Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 5 - Exercises and Problems - Page 87: 41


a) 307.5 Newtons b) $F_{seat}=\frac{-m_{seat}v^2}{r}$ c) Nothing would happen

Work Step by Step

a) We know that the centripetal forces are the force of gravity and the force of the seat. Thus, we find the centripetal force first: $F_c = \frac{mv^2}{r}=\frac{(60)(9.7)^2}{6.3}=896.10N$ We subtract the force of gravity to find the force of the seat: $F_{seat}=896.10-mg=896.10-(60)(9.81)=\fbox{307.5 Newtons}$ b) We know that the force of the seat belt is equal to the centripetal force on the seat belt. This is opposite the other force, so we see that it is: $F_{seat}=\frac{-m_{seat}v^2}{r}$ c) Nothing would happen, for the centripetal force keeps the person in circular motion instead of letting them fall out of circular motion.
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