Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems: 43e

Answer

$ F = 6. 15$ $N$

Work Step by Step

The force exerted on the top link can be calculated by adding the weights of all links to the force required to accelerate all the links at $2.50m/s^{2}$, $F=(m_{1}+m_{2}+m_{3}+m_{4}+m_{5})g + (m_{1}+m_{2}+m_{3}+m_{4}+m_{5})a$ $F=(m_{1}+m_{2}+m_{3}+m_{4}+m_{5} ) (g +a)$ $F =[(0.500kg) (9.8m/s^{2}+2.5m/s^{2})]=6.15$ $N$
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