Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 19, Problem 72CP

(a)

To determine

The tension in the wire.

(a)

Expert Solution
Check Mark

Answer to Problem 72CP

The tension in the wire is 125N_.

Explanation of Solution

Write the equation length under tension for steel.

  Ls=Ls[1+FYsAs]                                                                                             (I)

Here, Ls is the new length under tension for steel, Ls is the unstretched length at 20°C, F is the tension force, Ys is the Young modulus for steel, and As is the cross-sectional area.

Write the equation length under tension for copper.

  Lc=Lc[1+FYcAc]                                                                                              (II)

Here, Lc is the new length under tension for copper, Lc is the unstretched length at 20°C, Yc is the Young modulus for copper, and Ac is the cross-sectional area.

Write the equation length at 20°C for steel.

  Ls=L0s[1+αs(T2T1)]                                                                                (III)

Here, L0s is the length at 40°C, αs is the expansion coefficient for steel, T1 is the initial temperature and T2 is the final temperature.

Write the equation length at 20°C for copper.

  Lc=L0c[1+αc(T2T1)]                                                                                (IV)

Here, L0c is the length at 40°C and αc is the expansion coefficient of copper.

Write the equation cross-sectional area for steel.

  As=[(π)ds2]2[1+αs(T2T1)]2                                                                (V)

Here, d is the diameter of steel wire.

Write the equation cross-sectional area for copper.

  Ac=[(π)dc2]2[1+αc(T2T1)]2                                      (VI)

Here, d is the diameter of steel wire.

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

  F=[Ls+Lc][Ls+Lc](LsYsAs)+(LcYcAc)                                                                              (VII)

Conclusion:

Substitute 2.000m for L0s, 11.0×106/°C for αs, 20°C for T2 and 40°C for T1 in Equation (III) to find Ls.

  Ls=(2.000m)[1+(11.0×106/°C)(20°C40°C)]=1.99956m

Substitute 2.000m for L0c, 17.0×106/°C for αc, 20°C for T2 and 40°C for T1 in Equation (IV) to find Lc.

  Lc=(2.000m)[1+(17.0×106/°C)(20°C40°C)]=1.99932m

Substitute 2.000mm for ds, 11.0×106/°C for αs, 20°C for T2 and 40°C for T1 in Equation (V) to find As.

  As=[(π)[(2.000mm)(1×103m1mm)]2]2[1+(11.0×106/°C)(20°C40°C)]2=3.140×106m2

Substitute 2.000mm for dc, 17.0×106/°C for αc, 20°C for T2 and 40°C for T1 in Equation (VI) to find Ac.

  Ac=[(π)[(2.000mm)(1×103m1mm)]2]2[1+(17.0×106/°C)(20°C40°C)]2=3.139×106m2

Substitute 2.000m for Ls, 2.000m for Lc, 1.99956m for Ls, 1.99932m for Lc, 20.0×1010Pa for Ys, 11.0×1010Pa for Yc, 3.140×106m2 for As and 3.139×106m2 for Ac in Equation (VII) to find F.

  F=[2.000m+2.000m][1.99956m+1.99932m](1.99956m(20.0×1010Pa)(3.140×106m2))+(1.99932m(11.0×1010Pa)(3.139×106m2))=125N

Thus, the tension in the wire is 125N_.

(b)

To determine

The x-coordinate of the junction between the wires.

(b)

Expert Solution
Check Mark

Answer to Problem 72CP

The x-coordinate of the junction between the wires is 4.20×105m_.

Explanation of Solution

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

  x=x+Ls[1+FYsAs]                                                                                   (VIII)

Here, x is the x-coordinate and x is the extension of steel wire.

Conclusion:

Substitute 2.000m for x, 1.99956m for Ls, 20.0×1010Pa for Ys, 3.140×106m2 for As and 125N for F in Equation (VIII) to find x.

  x=(2.000m)+{(1.99956m)[1+(125N)(20.0×1010Pa)(3.140×106m2)]}=(2.000m)+(1.999958)=4.20×105m

Thus, the x-coordinate of the junction between the wires is 4.20×105m_.

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Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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