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 13.6 - Equations of Motion: Cyliindrical Coordinates - Problems - Page 158: 85



Work Step by Step

We can determine the required force as follows: $z=0.1sin2\theta$ Taking the derivative, we have $z^{\cdot}=0.2\theta^{\cdot}cos 2\theta$ $\implies z^{\cdot \cdot}=-0.4\theta^{\cdot2}sin2\theta+0.2\theta^{\cdot \cdot}cos2\theta$ We plug in the known values to obtain: $z^{\cdot \cdot}=-0.4(6)^2sin90+0=-14.4ft/s^2$ We know that $F_A-F_s=\frac{W}{g}z^{\cdot \cdot}$ $\implies F_A-k(z+s)=\frac{0.75}{32.2}(-14.4)$ $\implies F_A-12(0.1sin90+0.3)=\frac{0.75}{32.2}(-14.4)$ This simplifies to: $F_A=4.46lb$
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.