Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 147: 61


$v=9.29ft/s$, $T=38lb$

Work Step by Step

We can determine the required speed and the tension as follows: $Wcos\theta=ma_t$ $\implies 60cos\theta=\frac{60}{32.2}a_t$ $\implies a_t=32.2cos\theta$ As $a_t=\frac{dv}{dt}=v\frac{dv}{ds}$ $\implies ds=10d\theta$ $\implies 32.2cos\theta=v\frac{dv}{10d\theta}$ $\implies 322cos\theta d\theta=vdv$ Integrating both sides, we obtain: $\int_0^v vdv=\int_{60}^{90} 322cos\theta d\theta$ This simplifies to: $v=9.29ft/s$ Now $\Sigma F_n=ma_n$ $\implies 2T-Wsin\theta=m\frac{v^2}{\rho}$ We plug in the known values to obtain: $2T-(60 sin90)=\frac{60}{32.2}\times \frac{(9.288)^2}{10}$ This simplifies to: $T=38lb$
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