A flexible chain of weight W hangs between two fixed points A and B at the same level. The inclination of the chain with the horizontal at the two points of support is `theta`. Calculated the tension of the chain at end points and also at the lower most point .
[Hints : Consider equilibrium of each half of the chain. ]

Answer

To solve the problem of a flexible chain hanging between two fixed points A and B at the same level with an inclination of `theta` at the endpoints, we will analyze the forces acting on the chain and apply the principles of equilibrium.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The chain has a total weight \( W \) and is suspended between two fixed points A and B.
- The angle of inclination with the horizontal at points A and B is \( \theta \).
- The chain can be considered to be in static equilibrium.

2. **Dividing the Chain**:
- We can divide the chain into two halves for analysis. Let’s consider one half of the chain from point A to the midpoint of the chain.

3. **Forces Acting on the Half Chain**:
- The weight of the half chain can be represented as \( \frac{W}{2} \) acting downwards at the midpoint.
- The tension at point A is denoted as \( T_1 \) and at the midpoint as \( T_2 \).

4. **Resolving Tension into Components**:
- The tension \( T_1 \) at point A can be resolved into two components:
- Horizontal component: \( T_1 \cos(\theta) \)
- Vertical component: \( T_1 \sin(\theta) \)

5. **Applying Equilibrium Conditions**:
- For vertical equilibrium, the upward force must equal the downward force:
\[ T_1 \sin(\theta) = \frac{W}{2} \]
- For horizontal equilibrium, the horizontal components must balance:
\[ T_2 = T_1 \cos(\theta) \]

6. **Calculating Tension at Point A**:
- From the vertical equilibrium equation, we can solve for \( T_1 \):
\[ T_1 = \frac{W}{2 \sin(\theta)} \]

7. **Calculating Tension at the Midpoint**:
- Substitute \( T_1 \) into the horizontal equilibrium equation to find \( T_2 \):
\[ T_2 = T_1 \cos(\theta) = \left(\frac{W}{2 \sin(\theta)}\right) \cos(\theta) = \frac{W \cos(\theta)}{2 \sin(\theta)} \]

8. **Final Results**:
- The tension at the endpoints (point A and point B) is:
\[ T_1 = \frac{W}{2 \sin(\theta)} \]
- The tension at the lowest point (midpoint) is:
\[ T_2 = \frac{W \cos(\theta)}{2 \sin(\theta)} \]

### Summary of Results:
- Tension at endpoints \( T_1 = \frac{W}{2 \sin(\theta)} \)
- Tension at the lowest point \( T_2 = \frac{W \cos(\theta)}{2 \sin(\theta)} \)