Chapter 19, Problem 19.72CP

Review. A steel wire and a copper wire, each of diameter 2.000 mm, are joined end to end. At 40.0°C, each has an unstretched length of 2.000 m. The wires are connected between two fixed supports 4.000 m apart on a tabletop. The steel wire extends from x = –2.000 m to x = 0, the copper wire extends from x = 0 to x = 2.000 m, and the tension is negligible. The temperature is then lowered to 20.0°C. Assume the average coefficient of linear expansion of steel is 11.0 × 10^–6 (°C)^–1 and that of copper is 17.0 × 10^–6 (°C)^–1. Take Youngs modulus for steel to be 20.0 × 10¹⁰ N/m² and that for copper to be 11.0 × 10¹⁰ N/m². At this lower temperature, find (a) the tension in the wire and (b) the x coordinate of the junction between the wires.

To determine

The tension in the wire.

Answer to Problem 19.72CP

The tension in the wire is $125\text{N}$.

Explanation of Solution

Given Info: The diameter of both the wires is $2.000\text{mm}$, the unstretched length of each wire is $2.000\text{m}$, the initial temperature is $40.0°\text{C}$, the wires are connected between two fixed supports $4.000\text{m}$ apart on a tabletop, the steel wire extended from $x=-2.000\text{m}$ to $x=0$, the copper wire extended from $x=0$ to $x=2.000\text{m}$, the final temperature of the system is $20.0°\text{C}$, the average coefficient of linear expansion of steel is $11.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$, the average coefficient of linear expansion of copper is $17.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$, the young modulus for steel is $20.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the young modulus for copper is $11.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$.

Formula to calculate the radius of the wire is,

$r=\frac{d}{2}$

Here,

$d$ is the diameter of the wire.

Substitute $2.000\text{mm}$ for $d$ in the above expression.

$\begin{array}{c}r=\frac{2.000\text{mm}}{2}\left(\frac{{10}^{-3}}{1\text{mm}}\right)\\ =1\times {10}^{-3}\text{m}\end{array}$

Thus, the value of the radius is $1\times {10}^{-3}\text{m}$.

The initial area of cross section of the steel wire is,

$A_{\text{s1}}=\pi r^2$ (1)

Substitute $1\times {10}^{-3}\text{m}$ fort $r$ in the above expression.

$\begin{array}{c}{A}_{\text{s1}}=\pi {\left(1\times {10}^{-3}\text{m}\right)}^2\\ =1\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Thus, the value of the initial area of cross section of the steel wire is $1\times {10}^{-6}{\text{m}}^{\text{2}}$.

Substitute $1\times {10}^{-3}\text{m}$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$\begin{array}{c}{A}_{\text{c1}}=\pi {\left(1\times {10}^{-3}\text{m}\right)}^2\\ =1\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Thus, the value of the initial area of cross section of the copper wire is $1\times {10}^{-6}{\text{m}}^{\text{2}}$.

When the wire is stretched its length and its area of cross section both have changed.

Formula to calculate the new area of cross section of the steel wire is,

$A_{\text{s}}=A_{\text{s1}}\left(1+\alpha \left(T_2-T_1\right)\right)$

Substitute $1\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{s1}}$, $11.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for $\alpha$, $20°\text{C}$ for $T_2$ and $40°\text{C}$ in the above expression.

$\begin{array}{c}{A}_{\text{s}}=\left(1\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(1+\left(11.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\right)\left(20°\text{C}-40°\text{C}\right)\right)\\ =3.14\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Thus, the value of the final area of cross section of the steel wire is $3.14\times {10}^{-6}{\text{m}}^{\text{2}}$.

Formula to calculate the new area of cross section of the copper wire is,

$A_{\text{c}}=A_{\text{c1}}\left(1+\alpha \left(T_2-T_1\right)\right)$

Substitute $1\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{s1}}$, $17.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$ for $\alpha$, $20°\text{C}$ for $T_2$ and $40°\text{C}$ in the above expression.

$\begin{array}{c}{A}_{\text{c}}=\left(1\times {10}^{-6}{\text{m}}^{\text{2}}\right)\left(1+\left(17.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}\right)\left(20°\text{C}-40°\text{C}\right)\right)\\ =3.139\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Thus, the value of the final area of cross section of the copper wire is $3.139\times {10}^{-6}{\text{m}}^{\text{2}}$.

