Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.3 - Principle of Work and Energy for a System of Particles - Problems - Page 202: 35


$v_B=42.2ft/s$, $N=50.6lb$, $a_t=26.2ft/s^2$

Work Step by Step

According to the principle of work and energy $\frac{1}{2}mv_A^2+\Sigma U_{A\rightarrow B}=\frac{1}{2}mv_B^2$ $\implies \frac{1}{2}mv_A^2+W(y_A-y_B)=\frac{1}{2}mv_B^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{150}{32.2})(5)^2+150(50-22.7)=\frac{1}{2}(\frac{150}{32.2})v_B^2$ This simplifies to: $v_B=42.2ft/s$ We know that $\Sigma F_n=ma_n=m\frac{v^2}{\rho}$ $\implies -N+150cos(54.44^{\circ})=(\frac{150}{32.2})(\frac{(42.2)^2}{227.024})$ This simplifies to: $N=50.6lb$ We can find the acceleration as $\Sigma F_t=ma_t$ $\implies 150sin(54.44)=(\frac{150}{32.2})a_t$ This simplifies to: $a_t=26.2ft/s^2$
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