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 308: 128

Answer

$F=858N$

Work Step by Step

We know that: $\frac{dm}{dt}=\rho v A$ We plug in the known values to obtain: $\frac{dm}{dt}=1000(20)\pi(0.02)^2=25.13Kg/s$ Now, $\Sigma F_y=\frac{dm}{dt}v(v_B-v)$ We plug in the known values to obtain: $F=(25.13)(20sin 45-(-20))$ This simplifies to: $F=858N$
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