## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 8 - Problems - Page 314: 44

#### Answer

The tension in the Achilles tendon is $2087~N$ The tibia exerts a force of $2837~N$

#### Work Step by Step

We can consider the torques about a rotation axis at a point on the floor directly below the force from the tibia. The normal force from the floor exerts a counterclockwise torque about this axis. The force from the Achilles tendon exerts a clockwise torque about this axis. Since the system is in equilibrium, the clockwise torque is equal in magnitude to the counterclockwise torque. We can find the force that the Achilles tendon exerts: $\tau_{cw} = \tau_{ccw}$ $(4.60~cm)~(F_A) = (12.8~cm)(750~N)$ $F_A = \frac{(12.8~cm)(750~N)}{4.60~cm}$ $F_A = 2087~N$ The tension in the Achilles tendon is $2087~N$ Since the system is in equilibrium, the downward force is equal in magnitude to the sum of the upward forces. We can find the force that the tibia exerts: $F_{tib} = (2087~N)+(750~N) = 2837~N$ The tibia exerts a force of $2837~N$

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