Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 149: 71


$N=277lb$, $F=13.4lb$

Work Step by Step

We know that $\Sigma F_y=ma_n$ $N-Wcos\theta=ma_nsin\theta$ $\implies N=Wcos\theta+m\frac{v^2}{\rho}sin\theta$ We plug in the known values to obtain: $N=150cos60+(\frac{150}{32.2})[\frac{(20)^2}{8}]sin60$ $\implies N=277lb$ We also know that $\Sigma F_x=ma_x$ $\implies Wsin\theta-F=ma_ncos\theta$ $\implies F=Wsin\theta-m\frac{v^2}{\rho}cos\theta$ We plug in the known values to obtain: $F=150sin60-(\frac{150}{32.2})[\frac{(20)^2}{8}]cos60$ $\implies F=13.44lb$
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