A cable used to lift heavy materials like steel I-beams must be strong enough to resist breaking even under a load of 1.3  106 N. For safety, the cable must support twice that load.

(a) What cross-sectional area should the cable have if it's to be made of steel?
 
Make sure you are looking up the appropriate constant and using the proper relationship. In this case, you should be using the tensile strength of steel, because that is the force per cross-sectional area at which steel breaks. m2

(b) By how much will a 6.0-m length of this cable stretch when subject to the 1.3  106-N load?
 
Make sure you are using the appropriate constant and the correct relationship. In this case, you should be using Young's modulus for steel, because that is telling you how far the steel will stretch when a load is placed on it. m

To solve the problem, we’ll start with part (a) to find the required cross-sectional area of the cable, and then proceed to part (b) to determine how much it stretches under the load.

### Part (a): Cross-sectional Area of the Cable

**Given:**
- Load required to support: \( F = 2 \times 1.3 \times 10^6 \, \text{N} = 2.6 \times 10^6 \, \text{N} \)
- Tensile strength of steel: \( \sigma \) (approximately \( 4.0 \times 10^8 \, \text{N/m}^2 \))

**Formula:**
The tensile strength is defined as:
\[
\sigma = \frac{F}{A}
\]
Where:
- \( \sigma \) = tensile strength
- \( F \) = force (load)
- \( A \) = cross-sectional area

Rearranging the formula to solve for \( A \):
\[
A = \frac{F}{\sigma}
\]

**Calculation:**
\[
A = \frac{2.6 \times 10^6 \, \text{N}}{4.0 \times 10^8 \, \text{N/m}^2} = 0.0065 \, \text{m}^2
\]

### Part (b): Stretch of the Cable

**Given:**
- Length of the cable: \( L = 6.0 \, \text{m} \)
- Young's modulus for steel: \( E \) (approximately \( 2.0 \times 10^{11} \, \text{N/m}^2 \))

**Formula:**
The elongation (stretch) of the cable can be calculated using:
\[
\Delta L = \frac{F L}{A E}
\]
Where:
- \( \Delta L \) = change in length (stretch)
- \( F \) = force (load)
- \( L \) = original length
- \( A \) = cross-sectional area
- \( E \) = Young's modulus

**Calculation:**
\[
\Delta L = \frac{(2.6 \times 10^6 \, \text{N}) (6.0 \, \text{m})}{(0.0065 \, \text{m}^2)(2.0 \times 10^{11} \, \text{N/m}^2)}
\]
\[
\Delta L = \frac{15.6 \times 10^6}{1.3 \times 10^6} = 0.012 \, \text{m} = 12 \, \text{mm}
\]

### Summary of Results
(a) The cross-sectional area of the cable should be approximately **0.0065 m²**.

(b) The stretch of the cable when subject to a 1.3 million N load is approximately **0.012 m** or **12 mm**.

Would you like any further details on these calculations or have any questions? Here are some related questions to consider:

1. What are the implications of using a cable with insufficient cross-sectional area?
2. How do different materials compare in terms of tensile strength and Young's modulus?
3. What factors should be considered when designing a system that utilizes such cables?
4. How does temperature affect the tensile strength and Young's modulus of steel?
5. What safety factors are typically applied in engineering designs involving cables? 

**Tip:** Always consider including a safety factor in your designs to account for unexpected loads or material defects.

A cable used to lift heavy materials like steel I-beams must be strong enough to resist breaking even under a load of 1.3 106 N. For safety, the cable must support twice that load. (a) What cross-sectional area should the cable have if it's to be made of steel? (b) By how much will a 6.0-m length of this cable stretch when subject to the 1.3 106-N load?

Explore how to calculate the necessary cross-sectional area and stretch of a steel cable used to lift heavy materials. This problem involves a load of 2.6 million N, the tensile strength of steel, and Young's modulus, providing a comprehensive understanding for advanced students.

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