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The problem you’ve provided involves a bar suspended by two cables, one made of bronze and the other made of steel. The key task is to calculate the smallest allowable area for each cable, given that the bar has a mass of 1800 kg and the stress limits for the cables are provided as 60 MPa for the bronze and 120 MPa for the steel.

### Given information:
1. **Mass of the bar**: 1800 kg
2. **Stress limits**:
   - Bronze: \( \sigma_{\text{bronze}} = 60 \text{ MPa} \)
   - Steel: \( \sigma_{\text{steel}} = 120 \text{ MPa} \)
3. **Cable lengths**:
   - Length of the bronze section: \( L_{\text{bronze}} = 4 \text{ m} \)
   - Length of the steel section: \( L_{\text{steel}} = 3 \text{ m} \)
4. **Horizontal distance between the supports**: \( 10 \text{ m} \)

### Assumptions:
- The total weight of the bar is evenly distributed between the two supports.
- The cables are under tension, and we want to ensure that the tension in each cable does not exceed the stress limits given for each material.

### Step-by-Step Calculation:

1. **Weight of the bar**:
   The total weight of the bar is given by \( W = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity:
   \[
   W = 1800 \times 9.81 = 17658 \text{ N}
   \]
   This is the total downward force.

2. **Force distribution**:
   Let’s assume the forces in the bronze cable (\( F_{\text{bronze}} \)) and steel cable (\( F_{\text{steel}} \)) are distributed according to the geometry of the system. Since we are given the lengths of the segments and the horizontal distance, we can resolve the forces in both the vertical and horizontal directions.

I will now calculate these forces and the required areas for each cable based on the stress limits.The smallest allowable cross-sectional areas for the cables are:

- **Bronze cable**: 147.15 mm²
- **Steel cable**: 73.575 mm²

These values ensure that the tension in each cable does not exceed the stress limits of 60 MPa for the bronze and 120 MPa for the steel.

Would you like further details on how these values were calculated, or do you have any additional questions?

Here are five related questions for further exploration:
1. How would the required areas change if the mass of the bar increased to 2000 kg?
2. What would be the tension in each cable if only one of the cables was used?
3. How does the horizontal distance between the supports affect the force distribution?
4. What would happen to the cable areas if the stress limits for the materials were halved?
5. Can you calculate the elongation of each cable if the materials' Young's moduli were given?

**Tip:** The cross-sectional area is directly related to the stress limit and the force exerted, so doubling the stress limit will halve the required area.

A homogeneous 1800 kg bar AB is suspended by the end cables as shown. Calculate the smallest area of each cable if the tension is not to exceed 60 MPa for the bronze and 120 MPa for the steel.

Learn how to calculate the smallest cross-sectional areas for bronze and steel cables suspending a 1800 kg bar, ensuring that the stress limits of 60 MPa for bronze and 120 MPa for steel are not exceeded. Detailed calculations of force distribution and stress limits are provided.

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