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: 54

Answer

$A_A=0.16\times10^{-3}m^2$ $A_B=9.14\times10^{-3}m^2$

Work Step by Step

$F_A+F_B=2900g$ $-F_A(20.0m)-2900g(5.0m)=0$ $F_A=-7105N$ $F_B=35525N$ tensile strength $A_A=\frac{F_A}{TS}=\frac{9\times7105N}{40\times10^6\frac{N}{m^2}}=0.16\times10^{-3}m^2$ compressive strength $A_B=\frac{F_B}{CS}=\frac{9\times35525N}{35\times10^6\frac{N}{m^2}}=9.14\times10^{-3}m^2$
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