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 305: 114



Work Step by Step

We know that: $\Sigma F_x=\frac{dm_A}{dt}v_{A_x}+\frac{dm_B}{dt}vB_x$ $\implies Tcos60=\frac{dm_A}{dt}v_{A_x}+\frac{dm_B}{dt}vB_x$ $\implies Tcos60=\rho_w Qvcos30+\rho_wQvcos45$ We plug in the known values to obtain: $Tcos60=(1020)(0.25)(50)cos30+(1020)(0.25)(50)cos 45$ This simplifies to: $T=40.1KN$
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