Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 200: 29


(a) $F = 2.2\times 10^6~N$ (b) $\theta = -27.3^{\circ}$

Work Step by Step

The force on the plane has a vertical component $F_y$ directed toward the ground and a horizontal component $F_x$ directed toward the center of the circle. We can find the magnitude of $F_y$. $F_y = m~a_y$ $F_y = (85000~kg)(12~m/s)$ $F_y = 1.02\times 10^6~N$ We can find the magnitude of $F_x$ which is equal to the centripetal force on the plane. $F_x = \frac{mv^2}{r}$ $F_x = \frac{(85000~kg)(55~m/s)^2}{130~m}$ $F_x = 1.98\times 10^6~N$ (a) We can find the magnitude of the net force. $F = \sqrt{(F_x)^2+(F_y)^2}$ $F = \sqrt{(1.98\times 10^6~N)^2+(1.02\times 10^6~N)^2}$ $F = 2.2\times 10^6~N$ (b) We can find the angle below the horizontal. $tan(\theta) = \frac{F_y}{F_x}$ $\theta = arctan(\frac{1.02\times 10^6~N}{1.98\times 10^6~N})$ $\theta = -27.3^{\circ}$
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