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 146: 54



Work Step by Step

We can determine the required speed as follows: $\Sigma F_y=0$ $\implies T-W_A=0$ $\implies T=W_A=m_A g~~~$eq(1) But we know that $T=m_b\frac{v^2}{r}~~~$eq(2) Comparing eq(1) and eq(2), we obtain: $m_Ag=m_B\frac{v^2}{r}$ This can be rearranged as: $v=\sqrt{(\frac{m_A}{m_B})gr}$ We plug in the known values to obtain: $v=\sqrt{(\frac{15}{2})(9.81)(1.5)}$ This simplifies to: $v=10.5m/s$
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