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Alright, let’s solve the given problem step-by-step:
### Problem:
A steel wire having a length of 8 meters and a diameter of 4 millimeters is fixed between two rigid supports. Calculate the increase in the tension of the wire when the temperature drops by [tex]\( 10^\circ \text{C} \)[/tex]. The Young's modulus [tex]\( Y \)[/tex] of steel is [tex]\( 2 \times 10^{11} \, \text{N/m}^2 \)[/tex] and the coefficient of thermal expansion [tex]\( \alpha_{\text{steel}} \)[/tex] is [tex]\( 12 \times 10^{-5} \, \text{K}^{-1} \)[/tex].
### Step-by-Step Solution:
#### Step 1: Convert the Given Dimensions to Compatible Units
1. Length of the wire [tex]\( L = 8 \, \text{m} \)[/tex].
2. Diameter of the wire [tex]\( d = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)[/tex].
#### Step 2: Calculate the Cross-Sectional Area
The cross-sectional area [tex]\( A \)[/tex] of the wire, assuming it is circular, is given by the formula:
[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]
Plugging in the diameter:
[tex]\[ A = \pi \left( \frac{4 \times 10^{-3}}{2} \right)^2 = \pi \left( 2 \times 10^{-3} \right)^2 \][/tex]
[tex]\[ A = \pi \times 4 \times 10^{-6} \][/tex]
[tex]\[ A = 4\pi \times 10^{-6} \, \text{m}^2 \][/tex]
Using [tex]\( \pi \approx 3.141592653589793 \)[/tex]:
[tex]\[ A \approx 4 \times 3.141592653589793 \times 10^{-6} \][/tex]
[tex]\[ A \approx 1.2566370614359172 \times 10^{-5} \, \text{m}^2 \][/tex]
#### Step 3: Calculate the Thermal Strain
The thermal strain [tex]\( \Delta L / L \)[/tex] is given by the product of the coefficient of thermal expansion [tex]\( \alpha \)[/tex] and the change in temperature [tex]\( \Delta T \)[/tex]:
[tex]\[ \text{Thermal Strain} = \alpha \times \Delta T \][/tex]
Given:
[tex]\[ \alpha_{\text{steel}} = 12 \times 10^{-5} \, \text{K}^{-1} \][/tex]
[tex]\[ \Delta T = -10 \, \text{K} \][/tex]
Thus:
[tex]\[ \text{Thermal Strain} = 12 \times 10^{-5} \times (-10) \][/tex]
[tex]\[ \text{Thermal Strain} = -0.0012 \][/tex]
This negative result indicates contraction.
#### Step 4: Calculate the Tension Increase Using Hooke's Law
The increase in tension [tex]\( T \)[/tex] due to thermal contraction can be found using Hooke’s law for thermal stress:
[tex]\[ T = Y \times A \times \text{Thermal Strain} \][/tex]
Given:
[tex]\[ Y = 2 \times 10^{11} \, \text{N/m}^2 \][/tex]
So:
[tex]\[ T = 2 \times 10^{11} \times 1.2566370614359172 \times 10^{-5} \times (-0.0012) \][/tex]
[tex]\[ T = 2 \times 10^{11} \times 1.2566370614359172 \times 10^{-5} \times -0.0012 \][/tex]
[tex]\[ T = 2 \times 1.2566370614359172 \times 10^{6} \times -0.0012 \, \text{N} \][/tex]
[tex]\[ T \approx -3.0159289474462016 \times 10^3 \, \text{N} \][/tex]
Since tension can’t be negative, we take the absolute value, indicating an increase in tension:
[tex]\[ T \approx 3015.9289474462016 \, \text{N} \][/tex]
### Conclusion:
The increase in the tension of the steel wire when the temperature falls by [tex]\( 10^\circ \text{C} \)[/tex] is approximately [tex]\( 3015.93 \)[/tex] N.
### Problem:
A steel wire having a length of 8 meters and a diameter of 4 millimeters is fixed between two rigid supports. Calculate the increase in the tension of the wire when the temperature drops by [tex]\( 10^\circ \text{C} \)[/tex]. The Young's modulus [tex]\( Y \)[/tex] of steel is [tex]\( 2 \times 10^{11} \, \text{N/m}^2 \)[/tex] and the coefficient of thermal expansion [tex]\( \alpha_{\text{steel}} \)[/tex] is [tex]\( 12 \times 10^{-5} \, \text{K}^{-1} \)[/tex].
### Step-by-Step Solution:
#### Step 1: Convert the Given Dimensions to Compatible Units
1. Length of the wire [tex]\( L = 8 \, \text{m} \)[/tex].
2. Diameter of the wire [tex]\( d = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)[/tex].
#### Step 2: Calculate the Cross-Sectional Area
The cross-sectional area [tex]\( A \)[/tex] of the wire, assuming it is circular, is given by the formula:
[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]
Plugging in the diameter:
[tex]\[ A = \pi \left( \frac{4 \times 10^{-3}}{2} \right)^2 = \pi \left( 2 \times 10^{-3} \right)^2 \][/tex]
[tex]\[ A = \pi \times 4 \times 10^{-6} \][/tex]
[tex]\[ A = 4\pi \times 10^{-6} \, \text{m}^2 \][/tex]
Using [tex]\( \pi \approx 3.141592653589793 \)[/tex]:
[tex]\[ A \approx 4 \times 3.141592653589793 \times 10^{-6} \][/tex]
[tex]\[ A \approx 1.2566370614359172 \times 10^{-5} \, \text{m}^2 \][/tex]
#### Step 3: Calculate the Thermal Strain
The thermal strain [tex]\( \Delta L / L \)[/tex] is given by the product of the coefficient of thermal expansion [tex]\( \alpha \)[/tex] and the change in temperature [tex]\( \Delta T \)[/tex]:
[tex]\[ \text{Thermal Strain} = \alpha \times \Delta T \][/tex]
Given:
[tex]\[ \alpha_{\text{steel}} = 12 \times 10^{-5} \, \text{K}^{-1} \][/tex]
[tex]\[ \Delta T = -10 \, \text{K} \][/tex]
Thus:
[tex]\[ \text{Thermal Strain} = 12 \times 10^{-5} \times (-10) \][/tex]
[tex]\[ \text{Thermal Strain} = -0.0012 \][/tex]
This negative result indicates contraction.
#### Step 4: Calculate the Tension Increase Using Hooke's Law
The increase in tension [tex]\( T \)[/tex] due to thermal contraction can be found using Hooke’s law for thermal stress:
[tex]\[ T = Y \times A \times \text{Thermal Strain} \][/tex]
Given:
[tex]\[ Y = 2 \times 10^{11} \, \text{N/m}^2 \][/tex]
So:
[tex]\[ T = 2 \times 10^{11} \times 1.2566370614359172 \times 10^{-5} \times (-0.0012) \][/tex]
[tex]\[ T = 2 \times 10^{11} \times 1.2566370614359172 \times 10^{-5} \times -0.0012 \][/tex]
[tex]\[ T = 2 \times 1.2566370614359172 \times 10^{6} \times -0.0012 \, \text{N} \][/tex]
[tex]\[ T \approx -3.0159289474462016 \times 10^3 \, \text{N} \][/tex]
Since tension can’t be negative, we take the absolute value, indicating an increase in tension:
[tex]\[ T \approx 3015.9289474462016 \, \text{N} \][/tex]
### Conclusion:
The increase in the tension of the steel wire when the temperature falls by [tex]\( 10^\circ \text{C} \)[/tex] is approximately [tex]\( 3015.93 \)[/tex] N.
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