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: 57


$N=7.36N$, $v=1.63m/s$

Work Step by Step

We can determine the required normal force and speed as follows: $K_s=m\frac{mv^2}{\rho}$ We plug in the known values to obtain: $200(0.1)=0.75\frac{v^2}{0.1}$ This simplifies to: $v=1.63m/s$ As $\Sigma F_b=0$ $\implies N_b=mg=0.75\times 9.81=7.36N$ Now $N=\sqrt{N^2_b+N^2_t}$ We plug in the known values to obtain: $N=\sqrt{(7.36)^2+(0)^2}$ $\implies N=7.36N$
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