Fundamentals of Physics Extended (10th Edition)

$F = 6. 15$ $N$
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 $2.50m/s^{2}$, $F=(m_{1}+m_{2}+m_{3}+m_{4}+m_{5})g + (m_{1}+m_{2}+m_{3}+m_{4}+m_{5})a$ $F=(m_{1}+m_{2}+m_{3}+m_{4}+m_{5} ) (g +a)$ $F =[(0.500kg) (9.8m/s^{2}+2.5m/s^{2})]=6.15$ $N$