## Fundamentals of Physics Extended (10th Edition)

Published by Wiley

# Chapter 5 - Force and Motion-I - Problems: 43c

#### Answer

$F=3.69N$

#### Work Step by Step

The force that link 4 exerts on link 3 can be calculated by adding the weights of link 1 through link 3 to the force required to accelerate the links 1 through 3 together at $2.50m/s^{2}$, $F_{43}=(m_{1}+m_{2}+m_{3})g + (m_{1}+m_{2}+m_{3})a$ $F_{43}=(m_{1}+m_{2}+m_{3} ) (g +a)$ $F_{43}=[(0.300kg) (9.8m/s^{2}+2.5m/s^{2})]=3.69 N$

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