Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 9 - Static Equilibrium; Elasticity and Fracture - Problems - Page 256: 47

Answer

The suspended mass is about 25kg

Work Step by Step

We can solve for the mass using the following formula: $\Delta l=\frac{1}{E} \frac{F}{A}l_0$ $F=\Delta l *E*A*\frac{1}{l_0}$ $F=\Delta l *E_{steel}*A*\frac{1}{l_0}$ $F=0.00030m *200*10^9N/m^2*\pi (1/2*0.0023m)^2*\frac{1}{1m}$ $F=249N$ Now we can solve for the mass. Therefore; $m=F/g$ $m=249N/(9.8m/s^2)$ $m\approx25kg$ The suspended mass is about 25kg.
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