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We are tasked with determining the required diameter \(d\) of the steel wire so that the stress does not exceed 140 MPa, given a tensile load of 2000 N and other conditions from the image. Let's work through the problem step by step.

### Given Data:
- Length of the wire \(L = 15 \, \text{m}\)
- Load \(P = 2000 \, \text{N}\)
- Maximum allowable stress \( \sigma = 140 \, \text{MPa} = 140 \times 10^6 \, \text{Pa} \)
- Maximum allowable elongation \( \Delta L = 6 \, \text{mm} = 0.006 \, \text{m} \)
- Young’s modulus \( E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa} \)

### Step 1: Relationship for Stress
Stress \( \sigma \) is related to the force \( P \) and the cross-sectional area \( A \) by the formula:
\[
\sigma = \frac{P}{A}
\]
For a circular cross-section, the area \( A \) is:
\[
A = \frac{\pi d^2}{4}
\]
where \(d\) is the diameter.

Substitute \(A\) into the stress equation:
\[
\sigma = \frac{4P}{\pi d^2}
\]
Rearranging for \(d\):
\[
d = \sqrt{\frac{4P}{\pi \sigma}}
\]
Now, let's plug in the values:
\[
d = \sqrt{\frac{4 \times 2000}{\pi \times 140 \times 10^6}}
\]

### Step 2: Elongation Condition
The elongation \( \Delta L \) of the wire can be calculated using Hooke's Law:
\[
\Delta L = \frac{PL}{AE}
\]
Again, using the area \( A = \frac{\pi d^2}{4} \), we substitute it into the elongation equation:
\[
\Delta L = \frac{4PL}{\pi d^2 E}
\]
Rearranging to solve for \( d \) using the maximum elongation:
\[
d = \sqrt{\frac{4PL}{\pi E \Delta L}}
\]
Substitute the values:
\[
d = \sqrt{\frac{4 \times 2000 \times 15}{\pi \times 200 \times 10^9 \times 0.006}}
\]

### Conclusion:
I'll compute both values for the diameter based on the stress and elongation conditions. Let me calculate it for you.The required diameters based on both conditions are:

1. **Stress condition:** \( d = 4.26 \, \text{mm} \)
2. **Elongation condition:** \( d = 5.64 \, \text{mm} \)

Since the problem specifies that the total elongation must not exceed 6 mm, we use the larger diameter from the elongation condition, which is **5.64 mm**. Therefore, the required diameter is approximately **5.64 mm** to satisfy both conditions.

Would you like further details or clarifications? Here are some related questions you might find useful:

1. How does Young's modulus affect material elongation?
2. What happens if the diameter chosen is below the calculated value for elongation?
3. Can you explain how stress and strain are related in materials?
4. How does the length of the wire influence its elongation?
5. What safety factors should be considered in such stress-strain analysis?

**Tip:** Always check both stress and strain limits to ensure material safety and functionality under load.

Determine the required diameter if the stress is not to exceed 140 MPa.

Calculate the required diameter of a 15 m steel wire supporting a 2000 N load under a maximum stress of 140 MPa. This example demonstrates the use of Hooke's Law, stress-strain analysis, and circular cross-sectional area formulas. Explore the relationship between stress, elongation, and material properties.

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