Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - General Problems - Page 135: 70

Answer

$F_f=6180N$ in the same direction as the x component of the normal force.

Work Step by Step

$85\frac{km}{h}\times \frac{1000m}{1km} \times \frac{1h}{3600s}=23.6\frac{m}{s}$ We derived the following formula in example 5-7 $\tan(\theta)=\frac{v^2}{rg}$ From here, we can find the velocity where no friction is necessary $=\sqrt{\tan(\theta)rg}=\sqrt{\tan(14^o)\times72m\times9.8\frac{m}{s^2}}=13.3\frac{m}{s}$ $\sum F_y=F_N\cos(\theta)-F_f\sin(\theta)-F_G=0$ $F_G=mg=1050kg\times9.8\frac{m}{s^2}=10290N$ $10290N=F_N\cos(14^o)-F_f\sin(14^o)$ $F_N\cos(14^o)=10290N-F_f\sin(14^o)$ $\sum F_x=\frac{mv^2}{r}=F_N\sin(\theta)+F_f\cos(\theta)$ $\frac{mv^2}{r}=\frac{1050kg\times(23.6\frac{m}{s})^2}{72m}=8120N$ $8120N=F_N\sin(14^o)+F_f\cos(14^o)$ $8120N=(10290N-F_f\sin(14^o))\sin(14^o)$ $+F_f\cos(14^o)$ $F_f=\frac{8120N-10280N\sin(14^o)}{-\sin^2(14^o)+\cos(14^o)}=6180N$ The radial force is greater than the x component of the normal force, so the force of friction is acting in the same direction as the x component of the normal force.
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