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 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 137: 49



Work Step by Step

We can determine the required acceleration as follows: We know that $\Sigma F_x=m_A a_x$ $12-N_B sin 15=(\frac{8}{32.2})a_A~~~$eq(1) $\Sigma F_y=m_Ba_y$ $\implies N_Bcos15-15=(\frac{15}{32.2})a_B~~~$eq(2) As $S_B=S_Atan 15$ $\implies a_B=a_A tan15~~~$eq(3) Solving eq(1), eq(2) and eq(3), we obtain: $a_A=28.3ft/s^2$, $N_B=19.2lb$ and $a_B=7.59ft/s^2$
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.