Question

Review. A steel wire and a copper wire, each of diameter 2.000 mm, are joined end to end. At ${40.0}^{0}C$, 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 $x=-2.000m$ to $x=0$, the copper wire extends from $x=0$ to $x=2.000m$, and the tension is negligible. The temperature is then lowered to ${20.0}^{0}C$. Assume the average coefficient of linear expansion of steel is $11.0\times {10}^{-6}{(}^{0}C{)}^{-1}$ and that of copper is $17.0\times {10}^{-6}{(}^{0}C{)}^{-1}$. Take Young’s modulus for steel to be $20.0\times {10}^{10}N/m^2$ and that for copper to be $11.0\times {10}^{10}N/m^2$. At this lower temperature, find (a) the tension in the wire and (b) the x coordinate of the junction between the wires.

Answer 1

(a) At ${20.0}^{0}C$, the unstretched lengths of the steel and copper wires are

$L_s\left({20.0}^{0}C\right)$=$\left(2.000m\right)[1+(11.0\times {10}^{-6}{}^0{C}^{-1}\left)\right(-{20.0}^{0}C\left)\right]$
$=1.99956m$

$L_c\left({20.0}^{0}C\right)=\left(2.000m\right)[1+(17.0\times {10}^{-6}{}^0{C}^{-1}\left)\right(-{20.0}^{0}C\left)\right]$
$=1.99932m$

Under a tension F, the length of the steel and copper wires are
$L_s^{\prime }=L_s[1+\frac{F}{Ya}{]}_s$ and $L_c^{\prime }=L_c[1+\frac{F}{Ya}{]}_c$
where $L_s^{\prime }+L_c^{\prime }:=4.000m$

Since the tension F must be the same in each wire, we sol ve for F:

$F=\frac{(L_s^{\prime }+L_c^{\prime })-(L_s+L_c)}{L_s/Y_s{A}_s+L_c/Y_c{A}_c}$

When the wires are stretched, their areas become

$A_s=\pi (1.000\times {10}^{-3}m{)}^2[1+(11.0\times {10}^{-6}{}^0{C}^{-1})(-20.0){]}^2$

$=3.140\times {10}^{-6}{m}^2$

$A_c=\pi (1.000\times {10}^{-3}m{)}^2[1+(17.0\times {10}^{-6}{}^0{C}^{-1})(-20.0){]}^2$

$=3.139\times {10}^{-6}{m}^2$

Recall $Y_s=20.0\times {10}^{10}Pa$ and $Y_c=11.0\times {10}^{10}Pa$. Substituting into the equation for F, we obtain

$F=[4.000m-(1.99956m+1.99932m\left)\right]\times \frac{1}{\frac{1.99956m}{(20.0\times {10}^{10}Pa)(3.140\times {10}^{-6})m^2}+\frac{1.99932m}{(11.0\times {10}^{10}Pa(3.139\times {10}^{-6}\left)\right)m^2}}$

$F=125N$

(b) To find the x coordinate of the junction,
$L_s^{\prime }=\left(1.99956m\right)\left[a6\frac{125N}{(20.0\times {10}^{10}N/m^2)(3.140\times {10}^{-6}{m}^2)}\right]$

$=1.999958m$

Thus the x coordinate is $-2.000+1.999958=-4.20\times {10}^{-5}m$