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 146: 56

Answer

$\rho=9.32m$

Work Step by Step

We can determine the required radius of curvature as follows: $\Sigma F_n=ma_n$ $\implies 0.7N_c=5(\frac{(8)^2}{\rho})~~~$eq(1) and $\Sigma F_b=0$ $\implies N_C-49.05=0$ $\implies N_C=49.05$ We plug in this values in eq(1) to obtain: $0.7(49.05)=5(\frac{(8)^2}{\rho})$ This simplifies to: $\rho=9.32m$
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