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 × 10 10 N/m 2 and that for copper to be 11.0 × 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.

Question

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Answer 1

Textbook Question

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.

(a)

Expert Solution

To determine

The tension in the wire.

Answer to Problem 19.72CP

The tension in the wire is $125 N$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

Formula to calculate the radius of the wire is,

$r=d2$

Here,

$d$ is the diameter of the wire.

Substitute $2.000 mm$ for $d$ in the above expression.

$r=2.000 mm2(10−31 mm)=1×10−3 m$

Thus, the value of the radius is $1×10−3 m$ .

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

$As1=πr2$ (1)

Substitute $1×10−3 m$ fort $r$ in the above expression.

$As1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the steel wire is $1×10−6 m2$ .

Substitute $1×10−3 m$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$Ac1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the copper wire is $1×10−6 m2$ .

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,

$As=As1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $11.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$As=(1×10−6 m2)(1+(11.0×10−6(°C)−1)(20°C−40°C))=3.14×10−6 m2$

Thus, the value of the final area of cross section of the steel wire is $3.14×10−6 m2$ .

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

$Ac=Ac1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $17.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$Ac=(1×10−6 m2)(1+(17.0×10−6(°C)−1)(20°C−40°C))=3.139×10−6 m2$

Thus, the value of the final area of cross section of the copper wire is $3.139×10−6 m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

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

$Lc'=Lc[1+TYcAc]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=(Ls'+Lc')−(Ls+Lc)LsYsAs+LcYcAc$

Substitute $1.99956 m$ for $Ls$ , $1.99932 m$ for $Lc$ , $4 m$ for $Ls'+Lc'$ , $20.0×1010 N/m2$ for $Ys$ , $11.0×1010 N/m2$ for $Yc$ , $3.14×10−6 m2$ for $As$ , $3.139×10−6 m2$ for $Ac$ in the above expression.

$T=(4 m)−(1.99956 m+1.99932 m)[1.99956 m(20.0×1010 N/m2)(3.14×10−6 m2)+1.99932 m(20.0×1010 N/m2)(3.139×10−6 m2)]=125 N$

Conclusion:

Thus, the tension in the wire is $125 N$ .

(b)

Expert Solution

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 19.72CP

The final x-coordinate is $−4.2×10−5 m$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the Unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

Substitute $1.99956 m$ for $Ls$ , $125 N$ for $T$ , $20.0×1010 N/m2$ for $Ys$ , , $3.14×10−6 m2$ for $As$ in the above expression.

$Ls'=(1.99956 m)[1+125 N(20.0×1010 N/m2)(3.14×10−6 m2)]=1.999958 m$

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

Formula to find final x coordinate is,

$xf=x+Ls'$

Here,

$x$ is the initial x-coordinate.

Substitute $−2$ for $x$ and $1.999958 m$ for $Ls'$ in the above expression.

$xf=−2+1.999958=−4.2×10−5 m$

Conclusion:

Therefore, the final x-coordinate is $−4.2×10−5 m$ .

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Answer 2

Textbook Question

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.

(a)

Expert Solution

To determine

The tension in the wire.

Answer to Problem 19.72CP

The tension in the wire is $125 N$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

Formula to calculate the radius of the wire is,

$r=d2$

Here,

$d$ is the diameter of the wire.

Substitute $2.000 mm$ for $d$ in the above expression.

$r=2.000 mm2(10−31 mm)=1×10−3 m$

Thus, the value of the radius is $1×10−3 m$ .

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

$As1=πr2$ (1)

Substitute $1×10−3 m$ fort $r$ in the above expression.

$As1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the steel wire is $1×10−6 m2$ .

Substitute $1×10−3 m$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$Ac1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the copper wire is $1×10−6 m2$ .

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,

$As=As1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $11.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$As=(1×10−6 m2)(1+(11.0×10−6(°C)−1)(20°C−40°C))=3.14×10−6 m2$

Thus, the value of the final area of cross section of the steel wire is $3.14×10−6 m2$ .

