergraduate engineering students.

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