Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.4 - Power and Efficiency - Problems - Page 209: 42

Answer

$\theta=47.2^{\circ}$

Work Step by Step

The required angle can be determined as follows: $P=Fv$ We plug in the known values to obtain: $(100hp)(550lb\cdot ft)=2500sin\theta(30)$ This simplifies to: $\theta=sin^{-1}(\frac{55000}{75000})$ $\implies \theta=47.2^{\circ}$
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