University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 11 - Equilibrium and Elasticity - Problems - Exercises - Page 357: 11.5

Answer

$$F=6612N$$

Work Step by Step

$w=mg=3400$ $\Sigma\tau=FL-wL=0$ $FrSin\theta - mgL=0$ Find the force needed to lift the ladder: $F=\frac{mgL}{rSin\theta}$ $F=\frac{(3400)(10)}{8Sin(40)}= 6612 N$
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