Question 7

A copper cylinder and a brass cylinder are stacked end to end, as in the drawing. Each cylinder has a radius of [tex]0.25 \, \text{cm}[/tex]. A compressive force of [tex]F = 6500 \, \text{N}[/tex] is applied to the right end of the brass cylinder. Find the amount by which the length of the stack decreases.

Both cylinders experience the same force [tex]F[/tex]. The magnitude of this force is related to the change in length of each cylinder according to [tex]F = Y \left(\frac{\Delta L}{L_0}\right) A[/tex]. Each cylinder decreases in length; the total decrease is the sum of the decreases for each cylinder.

[tex]\[ F = Y \left(\frac{\Delta L}{L_0}\right) A \Rightarrow \Delta L_{\text{copper}} = \frac{F L_0}{Y A} = \frac{6500 \times 3.0 \times 10^{-2}}{-\left(0.25 \times 10^{-2}\right)^2} = 9.2 \times 10^{-5} \, \text{m} \][/tex]

(Note: Ensure the units and values in the equations are correctly applied according to the specific properties of copper and brass, such as Young's modulus [tex]Y[/tex] and original length [tex]L_0[/tex]).



Answer :

Sure! Let's solve the problem step-by-step:

### Given Data:
1. Radius of each cylinder: [tex]\( r = 0.25 \, \text{cm} = 0.25 \times 10^{-2} \, \text{m} = 0.0025 \, \text{m} \)[/tex]
2. Compressive force: [tex]\( F = 6500 \, \text{N} \)[/tex]
3. Young's Modulus for Copper (Y_copper): [tex]\( Y_{\text{copper}} = 1.1 \times 10^{11} \, \text{Pa} \)[/tex]
4. Young's Modulus for Brass (Y_brass): [tex]\( Y_{\text{brass}} = 9.0 \times 10^{10} \, \text{Pa} \)[/tex]
5. Original length of the Copper cylinder (L0_copper): [tex]\( L0_{\text{copper}} = 0.03 \, \text{m} \)[/tex]
6. Original length of the Brass cylinder (L0_brass): [tex]\( L0_{\text{brass}} = 0.03 \, \text{m} \)[/tex]

### Step-by-Step Solution:

1. Calculate the cross-sectional area (A) of each cylinder:

[tex]\[ A = \pi r^2 = \pi (0.0025 \, \text{m})^2 = \pi \times 6.25 \times 10^{-6} \, \text{m}^2 = 1.963495408 \times 10^{-5} \, \text{m}^2. \][/tex]

2. Calculate the deformation (ΔL) for the copper cylinder:

[tex]\[ \Delta L_{\text{copper}} = \frac{F \cdot L0_{\text{copper}}}{Y_{\text{copper}} \cdot A} \][/tex]

Substitute the values:

[tex]\[ \Delta L_{\text{copper}} = \frac{6500 \, \text{N} \times 0.03 \, \text{m}}{1.1 \times 10^{11} \, \text{Pa} \times 1.963495408 \times 10^{-5} \, \text{m}^2}. \][/tex]

[tex]\[ \Delta L_{\text{copper}} \approx 9.028425862667518 \times 10^{-5} \, \text{m} = 0.000090284 \, \text{m}. \][/tex]

3. Calculate the deformation (ΔL) for the brass cylinder:

[tex]\[ \Delta L_{\text{brass}} = \frac{F \cdot L0_{\text{brass}}}{Y_{\text{brass}} \cdot A} \][/tex]

Substitute the values:

[tex]\[ \Delta L_{\text{brass}} = \frac{6500 \, \text{N} \times 0.03 \, \text{m}}{9.0 \times 10^{10} \, \text{Pa} \times 1.963495408 \times 10^{-5} \, \text{m}^2}. \][/tex]

[tex]\[ \Delta L_{\text{brass}} \approx 0.00011034742721038078 \, \text{m} = 0.000110347 \, \text{m}. \][/tex]

4. Calculate the total decrease in length of the stack:

[tex]\[ \Delta L_{\text{total}} = \Delta L_{\text{copper}} + \Delta L_{\text{brass}} \][/tex]

[tex]\[ \Delta L_{\text{total}} = 0.000090284 \, \text{m} + 0.000110347 \, \text{m}. \][/tex]

[tex]\[ \Delta L_{\text{total}} \approx 0.00020063168583705595 \, \text{m} = 0.000200632 \, \text{m}. \][/tex]

### Conclusion:
The amount by which the length of the stack decreases is approximately [tex]\(0.000200632 \, \text{m}\)[/tex] or [tex]\(0.20063 \, \text{mm}\)[/tex].