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

$Ac=Ac1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $17.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$Ac=(1×10−6 m2)(1+(17.0×10−6(°C)−1)(20°C−40°C))=3.139×10−6 m2$

Thus, the value of the final area of cross section of the copper wire is $3.139×10−6 m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

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

$Lc'=Lc[1+TYcAc]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=(Ls'+Lc')−(Ls+Lc)LsYsAs+LcYcAc$

Substitute $1.99956 m$ for $Ls$ , $1.99932 m$ for $Lc$ , $4 m$ for $Ls'+Lc'$ , $20.0×1010 N/m2$ for $Ys$ , $11.0×1010 N/m2$ for $Yc$ , $3.14×10−6 m2$ for $As$ , $3.139×10−6 m2$ for $Ac$ in the above expression.

$T=(4 m)−(1.99956 m+1.99932 m)[1.99956 m(20.0×1010 N/m2)(3.14×10−6 m2)+1.99932 m(20.0×1010 N/m2)(3.139×10−6 m2)]=125 N$

Conclusion:

Thus, the tension in the wire is $125 N$ .

(b)

Expert Solution

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 19.72CP

The final x-coordinate is $−4.2×10−5 m$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the Unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

Substitute $1.99956 m$ for $Ls$ , $125 N$ for $T$ , $20.0×1010 N/m2$ for $Ys$ , , $3.14×10−6 m2$ for $As$ in the above expression.

$Ls'=(1.99956 m)[1+125 N(20.0×1010 N/m2)(3.14×10−6 m2)]=1.999958 m$

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

Formula to find final x coordinate is,

$xf=x+Ls'$

Here,

$x$ is the initial x-coordinate.

Substitute $−2$ for $x$ and $1.999958 m$ for $Ls'$ in the above expression.

$xf=−2+1.999958=−4.2×10−5 m$

Conclusion:

Therefore, the final x-coordinate is $−4.2×10−5 m$ .

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Answer 3

Textbook Question

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.

(a)

Expert Solution

To determine

The tension in the wire.

Answer to Problem 19.72CP

The tension in the wire is $125 N$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

Formula to calculate the radius of the wire is,

$r=d2$

Here,

$d$ is the diameter of the wire.

Substitute $2.000 mm$ for $d$ in the above expression.

$r=2.000 mm2(10−31 mm)=1×10−3 m$

Thus, the value of the radius is $1×10−3 m$ .

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

$As1=πr2$ (1)

Substitute $1×10−3 m$ fort $r$ in the above expression.

$As1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the steel wire is $1×10−6 m2$ .

Substitute $1×10−3 m$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$Ac1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the copper wire is $1×10−6 m2$ .

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,

$As=As1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $11.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$As=(1×10−6 m2)(1+(11.0×10−6(°C)−1)(20°C−40°C))=3.14×10−6 m2$

Thus, the value of the final area of cross section of the steel wire is $3.14×10−6 m2$ .

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

$Ac=Ac1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $17.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$Ac=(1×10−6 m2)(1+(17.0×10−6(°C)−1)(20°C−40°C))=3.139×10−6 m2$

Thus, the value of the final area of cross section of the copper wire is $3.139×10−6 m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

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

$Lc'=Lc[1+TYcAc]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=(Ls'+Lc')−(Ls+Lc)LsYsAs+LcYcAc$

Substitute $1.99956 m$ for $Ls$ , $1.99932 m$ for $Lc$ , $4 m$ for $Ls'+Lc'$ , $20.0×1010 N/m2$ for $Ys$ , $11.0×1010 N/m2$ for $Yc$ , $3.14×10−6 m2$ for $As$ , $3.139×10−6 m2$ for $Ac$ in the above expression.

$T=(4 m)−(1.99956 m+1.99932 m)[1.99956 m(20.0×1010 N/m2)(3.14×10−6 m2)+1.99932 m(20.0×1010 N/m2)(3.139×10−6 m2)]=125 N$

Conclusion:

Thus, the tension in the wire is $125 N$ .

(b)

Expert Solution

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 19.72CP

The final x-coordinate is $−4.2×10−5 m$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the Unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

Substitute $1.99956 m$ for $Ls$ , $125 N$ for $T$ , $20.0×1010 N/m2$ for $Ys$ , , $3.14×10−6 m2$ for $As$ in the above expression.

