Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.9 - Propulsion with Variable Mass - Problems - Page 311: 142



Work Step by Step

We know that $F_x=\frac{dm}{dt}(v_{Bx}-v_{Ax})$ $\implies B_x=50(1.5cos30-0)$ $\implies B_x=64.952N$ Now $\Sigma F_s=m\frac{dv}{dt}-\frac{dm}{dt}v_{D/e}+\frac{dm_i}{dt}v_{D/i}$ $\implies-F_D=0-v_{D/e}(\frac{dm}{dt}+q\rho_{\circ})+v_{D/i} (q\rho_{\circ})$ We plug in the known values to obtain: $F_D=450(0.4+1.22(50))-263.9(1.22)(50)$ This simplifies to: $F_D=11.5KN$
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