College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 398: 12


The minimum diameter of the wire is $1.7~mm$

Work Step by Step

We can find the acrobat's weight: $weight = mg = (55~kg)(9.80~m/s^2) = 539~N$ We can find the minimum radius: $\frac{F}{A} = 2.5\times 10^8~N/m^2$ $A = \frac{F}{2.5\times 10^8~N/m^2}$ $\pi~r^2 = \frac{F}{2.5\times 10^8~N/m^2}$ $r = \sqrt{\frac{F}{(\pi)(2.5\times 10^8~N/m^2)}}$ $r = \sqrt{\frac{539~N}{(\pi)(2.5\times 10^8~N/m^2)}}$ $r = 8.28\times 10^{-4}~m$ $r = 0.828~mm$ Since the diameter is twice the radius, the minimum diameter of the wire is $1.7~mm$
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