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 151: 79


$N_p=2.65KN$, $\rho=68.3m$

Work Step by Step

We apply Newton's second law to determine the unknown values $\Sigma F_b=ma_b=0$ $\implies N_psin\theta=0$ $\implies N_p sin15=70\times 9.81$ This simplifies to: $N_p=2653.2=2.65KN$ and $\Sigma F_n=ma_n$ $\implies N_p cos\theta=m\frac{v^2}{\rho}$ We plug in the known values to obtain: $2653.2cos15=70\times \frac{(50)^2}{\rho}$ This simplifies to: $\rho=68.3m$
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