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 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 132: 21



Work Step by Step

We can determine the smallest incline $\theta$ as follows: $v^2=v_{A_{\circ}}^2+2a(s_B-s_A)$ We plug in the known values to obtain: $0=(2.5)^2+2a(3-0)$ $\implies a=1.042m/s^2$ We know that $\Sigma F_y=0$ $\implies N=Wcos\theta=mgcos\theta$ and $\Sigma F_x=ma_x$ $\implies Wsin\theta-\mu_k N=ma$ $\implies ma=mgsin\theta-\mu_k cos\theta$ This can be rearranged as: $\frac{a}{g}=\frac{1.042}{9.81}=sin\theta-0.3cos\theta$ This simplifies to: $\theta=22.6^{\circ}$
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