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Physics / Sears and Zemansky's University Physics with Modern Physics 13 / Chapter 11 / Problem 26E

Sears and Zemansky's University Physics with Modern Physics | 13th Edition | ISBN: 9780321696861 | Authors: Hugh D. Young; Roger A. Freedman; A. Lewis Ford

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Answer: Two circular rods, one steel and the other copper,

Chapter 11, Problem 26E

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QUESTION:

Problem 26E

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

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QUESTION:

Problem 26E

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

ANSWER:

Solution 26E

Introduction

We have to calculate the strain and elongation for both steel and copper rod.

Step 1

(a) The diameter of each rod is $d=1.50\ cm=0.0150\ m$

Hence the area of cross section is $A=\pi{}\frac{{d}^{2}}{4}=\pi{}\frac{(0.0150\ m{)}^{2}}{4}=1.77\times{}1{0}^{-4}\ {m}^{2}$

The young modulus of copper wire is

$Y=11\times{}1{0}^{10}\ Pa$

Hence the strain in copper wire is

$\frac{\Delta{}L}{L}=\frac{F/A}{Y}=\frac{F}{YA}$

$=\frac{4000\ N}{(11\times{}1{0}^{10}\ Pa)(1.77\times{}1{0}^{-4}\ {m}^{2})}=2.05\times{}1{0}^{-4}$

Hence strain for copper wire si $2.05\times{}1{0}^{-4}$ .

Now the Young’s modulus for the steel wire is

$Y=20\times{}1{0}^{10}\ Pa$

$\frac{\Delta{}L}{L}=\frac{F}{AY}=\frac{4000\ N}{(20\times{}1{0}^{10}\ Pa)(1.77\times{}1{0}^{-4}\ {m}^{2})}=1.13\times{}1{0}^{-4}$

Hence the strain in the steel wire is $1.13\times{}1{0}^{-4}$ .

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