A 6-meter cable is attached to two points of the same elevation of the ceiling room. The two points are three meters apart. A 100 N load is tied to the center of the cable. Determine the stresses in Pascals of each cable if their cross-sectional area is equal to 5 mm^2.

To determine the stress in each cable, we need to follow these steps:

### 1. Geometry and Forces Analysis

Given:
- Length of the cable: \( L = 6 \, \text{meters} \)
- Distance between the two points on the ceiling: \( d = 3 \, \text{meters} \)
- Load attached at the center of the cable: \( W = 100 \, \text{N} \)
- Cross-sectional area of each cable: \( A = 5 \, \text{mm}^2 = 5 \times 10^{-6} \, \text{m}^2 \)

When the load is attached at the center, the cable forms a symmetrical shape, and the vertical displacement of the load creates two symmetrical segments of the cable.

Let \( T_1 \) and \( T_2 \) be the tensions in the left and right parts of the cable, respectively. Since the problem is symmetrical, \( T_1 = T_2 = T \).

### 2. Horizontal and Vertical Force Equilibrium

- **Horizontal equilibrium**: The horizontal components of the tensions must balance out.
  
  \[
  T \cdot \cos(\theta) + T \cdot \cos(\theta) = 0 \implies 2T \cdot \cos(\theta) = 0
  \]

- **Vertical equilibrium**: The vertical components must support the load.

  \[
  2T \cdot \sin(\theta) = W
  \]

  \[
  T = \frac{W}{2 \sin(\theta)}
  \]

Where \( \theta \) is the angle that the cable makes with the horizontal.

### 3. Geometry to Find \( \sin(\theta) \)

The geometry of the problem gives us:

\[
\cos(\theta) = \frac{d/2}{L/2} = \frac{3/2}{6/2} = \frac{3}{6} = 0.5
\]

\[
\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - 0.5^2} = \sqrt{1 - 0.25} = \sqrt{0.75} = \frac{\sqrt{3}}{2}
\]

### 4. Calculate the Tension \( T \)

Now, substitute \( \sin(\theta) \):

\[
T = \frac{100 \, \text{N}}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{100}{\sqrt{3}} \approx 57.74 \, \text{N}
\]

### 5. Calculate the Stress

Stress \( \sigma \) in each cable is given by:

\[
\sigma = \frac{T}{A} = \frac{57.74 \, \text{N}}{5 \times 10^{-6} \, \text{m}^2} = 11.548 \times 10^6 \, \text{Pa} = 11.548 \, \text{MPa}
\]

### Final Answer:

The stress in each cable is approximately **11.548 MPa**.

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Do you want further details or have any questions?

Here are 5 related questions for deeper understanding:

1. How would the stress change if the load was increased to 200 N?
2. What would happen to the tension if the points were 4 meters apart instead of 3 meters?
3. How would the stress change if the cross-sectional area was increased to 10 mm²?
4. Can you derive the general formula for stress given any load, distance, and cable length?
5. How would the system change if the load was not placed at the center but shifted to one side?

**Tip:** Always ensure that units are consistent when calculating stress to avoid errors in the final result.

Learn how to calculate stress in cables when a 100 N load is applied at the center of a 6-meter cable suspended 3 meters apart. Detailed step-by-step solution using statics and trigonometry to find tensions and stress. Cross-sectional area of cables is 5 mm².

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