Fundamentals of Engineering Thermodynamics 8th Edition

Published by Wiley
ISBN 10: 1118412931
ISBN 13: 978-1-11841-293-0

Chapter 1 - Problems: Developing Engineering Skills - Page 30: 1.12

Answer

$m=13.35 ~lb$

Work Step by Step

It is given that the spring deflects 0.14 in. for the applied force of every 1 lbf. So, the stiffness of the spring is $k=F/δ=1/0.14=7.143~ lbf/in.$ This force is equal to the weight of the object (W). $W=k×δ=7.143×1.8=12.857 ~lbf$ The mass of the object is $m=W/g=12.857/31×32.2$ $m=13.35 lb$
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