Chapter 19, Problem 72CP

(a)

To determine

The tension in the wire.

Answer to Problem 72CP

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

Explanation of Solution

Write the equation length under tension for steel.

  ${L^{\prime }}_s=L_s\left[1+\frac{F}{Y_s{A}_s}\right]$                                                                                             (I)

Here, ${L^{\prime }}_s$ is the new length under tension for steel, $L_s$ is the unstretched length at $20\text{°C}$, $F$ is the tension force, $Y_s$ is the Young modulus for steel, and $A_s$ is the cross-sectional area.

Write the equation length under tension for copper.

  ${L^{\prime }}_c=L_c\left[1+\frac{F}{Y_c{A}_c}\right]$                                                                                              (II)

Here, ${L^{\prime }}_c$ is the new length under tension for copper, $L_c$ is the unstretched length at $20\text{°C}$, $Y_c$ is the Young modulus for copper, and $A_c$ is the cross-sectional area.

Write the equation length at $20\text{°C}$ for steel.

  $L_s=L_{0s}\left[1+{\alpha }_s\left(T_2-T_1\right)\right]$                                                                                (III)

Here, $L_{0s}$ is the length at $40\text{°C}$, ${\alpha }_s$ is the expansion coefficient for steel, $T_1$ is the initial temperature and $T_2$ is the final temperature.

Write the equation length at $20\text{°C}$ for copper.

  $L_c=L_{0c}\left[1+{\alpha }_c\left(T_2-T_1\right)\right]$                                                                                (IV)

Here, $L_{0c}$ is the length at $40\text{°C}$ and ${\alpha }_c$ is the expansion coefficient of copper.

Write the equation cross-sectional area for steel.

  $A_s={\left[\left(\pi \right)\frac{d_s}{2}\right]}^2{\left[1+{\alpha }_s\left(T_2-T_1\right)\right]}^2$                                                                (V)

Here, $d$ is the diameter of steel wire.

Write the equation cross-sectional area for copper.

  $A_c={\left[\left(\pi \right)\frac{d_c}{2}\right]}^2{\left[1+{\alpha }_c\left(T_2-T_1\right)\right]}^2$                                      (VI)

Here, $d$ is the diameter of steel wire.

Write the expression of tension force from equation (I) and (II).

  $F=\frac{\left[{L^{\prime }}_s+{L^{\prime }}_c\right]-\left[L_s+L_c\right]}{\left(\frac{L_s}{Y_s{A}_s}\right)+\left(\frac{L_c}{Y_c{A}_c}\right)}$                                                                              (VII)

Conclusion:

Substitute $2.000\text{m}$ for $L_{0s}$, $11.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_s$, $20\text{°C}$ for $T_2$ and $40\text{°C}$ for $T_1$ in Equation (III) to find $L_s$.

  $\begin{array}{c}{L}_s=\left(2.000\text{m}\right)\left[1+\left(11.0\times {10}^{-6}\text{/°C}\right)\left(20\text{°C}-40\text{°C}\right)\right]\\ =1.99956\text{m}\end{array}$

Substitute $2.000\text{m}$ for $L_{0c}$, $17.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_c$, $20\text{°C}$ for $T_2$ and $40\text{°C}$ for $T_1$ in Equation (IV) to find $L_c$.

  $\begin{array}{c}{L}_c=\left(2.000\text{m}\right)\left[1+\left(17.0\times {10}^{-6}\text{/°C}\right)\left(20\text{°C}-40\text{°C}\right)\right]\\ =1.99932\text{m}\end{array}$

Substitute $2.000\text{mm}$ for $d_s$, $11.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_s$, $20\text{°C}$ for $T_2$ and $40\text{°C}$ for $T_1$ in Equation (V) to find $A_s$.

  $\begin{array}{c}{A}_s={\left[\left(\pi \right)\frac{\left[\left(2.000\text{mm}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right]}{2}\right]}^2{\left[1+\left(11.0\times {10}^{-6}\text{/°C}\right)\left(20\text{°C}-40\text{°C}\right)\right]}^2\\ =3.140\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Substitute $2.000\text{mm}$ for $d_c$, $17.0\times {10}^{-6}\text{/°C}$ for ${\alpha }_c$, $20\text{°C}$ for $T_2$ and $40\text{°C}$ for $T_1$ in Equation (VI) to find $A_c$.

  $\begin{array}{c}{A}_c={\left[\left(\pi \right)\frac{\left[\left(2.000\text{mm}\right)\left(\frac{1\times {10}^{-3}\text{m}}{1\text{mm}}\right)\right]}{2}\right]}^2{\left[1+\left(17.0\times {10}^{-6}\text{/°C}\right)\left(20\text{°C}-40\text{°C}\right)\right]}^2\\ =3.139\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Substitute $2.000\text{m}$ for ${L^{\prime }}_s$, $2.000\text{m}$ for ${L^{\prime }}_c$, $1.99956\text{m}$ for $L_s$, $1.99932\text{m}$ for $L_c$, $20.0\times {10}^{10}\text{Pa}$ for $Y_s$, $11.0\times {10}^{10}\text{Pa}$ for $Y_c$, $3.140\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_s$ and $3.139\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_c$ in Equation (VII) to find $F$.

  $\begin{array}{c}F=\frac{\left[2.000\text{m}+2.000\text{m}\right]-\left[1.99956\text{m}+1.99932\text{m}\right]}{\left(\frac{1.99956\text{m}}{\left(20.0\times {10}^{10}\text{Pa}\right)\left(3.140\times {10}^{-6}{\text{m}}^{\text{2}}\right)}\right)+\left(\frac{1.99932\text{m}}{\left(11.0\times {10}^{10}\text{Pa}\right)\left(3.139\times {10}^{-6}{\text{m}}^{\text{2}}\right)}\right)}\\ =125\text{N}\end{array}$

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

(b)

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 72CP

The x-coordinate of the junction between the wires is $\underset{\_}{-4.20\times {10}^{-5}\text{m}}$.

Explanation of Solution

Write the equation for the x-coordinate by using equation (I).

  $x^{\prime }=x+L_s\left[1+\frac{F}{Y_s{A}_s}\right]$                                                                                   (VIII)

Here, $x^{\prime }$ is the x-coordinate and $x$ is the extension of steel wire.

Conclusion:

Substitute $-2.000\text{m}$ for $x$, $1.99956\text{m}$ for $L_s$, $20.0\times {10}^{10}\text{Pa}$ for $Y_s$, $3.140\times {10}^{-6}{\text{m}}^{\text{2}}$ for $A_s$ and $125\text{N}$ for $F$ in Equation (VIII) to find $x^{\prime }$.

  $\begin{array}{c}{x}^{\prime }=\left(-2.000\text{m}\right)+\left\{\left(1.99956\text{m}\right)\left[1+\frac{\left(125\text{N}\right)}{\left(20.0\times {10}^{10}\text{Pa}\right)\left(3.140\times {10}^{-6}{\text{m}}^{\text{2}}\right)}\right]\right\}\\ =\left(-2.000\text{m}\right)+\left(1.999958\right)\\ =-4.20\times {10}^{-5}\text{m}\end{array}$

Thus, the x-coordinate of the junction between the wires is $\underset{\_}{-4.20\times {10}^{-5}\text{m}}$.