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A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration of 2.5 m/s2The force of interaction between the top link and the link immediately below it, will be??

pradeepta choudhury , 11 Years ago
Grade 12
anser 5 Answers
Akash Kumar Dutta

Last Activity: 11 Years ago

force applied by the first link on to the last=m(g+a)
where m=mass of the later part of string

putting the values,
force=.4(12.5)=5N

Bevkoof Singh

Last Activity: 11 Years ago

o.75 n i think

pradeepta choudhury

Last Activity: 11 Years ago

hey sry.....both the answers are wrong....................the options of the question were (a)6.15N (b)4.92N (c)3.69 N (d)20.46N

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shubham sharma

Last Activity: 11 Years ago

Ans is 4.92 !!!

Here it goes »

1st for chain as a system

F-5mg»5ma

Then let T be the interaction force b/t 1st & 2nd chain then downward force from 1st chain LL be (mg plusT)

Then, F - (mg plus T)=ma

Solve these for T !!!!;)

Satwik

Last Activity: 6 Years ago

M=0.4
F=m(a+g)
F=0.4(2.5+9.8)
F=0.4(12.3)
Therefore ,F=4.92N
Therefore, the force of interaction between
the first link and link immediately below of it is 4.92N....thank-you

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