Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 147: 62

Answer

$v=2.10m/s$

Work Step by Step

The required speed can be determined as follows: $F_n=F_f$ $\implies ma_N=mg\mu_s$ $\implies m(\frac{v^2}{\rho})=mg\mu_s$ This can be rearranged as: $v=\sqrt{\rho g\mu_s}$ We plug in the known values to obtain: $v=\sqrt{1.5(9.81)(0.3)}$ $\implies v=2.10m/s$

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