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 435: 28

Answer

$F_{CD}=288.87KN$

Work Step by Step

We can determine the required force as follows: We know that $s_B+2s_L=L$ Differentiating both sides, we obtain: $v_B+2v_L=0$ Again differentiating the above equation, we obtain: $a_B+2a_L=0$ $\implies a_B=-2a_L$ $\implies 2=-2a_L$ $\implies a_L=-1m/s^2$ The negative sign shows that its direction is upward. According to the equation of motion in y-direction $\Sigma F_y=ma_y$ $\implies T+T-W=ma_L$ We plug in the known values to obtain: $2T-8000(9.81)=8000\times 1$ This simplifies to: $T=43240N$ The sum of of moments about is given as $\Sigma M_A=0$ $\implies F_{CD}(4cos60)-W_B(6cos60)-2T(12cos60)=0$ $\implies F_{CD}(2)-(2000\times 9.81)(3)-2(43240)(6)=0$ This simplifies to: $F_{CD}=288870N=288.87KN$
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