P In constructing a large mobile, an artist hangs an

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

Problem 29E Chapter 11

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 29E

P In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 X 10-3 cm2 in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. For each wire, compute (a) the tensile strain and (b) the elongation.

Step-by-Step Solution:

Solution 29E Step 1: Introduction : In this question, we need to find the tensile strain and the elongation Data given Mass of sphere m = 6.0 kg S Length of the wire l = 0.50 m 3 2 5 2 Area of cross section of the wire A = 2.5 × 10 cm = 2.5 × 10 m Mass of brass cube m =c10.0 kg Step 2: We shall find on the top of the wire It is given by T = (m ×sg) + (m × c) 2 2 T = (6.0 kg × 9.8 m/s ) + (10.0 kg × 9.8 m/s ) T = 156.8 N We can find the tensile strain using TS = T/(A × E) top 9 Young’s modulus steel is 2.0 × 10 Substituting values we get TS top = 156.8 N/(2.5 × 10 5m × 2.0 × 10 )9 TS = 3.1 × 103 top 3 Hence we have tensile strain top of the wire as 3.1 × 10 Step 3 : Now to find the tensile strain at the bottom of the wire We have tension at the bottom of the wire T = m × g c 2 T = 10.0 kg × 9.8 m/s T = 98 N We can find the tensile strain using TS = T/(A × E) B 9 Young’s modulus steel is 2.0 × 10 Substituting values we get TS B 9.8 N/(2.5 × 10 5m × 2.0 × 10 )9 TS = 1.96 × 10 3 B 3 This can be approximated to 2.0 × 10 3 Hence we have tensile strain at the bottom of the wire as 2.0 × 10

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Chapter 11, Problem 29E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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