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 254: 24

Answer

The distance from his feet to his center of gravity is 90.5cm

Work Step by Step

Let x be the distance from his feet to his center of gravity. We are given that the weight of the person is $35.1kg+31.6kg=66.7kg$ Now we sum the torques about the scale on the right to solve for x. $(+ \circlearrowleft) \sum \tau=0$ $-35.1kg*g*1.72m+66.7kg*g*x=0$ $-35.1kg*1.72m+66.7kg*x=0$ $x=\frac{35.1kg*1.72m}{66.7kg}$ $x\approx.905m$ or $90.5cm$ The distance from his feet to his center of gravity is 90.5cm
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