$Ls'=(1.99956 m)[1+125 N(20.0×1010 N/m2)(3.14×10−6 m2)]=1.999958 m$

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

Formula to find final x coordinate is,

$xf=x+Ls'$

Here,

$x$ is the initial x-coordinate.

Substitute $−2$ for $x$ and $1.999958 m$ for $Ls'$ in the above expression.

$xf=−2+1.999958=−4.2×10−5 m$

Conclusion:

Therefore, the final x-coordinate is $−4.2×10−5 m$ .

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Answer 4

Textbook Question

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.

(a)

Expert Solution

To determine

The tension in the wire.

Answer to Problem 19.72CP

The tension in the wire is $125 N$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

Formula to calculate the radius of the wire is,

$r=d2$

Here,

$d$ is the diameter of the wire.

Substitute $2.000 mm$ for $d$ in the above expression.

$r=2.000 mm2(10−31 mm)=1×10−3 m$

Thus, the value of the radius is $1×10−3 m$ .

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

$As1=πr2$ (1)

Substitute $1×10−3 m$ fort $r$ in the above expression.

$As1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the steel wire is $1×10−6 m2$ .

Substitute $1×10−3 m$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$Ac1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the copper wire is $1×10−6 m2$ .

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,

$As=As1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $11.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$As=(1×10−6 m2)(1+(11.0×10−6(°C)−1)(20°C−40°C))=3.14×10−6 m2$

Thus, the value of the final area of cross section of the steel wire is $3.14×10−6 m2$ .

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

$Ac=Ac1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $17.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$Ac=(1×10−6 m2)(1+(17.0×10−6(°C)−1)(20°C−40°C))=3.139×10−6 m2$

Thus, the value of the final area of cross section of the copper wire is $3.139×10−6 m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

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

$Lc'=Lc[1+TYcAc]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=(Ls'+Lc')−(Ls+Lc)LsYsAs+LcYcAc$

Substitute $1.99956 m$ for $Ls$ , $1.99932 m$ for $Lc$ , $4 m$ for $Ls'+Lc'$ , $20.0×1010 N/m2$ for $Ys$ , $11.0×1010 N/m2$ for $Yc$ , $3.14×10−6 m2$ for $As$ , $3.139×10−6 m2$ for $Ac$ in the above expression.

$T=(4 m)−(1.99956 m+1.99932 m)[1.99956 m(20.0×1010 N/m2)(3.14×10−6 m2)+1.99932 m(20.0×1010 N/m2)(3.139×10−6 m2)]=125 N$

Conclusion:

Thus, the tension in the wire is $125 N$ .

(b)

Expert Solution

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 19.72CP

The final x-coordinate is $−4.2×10−5 m$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the Unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

Substitute $1.99956 m$ for $Ls$ , $125 N$ for $T$ , $20.0×1010 N/m2$ for $Ys$ , , $3.14×10−6 m2$ for $As$ in the above expression.

$Ls'=(1.99956 m)[1+125 N(20.0×1010 N/m2)(3.14×10−6 m2)]=1.999958 m$

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

Formula to find final x coordinate is,

$xf=x+Ls'$

Here,

$x$ is the initial x-coordinate.

Substitute $−2$ for $x$ and $1.999958 m$ for $Ls'$ in the above expression.

$xf=−2+1.999958=−4.2×10−5 m$

Conclusion:

Therefore, the final x-coordinate is $−4.2×10−5 m$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The tension in the wire.

Answer to Problem 19.72CP

The tension in the wire is $125 N$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

Formula to calculate the radius of the wire is,

$r=d2$

Here,

$d$ is the diameter of the wire.

Substitute $2.000 mm$ for $d$ in the above expression.

$r=2.000 mm2(10−31 mm)=1×10−3 m$

Thus, the value of the radius is $1×10−3 m$ .

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

$As1=πr2$ (1)

Substitute $1×10−3 m$ fort $r$ in the above expression.

$As1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the steel wire is $1×10−6 m2$ .

Substitute $1×10−3 m$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$Ac1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the copper wire is $1×10−6 m2$ .

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,

$As=As1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $11.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$As=(1×10−6 m2)(1+(11.0×10−6(°C)−1)(20°C−40°C))=3.14×10−6 m2$

Thus, the value of the final area of cross section of the steel wire is $3.14×10−6 m2$ .

