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
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:
4.92N
Explanation: T-mg=ma