Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - General Problems - Page 107: 89

Answer

The climber must exert a minimum normal force of 490 N.

Work Step by Step

We can assume that the force of static friction pushing up on the climber is equal in magnitude to the climber's weight $mg$: $F_N (0.80) + F_N (0.60) = mg$ $F_N (1.40) = (70.0 ~kg)(9.80 ~m/s^2)$ $F_N = \frac{(70.0 ~kg)(9.80 ~m/s^2)}{1.40} = 490 ~N$ Therefore, the climber must exert a minimum normal force of 490 N.
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