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 255: 36

Answer

The force on the leg bone is about 2100N and the force on the Achilles tendon is about 1400N

Work Step by Step

We sum the torques about the lower leg bone to solve for the force in the Achilles tendon, given that $D=2d$: $(+ \circlearrowleft) \sum \tau=0$ $-F_T*d+D*mg=0$ $-F_T*d+2d*mg=0$ $F_T=(2d*mg)/d$ $F_T=2mg=2*72kg*9.8m/s^2$ $F_T\approx1400N$ We then sum the torques about the Achilles tendon to solve for the force on the leg bone; $(+ \circlearrowleft) \sum \tau=0$ $-F*d+mg(D+d)=0$ $-F*d+mg(2d+d)=0$ $-F*d+mg(3d)=0$ $F=mg(3d)/d$ $F=mg3=72kg*9.8m/s^2*3$ $F\approx2100N$ The force on the leg bone is about 2100N and the force on the Achilles tendon is about 1400N.
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