Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 6 - Force and Motion-II - Problems - Page 141: 14b

Answer

$$ F=3.0 \times 10^{7} \mathrm{N} $$

Work Step by Step

Applying Newton's second law, we have $$ F+m g \sin \theta-f_{s, \max } =m a=0 $$$$ F_{N}-m g \cos \theta=0 $$ Along with Eq. $6-1\left(f_{s, \max }=\mu_{s} F_{N}\right)$ we have enough information to solve for $F .$ With $\theta=24^{\circ}$ and $m=1.8 \times 10^{7} \mathrm{kg}$ , we find $$ F=m g\left(\mu_{s} \cos \theta-\sin \theta\right)=3.0 \times 10^{7} \mathrm{N} $$
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