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Question

A chain consisting of 5 links each of mass 0.1 kg is lifted vertically up with a constant acceleration of 2.5 m/s2. The force of interaction between 1st and 2nd links as shown
1290146_effecdf4ca3144b9bb356d02b58507e3.png

A
5.20 N
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B
4.92 N
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C
9.84 N
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D
2.46 N
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Solution

The correct option is A 5.20 N
5.20N

For links of mass 0.1kg and upward acceleration a=2.5m/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=3.69N

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=5.20N

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