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 148: 67



Work Step by Step

We can determine the required tilt angle as follows: $\Sigma F_b=0$ $\implies Ncos\theta-mg=0$ $\implies N=\frac{mg}{cos\theta}$ We know that $\Sigma F_n=ma_n$ $\implies Nsin\theta=m\frac{v^2}{\rho}$ $\implies \frac{mgsin\theta}{cos\theta}=m\frac{v^2}{\rho}$ We plug in the known values to obtain: $9.81\times tan\theta=\frac{22.22m/s}{100}$ This simplifies to: $\theta=26.7^{\circ}$
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