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What is the vertical movement in the aluminum bar?

To solve for the vertical movement (or displacement) in the aluminum bar, we need to apply the concepts of **compatibility of deformation** and **equilibrium of forces**.

### Given Data:
- **Steel Bar:**
  - Length \( L_{\text{steel}} = 4 \, \text{m} \)
  - Cross-sectional area \( A_{\text{steel}} = 300 \, \text{mm}^2 = 300 \times 10^{-6} \, \text{m}^2 \)
  - Young’s Modulus \( E_{\text{steel}} = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa} \)

- **Aluminum Bar:**
  - Length \( L_{\text{al}} = 5 \, \text{m} \)
  - Cross-sectional area \( A_{\text{al}} = 500 \, \text{mm}^2 = 500 \times 10^{-6} \, \text{m}^2 \)
  - Young’s Modulus \( E_{\text{al}} = 70 \, \text{GPa} = 70 \times 10^9 \, \text{Pa} \)

- **Load Applied**: \( P = 70 \, \text{kN} = 70,000 \, \text{N} \)
  
- Distances:
  - Distance from the steel bar to the load point: \( 2 \, \text{m} \)
  - Distance from the aluminum bar to the load point: \( 3 \, \text{m} \)

### Step 1: Force distribution

Assuming the bar remains horizontal after the load is applied, the deformation in both the steel and aluminum rods must be equal (compatibility condition). The forces on each rod will vary based on their stiffness.

Let the forces in the steel and aluminum bars be \( F_{\text{steel}} \) and \( F_{\text{al}} \), respectively. By equilibrium of forces:

\[
F_{\text{steel}} + F_{\text{al}} = P = 70,000 \, \text{N}
\]

### Step 2: Deformation of the bars

The elongation (or deformation) of each rod is given by:

\[
\delta_{\text{steel}} = \frac{F_{\text{steel}} L_{\text{steel}}}{A_{\text{steel}} E_{\text{steel}}}
\]
\[
\delta_{\text{al}} = \frac{F_{\text{al}} L_{\text{al}}}{A_{\text{al}} E_{\text{al}}}
\]

Since both deformations are equal due to compatibility:

\[
\delta_{\text{steel}} = \delta_{\text{al}}
\]

Substitute the deformation equations:

\[
\frac{F_{\text{steel}} L_{\text{steel}}}{A_{\text{steel}} E_{\text{steel}}} = \frac{F_{\text{al}} L_{\text{al}}}{A_{\text{al}} E_{\text{al}}}
\]

Substitute the given values:

\[
\frac{F_{\text{steel}} \times 4}{300 \times 10^{-6} \times 200 \times 10^9} = \frac{F_{\text{al}} \times 5}{500 \times 10^{-6} \times 70 \times 10^9}
\]

Simplify this equation to find the relationship between \( F_{\text{steel}} \) and \( F_{\text{al}} \).

### Step 3: Solve for \( F_{\text{steel}} \) and \( F_{\text{al}} \)

Once the forces in each bar are found, we can substitute them back into either deformation equation to find the vertical movement \( \delta \).

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Here are some related questions that could help deepen your understanding:
1. How is compatibility of deformation applied in problems involving different materials?
2. Why does the stiffness of a material affect the distribution of forces between bars?
3. What would happen to the system if the materials were switched between the left and right bars?
4. How does Young's Modulus impact the elongation of a bar under load?
5. What is the significance of ensuring equilibrium in these kinds of problems?

**Tip**: Always check that your units are consistent throughout the calculations to avoid errors!

Learn how to calculate the vertical displacement of an aluminum bar subjected to a 70 kN load, considering the equilibrium of forces and compatibility of deformation between a steel and aluminum bar. This solution involves applying Hooke's Law and the deformation formula for materials.

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