Formula to calculate the final length of the steel wire under a tension $T$ is,

$L_{\text{s}}^{\text{'}}=L_{\text{s}}\left[1+\frac{T}{Y_{\text{s}}{A}_{\text{s}}}\right]$

Here,

$T$ is the tension in the wire.

$Y_{\text{s}}$ is the Young’s modulus of the steel wire.

Formula to calculate the final length of the copper wire under a tension $T$ is,

$L_{\text{c}}^{\text{'}}=L_{\text{c}}\left[1+\frac{T}{Y_{\text{c}}{A}_{\text{c}}}\right]$

Here,

$T$ is the tension in the wire.

$Y_{\text{s}}$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=\frac{\left(L_{\text{s}}^{\text{'}}+L_{\text{c}}^{\text{'}}\right)-\left(L_{\text{s}}+L_{\text{c}}\right)}{\frac{L_{\text{s}}}{Y_{\text{s}}{A}_{\text{s}}}+\frac{L_{\text{c}}}{Y_{\text{c}}{A}_{\text{c}}}}$

Substitute $1.99956\text{m}$ for $L_{\text{s}}$, $1.99932\text{m}$ for $L_{\text{c}}$, $4\text{m}$ for $L_{\text{s}}^{\text{'}}+L_{\text{c}}^{\text{'}}$, $20.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$ for $Y_{\text{s}}$, $11.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$ for $Y_{\text{c}}$, $3.14\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{s}}$, $3.139\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{c}}$ in the above expression.

$\begin{array}{c}T=\frac{\left(4\text{m}\right)-\left(1.99956\text{m}+1.99932\text{m}\right)}{\left[\begin{array}{l}\frac{1.99956\text{m}}{\left(20.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}\right)\left(3.14\times {10}^{-6}{\text{m}}^{\text{2}}\right)}\\ +\frac{1.99932\text{m}}{\left(20.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}\right)\left(3.139\times {10}^{-6}{\text{m}}^{\text{2}}\right)}\end{array}\right]}\\ =125\text{N}\end{array}$

Conclusion:

Thus, the tension in the wire is $125\text{N}$.

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 19.72CP

The final x-coordinate is $-4.2\times {10}^{-5}\text{m}$.

Explanation of Solution

Given Info: The diameter of both the wires is $2.000\text{mm}$, the Unstretched length of each wire is $2.000\text{m}$, the initial temperature is $40.0°\text{C}$, the wires are connected between two fixed supports $4.000\text{m}$ apart on a tabletop, the steel wire extended from $x=-2.000\text{m}$ to $x=0$, the copper wire extended from $x=0$ to $x=2.000\text{m}$, the final temperature of the system is $20.0°\text{C}$, the average coefficient of linear expansion of steel is $11.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$, the average coefficient of linear expansion of copper is $17.0\times {10}^{-6}{\left(°\text{C}\right)}^{-1}$, the young modulus for steel is $20.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$, the young modulus for copper is $11.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$.

Formula to calculate the final length of the steel wire under a tension $T$ is,

$L_{\text{s}}^{\text{'}}=L_{\text{s}}\left[1+\frac{T}{Y_{\text{s}}{A}_{\text{s}}}\right]$

Here,

$T$ is the tension in the wire.

$Y_{\text{s}}$ is the Young’s modulus of the steel wire.

Substitute $1.99956\text{m}$ for $L_{\text{s}}$, $125\text{N}$ for $T$, $20.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}$ for $Y_{\text{s}}$, , $3.14\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_{\text{s}}$ in the above expression.

$\begin{array}{c}{L}_{\text{s}}^{\text{'}}=\left(1.99956\text{m}\right)\left[1+\frac{125\text{N}}{\left(20.0\times {10}^{10}\text{N}/{\text{m}}^{\text{2}}\right)\left(3.14\times {10}^{-6}{\text{m}}^{\text{2}}\right)}\right]\\ =1.999958\text{m}\end{array}$

Thus, the final length of the steel wire under a tension $T$ is $1.999958\text{m}$.

Formula to find final x coordinate is,

$x_{\text{f}}=x+L_{\text{s}}^{\text{'}}$

Here,

$x$ is the initial x-coordinate.

Substitute $-2$ for $x$ and $1.999958\text{m}$ for $L_{\text{s}}^{\text{'}}$ in the above expression.

$\begin{array}{c}{x}_{\text{f}}=-2+1.999958\\ =-4.2\times {10}^{-5}\text{m}\end{array}$

Conclusion:

Therefore, the final x-coordinate is $-4.2\times {10}^{-5}\text{m}$.