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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/s2 The force of interaction between the top link and the link immediately below it, will be??

pradeepta choudhury , 12 Years ago

Grade 12

5 Answers

Akash Kumar Dutta

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

Last Activity: 12 Years ago

Bevkoof Singh

o.75 n i think

Approved

Last Activity: 12 Years ago

pradeepta choudhury

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

Last Activity: 12 Years ago

shubham sharma

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 !!!!;)

Last Activity: 12 Years ago

Satwik

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

Last Activity: 7 Years ago

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Mechanics> laws-of-motion...

A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration of 2.5 m/s2 The force of interaction between the top link and the link immediately below it, will be??

pradeepta choudhury , 12 Years ago

Akash Kumar Dutta

Bevkoof Singh

pradeepta choudhury

shubham sharma

Satwik

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