Question

A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with constant acceleration of 2.5 M per second square the force of interaction between the top link and the link immediately below it will be .... please someone answer the question

Answer 1

For links of mass 0.1 kg and upward acceleration a = 2.5 m/s²,

F = n*m*(g + a) = n*0.1kg*(9.8+2.5)m/s² = n*1.23N

where n = number of links below point of interest.

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

F = 3*1.23N = 3.69 N

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

F = (9.8+2)m/s² * (0.187+0.254)kg = 5.20 N

Answer 2

Answer:

4.92N

Explanation: T-mg=ma