Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 17 - Planar Kinetics of a Rigid Body: Force and Acceleration - Section 17.3 - Equations of Motion: Translation - Problems - Page 436: 32

Answer

$N_A=113.42N$ $N_B=325.18N$

Work Step by Step

We can determine the required normal reactions as follows: According to the equation of motion in x-direction $\Sigma F_x=ma_x$ $\implies -300 cos 30=60a$ This can be rearranged as: $a=\frac{-259.807}{60}=-4.33m/s^2$ The negative sign shows that it is directed leftward. Now the sum of moment about $A$ is given as $\Sigma M_A=ma(0.3)$ $\implies -W(0.3)+N_B(0.5)+Pcos 30(0.4)-Psin 30(0.08)=60\times 4.33(0.3)$ $\implies -60(9.81)(0.3)+0.5N_B+300cos 30(0.4)-300sin 30(0.08)=77.94$ This simplifies to: $N_B=325.18N$ We apply the equation of motion in y-direction $\Sigma F_y=ma_y$ $\implies N_A+N_B-W+Psin30=ma_y$ We plug in the known values to obtain: $N_A+325.18-(60\times 9.81)+300sin30=600(0)$ This simplifies to: $N_A=113.42N$
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