Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 4 - Forces and Newton's Laws of Motion - Problems - Page 115: 52


(a) The lift force $L=5.78\times10^4N$ (b) Air resistance $R=2.08\times10^4N$

Work Step by Step

Let's consider the lift force $\vec{L}$ - Vertical component: $L_y=L\cos21=0.93L$ - Horizontal component: $L_x=L\sin21=0.36L$ (a) To find $L$, we can rely on the fact that we already know $W=mg=53800N$ The vertical forces do not affect the movement of the helicopter here. In other words, $\sum F_y=0$ And since there are only 2 vertical forces $m\vec{g}$ and $\vec{L}_y$, they cancel out each other. $$L_y=0.93L=53800N$$ $$L=57849.5N=5.78\times10^4N$$ (b) As the helicopter moves at a constant velocity, the horizontal forces also balance each other. Here, there are 2 horizontal forces: $\vec{R}$ and $\vec{L}_x$ $$R=L_x=0.36L=20825.8N=2.08\times10^4N$$
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