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issacnewton
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Hi
I am trying to solve this problem. I will present my solution though I couldn't get the tension right.
At 40 Celsius , there is no tension in the wire. When the wires are cooled , they contract and the tension increases. But the tensile stress, which is force per unit area, will remain the same for both the wires, even if they are different in constitution.
I am neglecting the increase in radius, hence the area of the wires, because its negligible.
When the tensions develop , let the steel wire be stretched by amount \Delta x_s
from its reduced length Ls. Similarly, let copper wire be stretched by amount
\Delta x_c from its reduced length Lc. Let L = 2 m be the original length of each wire. Since
\mbox{Young's Modulus}\, =\frac{\mbox{Tensile Stress}}{\mbox{Tensile Strain}}
we have
\frac{F}{A}= Y_s \,\frac{\Delta x_s}{L_s} = Y_c \,\frac{\Delta x_c}{L_c}
where Ys and Yc are Young's Moduli for the steel and copper.
They are
Y_s=20 \times 10^{10} N/m^2
Y_c= 11 \times 10^{10} N/m^2
so from above equation, we get , rearranging
\Delta x_s = \left(\frac{Y_c}{Y_s}\right)\left(\frac{L_s}{L_c}\right) \Delta x_c
Since amount of stretching must be equal to the thermal contraction, we have
\Delta x_s + \Delta x_c= \Delta L_s + \Delta L_c
where \Delta L_s \cdots \Delta L_c are change in the length of the steel and copper rods due to cooling. Also note that
L_s = L\left[1-\alpha_s (\Delta T) \right]
L_c= L\left[1-\alpha_c(\Delta T) \right]
where \alpha_s \cdots \alpha_c are Linear thermal coefficients of the steel and the copper respectively.
So, we have two equations in variables, \Delta x_s and \Delta x_c
we can solve for one of them now.
\Delta x_c=\left[1+\frac{Y_c}{Y_s}\frac{(1-\alpha_s \Delta T)}{(1-\alpha_c \Delta T)} \right]^{-1} L (\alpha_s +\alpha_c) \Delta T
The values for \alpha_s \cdots \alpha_c are
\alpha_s = 11 \times 10^{-6} (^{\circ}C)^{-1}
\alpha_c=17 \times 10^{-6} (^{\circ}C)^{-1}
solving we get, for various quantities described above,
\Delta x_c= 7.2555 \times 10^{-4} m
\Delta L_c=\alpha_c L(\Delta T) =(17 \times 10^{-6})(2)(20)=6.8 \times 10^{-4} m
\Delta L_s=\alpha_s L(\Delta T) =(11 \times 10^{-6})(2)(20)=4.4 \times 10^{-4} m
\Delta x_s =\Delta L_s + \Delta L_c -\Delta x_c
\Delta x_s = 3.97 \times 10^{-4} m
and the junction shifts by the distance of , say k,
k= \Delta L_s - \Delta x_s=\Delta x_c - \Delta L_c = 0.43 \times 10^{-4} m
Now the tension , F is given by
F= Y_s(A)\frac{\Delta x_s}{L(1-\alpha_s \Delta T)} = 119.093 \, N
I got the value of k right, but the book's answer for F is 125 N. Is there anything wrong with
the calculations ?
I had taken the value of \Delta T = 20^{\circ}\, C, as positive.
I am trying to solve this problem. I will present my solution though I couldn't get the tension right.
At 40 Celsius , there is no tension in the wire. When the wires are cooled , they contract and the tension increases. But the tensile stress, which is force per unit area, will remain the same for both the wires, even if they are different in constitution.
I am neglecting the increase in radius, hence the area of the wires, because its negligible.
When the tensions develop , let the steel wire be stretched by amount \Delta x_s
from its reduced length Ls. Similarly, let copper wire be stretched by amount
\Delta x_c from its reduced length Lc. Let L = 2 m be the original length of each wire. Since
\mbox{Young's Modulus}\, =\frac{\mbox{Tensile Stress}}{\mbox{Tensile Strain}}
we have
\frac{F}{A}= Y_s \,\frac{\Delta x_s}{L_s} = Y_c \,\frac{\Delta x_c}{L_c}
where Ys and Yc are Young's Moduli for the steel and copper.
They are
Y_s=20 \times 10^{10} N/m^2
Y_c= 11 \times 10^{10} N/m^2
so from above equation, we get , rearranging
\Delta x_s = \left(\frac{Y_c}{Y_s}\right)\left(\frac{L_s}{L_c}\right) \Delta x_c
Since amount of stretching must be equal to the thermal contraction, we have
\Delta x_s + \Delta x_c= \Delta L_s + \Delta L_c
where \Delta L_s \cdots \Delta L_c are change in the length of the steel and copper rods due to cooling. Also note that
L_s = L\left[1-\alpha_s (\Delta T) \right]
L_c= L\left[1-\alpha_c(\Delta T) \right]
where \alpha_s \cdots \alpha_c are Linear thermal coefficients of the steel and the copper respectively.
So, we have two equations in variables, \Delta x_s and \Delta x_c
we can solve for one of them now.
\Delta x_c=\left[1+\frac{Y_c}{Y_s}\frac{(1-\alpha_s \Delta T)}{(1-\alpha_c \Delta T)} \right]^{-1} L (\alpha_s +\alpha_c) \Delta T
The values for \alpha_s \cdots \alpha_c are
\alpha_s = 11 \times 10^{-6} (^{\circ}C)^{-1}
\alpha_c=17 \times 10^{-6} (^{\circ}C)^{-1}
solving we get, for various quantities described above,
\Delta x_c= 7.2555 \times 10^{-4} m
\Delta L_c=\alpha_c L(\Delta T) =(17 \times 10^{-6})(2)(20)=6.8 \times 10^{-4} m
\Delta L_s=\alpha_s L(\Delta T) =(11 \times 10^{-6})(2)(20)=4.4 \times 10^{-4} m
\Delta x_s =\Delta L_s + \Delta L_c -\Delta x_c
\Delta x_s = 3.97 \times 10^{-4} m
and the junction shifts by the distance of , say k,
k= \Delta L_s - \Delta x_s=\Delta x_c - \Delta L_c = 0.43 \times 10^{-4} m
Now the tension , F is given by
F= Y_s(A)\frac{\Delta x_s}{L(1-\alpha_s \Delta T)} = 119.093 \, N
I got the value of k right, but the book's answer for F is 125 N. Is there anything wrong with
the calculations ?
I had taken the value of \Delta T = 20^{\circ}\, C, as positive.
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