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

$Ac=Ac1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $17.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$Ac=(1×10−6 m2)(1+(17.0×10−6(°C)−1)(20°C−40°C))=3.139×10−6 m2$

Thus, the value of the final area of cross section of the copper wire is $3.139×10−6 m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

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

$Lc'=Lc[1+TYcAc]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=(Ls'+Lc')−(Ls+Lc)LsYsAs+LcYcAc$

Substitute $1.99956 m$ for $Ls$ , $1.99932 m$ for $Lc$ , $4 m$ for $Ls'+Lc'$ , $20.0×1010 N/m2$ for $Ys$ , $11.0×1010 N/m2$ for $Yc$ , $3.14×10−6 m2$ for $As$ , $3.139×10−6 m2$ for $Ac$ in the above expression.

$T=(4 m)−(1.99956 m+1.99932 m)[1.99956 m(20.0×1010 N/m2)(3.14×10−6 m2)+1.99932 m(20.0×1010 N/m2)(3.139×10−6 m2)]=125 N$

Conclusion:

Thus, the tension in the wire is $125 N$ .

(b)

Expert Solution

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 19.72CP

The final x-coordinate is $−4.2×10−5 m$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the Unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

Substitute $1.99956 m$ for $Ls$ , $125 N$ for $T$ , $20.0×1010 N/m2$ for $Ys$ , , $3.14×10−6 m2$ for $As$ in the above expression.

$Ls'=(1.99956 m)[1+125 N(20.0×1010 N/m2)(3.14×10−6 m2)]=1.999958 m$

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

Formula to find final x coordinate is,

$xf=x+Ls'$

Here,

$x$ is the initial x-coordinate.

Substitute $−2$ for $x$ and $1.999958 m$ for $Ls'$ in the above expression.

$xf=−2+1.999958=−4.2×10−5 m$

Conclusion:

Therefore, the final x-coordinate is $−4.2×10−5 m$ .

Answer 8

(a)

Expert Solution

To determine

The tension in the wire.

Answer to Problem 19.72CP

The tension in the wire is $125 N$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

Formula to calculate the radius of the wire is,

$r=d2$

Here,

$d$ is the diameter of the wire.

Substitute $2.000 mm$ for $d$ in the above expression.

$r=2.000 mm2(10−31 mm)=1×10−3 m$

Thus, the value of the radius is $1×10−3 m$ .

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

$As1=πr2$ (1)

Substitute $1×10−3 m$ fort $r$ in the above expression.

$As1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the steel wire is $1×10−6 m2$ .

Substitute $1×10−3 m$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$Ac1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the copper wire is $1×10−6 m2$ .

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,

$As=As1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $11.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$As=(1×10−6 m2)(1+(11.0×10−6(°C)−1)(20°C−40°C))=3.14×10−6 m2$

Thus, the value of the final area of cross section of the steel wire is $3.14×10−6 m2$ .

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

$Ac=Ac1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $17.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$Ac=(1×10−6 m2)(1+(17.0×10−6(°C)−1)(20°C−40°C))=3.139×10−6 m2$

Thus, the value of the final area of cross section of the copper wire is $3.139×10−6 m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

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

$Lc'=Lc[1+TYcAc]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=(Ls'+Lc')−(Ls+Lc)LsYsAs+LcYcAc$

Substitute $1.99956 m$ for $Ls$ , $1.99932 m$ for $Lc$ , $4 m$ for $Ls'+Lc'$ , $20.0×1010 N/m2$ for $Ys$ , $11.0×1010 N/m2$ for $Yc$ , $3.14×10−6 m2$ for $As$ , $3.139×10−6 m2$ for $Ac$ in the above expression.

$T=(4 m)−(1.99956 m+1.99932 m)[1.99956 m(20.0×1010 N/m2)(3.14×10−6 m2)+1.99932 m(20.0×1010 N/m2)(3.139×10−6 m2)]=125 N$

Conclusion:

Thus, the tension in the wire is $125 N$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The tension in the wire.

Answer 13

Answer to Problem 19.72CP

The tension in the wire is $125 N$ .

Answer 14

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

Formula to calculate the radius of the wire is,

$r=d2$

Here,

$d$ is the diameter of the wire.

