An aluminum cylinder `10` cm long with a cross section area of `20cm^2` is used as a spacer between two steel walls. At `17.2^(@)C` it just slips in between the walls. When it warms to `22.3^(@)C` calculate the stress in the cylinder and the total force it exerts on each wall, assuming that the walls are perfectly rigid and a constant distance apart.

Explanation

To solve the problem step-by-step, we will calculate the thermal stress in the aluminum cylinder and then determine the total force it exerts on each wall.

### Step 1: Identify the given data
- Length of the aluminum cylinder, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Cross-sectional area, \( A = 20 \, \text{cm}^2 = 20 \times 10^{-4} \, \text{m}^2 = 0.002 \, \text{m}^2 \)
- Initial temperature, \( T_1 = 17.2^\circ C \)
- Final temperature, \( T_2 = 22.3^\circ C \)
- Young's modulus for aluminum, \( E = 7 \times 10^{10} \, \text{Pa} \)
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