A chain consisting of 5 links each of mass 0.1 kg is lifted vertically up with a constant acceleration of 2.5m/s². The force of interaction between 1st and 2nd links as shown

For links of mass 0.1kg and upward acceleration a = 2.5$m/s^2$,

F = n × m ×( g + a) = n × 0.1kg × (9.8 + 2.5)$\mathrm{m}/{\mathrm{s}}^2$= n × 1.23N

where n = number of links below the point of interest.

For instance, the force on link 2 from link 3 has n=3, and

F = 3 × 1.23N = 3.69N

You have a second set of data; for that set, the force on link 3 from link 4 is

F = (9.8 + 2)$\mathrm{m}/{\mathrm{s}}^2$ × (0.187 + 0.254)kg = 5.20N