Substitute $2.000 mm$ for $d$ in the above expression.

$r=2.000 mm2(10−31 mm)=1×10−3 m$

Thus, the value of the radius is $1×10−3 m$ .

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

$As1=πr2$ (1)

Substitute $1×10−3 m$ fort $r$ in the above expression.

$As1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the steel wire is $1×10−6 m2$ .

Substitute $1×10−3 m$ fort $r$ in the equation (1) to calculate the initial area of cross section of the steel wire.

$Ac1=π(1×10−3 m)2=1×10−6 m2$

Thus, the value of the initial area of cross section of the copper wire is $1×10−6 m2$ .

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,

$As=As1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $11.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$As=(1×10−6 m2)(1+(11.0×10−6(°C)−1)(20°C−40°C))=3.14×10−6 m2$

Thus, the value of the final area of cross section of the steel wire is $3.14×10−6 m2$ .

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

$Ac=Ac1(1+α(T2−T1))$

Substitute $1×10−6 m2$ for $As1$ , $17.0×10−6(°C)−1$ for $α$ , $20°C$ for $T2$ and $40°C$ in the above expression.

$Ac=(1×10−6 m2)(1+(17.0×10−6(°C)−1)(20°C−40°C))=3.139×10−6 m2$

Thus, the value of the final area of cross section of the copper wire is $3.139×10−6 m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

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

$Lc'=Lc[1+TYcAc]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the copper wire.

Formula to calculate the tension in the composite wire is,

$T=(Ls'+Lc')−(Ls+Lc)LsYsAs+LcYcAc$

Substitute $1.99956 m$ for $Ls$ , $1.99932 m$ for $Lc$ , $4 m$ for $Ls'+Lc'$ , $20.0×1010 N/m2$ for $Ys$ , $11.0×1010 N/m2$ for $Yc$ , $3.14×10−6 m2$ for $As$ , $3.139×10−6 m2$ for $Ac$ in the above expression.

$T=(4 m)−(1.99956 m+1.99932 m)[1.99956 m(20.0×1010 N/m2)(3.14×10−6 m2)+1.99932 m(20.0×1010 N/m2)(3.139×10−6 m2)]=125 N$

Conclusion:

Thus, the tension in the wire is $125 N$ .

Answer 15

(b)

Expert Solution

To determine

The x-coordinate of the junction between the wires.

Answer to Problem 19.72CP

The final x-coordinate is $−4.2×10−5 m$ .

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the Unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .

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

$Ls'=Ls[1+TYsAs]$

Here,

$T$ is the tension in the wire.

$Ys$ is the Young’s modulus of the steel wire.

Substitute $1.99956 m$ for $Ls$ , $125 N$ for $T$ , $20.0×1010 N/m2$ for $Ys$ , , $3.14×10−6 m2$ for $As$ in the above expression.

$Ls'=(1.99956 m)[1+125 N(20.0×1010 N/m2)(3.14×10−6 m2)]=1.999958 m$

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

Formula to find final x coordinate is,

$xf=x+Ls'$

Here,

$x$ is the initial x-coordinate.

Substitute $−2$ for $x$ and $1.999958 m$ for $Ls'$ in the above expression.

$xf=−2+1.999958=−4.2×10−5 m$

Conclusion:

Therefore, the final x-coordinate is $−4.2×10−5 m$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The x-coordinate of the junction between the wires.

Answer 20

Answer to Problem 19.72CP

The final x-coordinate is $−4.2×10−5 m$ .

Answer 21

Explanation of Solution

Given Info: The diameter of both the wires is $2.000 mm$ , the Unstretched length of each wire is $2.000 m$ , the initial temperature is $40.0°C$ , the wires are connected between two fixed supports $4.000 m$ apart on a tabletop, the steel wire extended from $x=−2.000 m$ to $x=0$ , the copper wire extended from $x=0$ to $x=2.000 m$ , the final temperature of the system is $20.0°C$ , the average coefficient of linear expansion of steel is $11.0×10−6(°C)−1$ , the average coefficient of linear expansion of copper is $17.0×10−6(°C)−1$ , the young modulus for steel is $20.0×1010 N/m2$ , the young modulus for copper is $11.0×1010 N/m2$ .