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 306: 121


$F_x=19.5lb$, $F_y=1.9lb$

Work Step by Step

We know that: $\frac{dm}{dt}=\rho\times v\times A$ $\frac{dm_A}{dt}=\frac{62.4}{32.2}\times 12\pi\times (\frac{0.25}{12})^2=0.03171slug/s$ Similarly $\frac{dm_B}{dt}=\frac{62.4}{32.2}\times 25\pi \times (\frac{0.25}{12})^2=0.06606slug/s$ and $\frac{dm_c}{dt}=\frac{dm_A}{dt}+\frac{dm_B}{dt}=0.03171+0.06606=0.097777slug/s$ Now the balancing force in the $x$ direction is given as $\Sigma F_x=\frac{dm_A}{dt}v_{Ax}+\frac{dm_B}{dt}v_{Bx}+\frac{dm_C}{dt}v_{Cx}$ We plug in the above known values to obtain: $F_x=19.5lb$ and the balancing force in the $y$ direction is $\Sigma F_y=\frac{dm_A}{dt}v_{Ay}+\frac{dm_B}{dt}v_{By}+\frac{dm_C}{dt}v_{Cy}$ We plug in the known values to obtain: $F_y=1.9lb$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.