So the first part of this question asks: A chain consisting of 5 links, each of mass .19kg, is lifted vertically with a constant acceleration of a=2.8 m/s^2. Find the magnitude of the force that link 3 exerts on link 2. I found this answer to be 4.7 N with the following formula: F(link 3 exerts on link 2) = ma + mg + F (link 2 exerts on link 1) I can't seem to get the second part of this question: What is the magnitude of the force F that must be exerted on the top link to achieve this acceleration? I used the same logic as above, namely: F = ma + mg + F(link 5 exerts on link 4) or, can you simply just use the formula F=ma, where m=(5)(.19kg) and a = 2.8m/s^2 ? Thanks!