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: 43d

Answer

$F = 4.92$ $N$

Work Step by Step

The force that link 5 exerts on link 4 can be calculated by adding the weights of link 1 through link 4 to the force required to accelerate the links 1 through 4 together at $2.50m/s^{2}$, $F_{54}=(m_{1}+m_{2}+m_{3}+m_{4})g + (m_{1}+m_{2}+m_{3}+m_{4})a$ $F_{54}=(m_{1}+m_{2}+m_{3}+m_{4}) (g +a)$ $F_{54}=[(0.400kg) (9.8m/s^{2}+2.5m/s^{2})]=4.92$ $N$
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