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 131: 13


$a_A=9.66ft/s^2$, $a_B=15ft/s^2$

Work Step by Step

The acceleration of each block can be determined as follows: We know that $\Sigma F_y=0$ $\implies N-W=0$ $\implies N=W=8lb$ and $\Sigma F_x=ma_x$ $\implies kx-\mu_k N=m_A a_A$ We plug in the known values to obtain: $20\times 0.2-(0.2\times 8)=\frac{8}{32.2}a_A$ This simplifies to: $a_A=9.66ft/s^2$ Now for the second block $\Sigma F_y=0$ $\implies N-W=0$ $\implies N=W=6lb$ and $\Sigma F_x=ma_x$ $\implies kx-\mu_k N=m_B a_B$ We plug in the known values to obtain: $20\times 0.2-0.2\times 6=\frac{6}{32.2}a_B$ This simplifies to: $a_B=15ft/s^2$
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