To solve the problem, we need to determine the stress required to produce a net elongation of 0.2 mm in a series arrangement of brass and steel wires.
### Given Data:
- Length of each wire, \( L = 1 \, \text{m} = 1000 \, \text{mm} \)
- Area of cross-section, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)
- Young’s Modulus for steel, \( Y_s = 120 \times 10^9 \, \text{N/m}^2 \)
- Young’s Modulus for brass, \( Y_b = 60 \times 10^9 \, \text{N/m}^2 \)
- Net elongation, \( \Delta L = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)
### Step-by-Step Solution:
1. **Understand the Series Connection**:
When two wires are connected in series, the total elongation (\( \Delta L \)) is the sum of the elongations of each wire:
\[
\Delta L = \Delta L_1 + \Delta L_2
\]
2. **Relate Elongation to Stress and Young's Modulus**:
The elongation of each wire can be expressed using the formula:
\[
\Delta L = \frac{F L}{A Y}
\]
Where \( F \) is the force applied, \( L \) is the length of the wire, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus.
3. **Calculate Elongation for Each Wire**:
For the brass wire:
\[
\Delta L_1 = \frac{F L}{A Y_b} = \frac{F \cdot 1}{1 \times 10^{-6} \cdot 60 \times 10^9}
\]
For the steel wire:
\[
\Delta L_2 = \frac{F L}{A Y_s} = \frac{F \cdot 1}{1 \times 10^{-6} \cdot 120 \times 10^9}
\]
4. **Substituting into the Total Elongation Equation**:
Substitute \( \Delta L_1 \) and \( \Delta L_2 \) into the total elongation equation:
\[
0.2 \times 10^{-3} = \frac{F \cdot 1}{1 \times 10^{-6} \cdot 60 \times 10^9} + \frac{F \cdot 1}{1 \times 10^{-6} \cdot 120 \times 10^9}
\]
5. **Factor Out Common Terms**:
Factor out \( F \) and the common area term:
\[
0.2 \times 10^{-3} = F \left( \frac{1}{1 \times 10^{-6} \cdot 60 \times 10^9} + \frac{1}{1 \times 10^{-6} \cdot 120 \times 10^9} \right)
\]
6. **Calculate the Right Side**:
Simplifying the right side:
\[
= F \left( \frac{1}{60 \times 10^3} + \frac{1}{120 \times 10^3} \right) = F \left( \frac{2 + 1}{120 \times 10^3} \right) = F \left( \frac{3}{120 \times 10^3} \right)
\]
7. **Setting Up the Equation**:
Now, equate and solve for \( F \):
\[
0.2 \times 10^{-3} = F \left( \frac{3}{120 \times 10^3} \right)
\]
\[
F = \frac{0.2 \times 10^{-3} \cdot 120 \times 10^3}{3} = \frac{24}{3} = 8 \, \text{N}
\]
8. **Calculate Stress**:
Stress (\( \sigma \)) is given by:
\[
\sigma = \frac{F}{A} = \frac{8}{1 \times 10^{-6}} = 8 \times 10^6 \, \text{N/m}^2 = 8 \times 10^8 \, \text{N/m}^2
\]
### Final Answer:
The stress required to produce a net elongation of 0.2 mm is \( 8 \times 10^8 \, \text{N/m}^2 \).
