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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-m-s-2-The-force-of-interaction-in-newton-between-the-top-link-and-the-link-immediately-belo




Question Number 22582 by Tinkutara last updated on 20/Oct/17
A chain consisting of 5 links each of  mass 0.1 kg is lifted vertically with a  constant acceleration of 2 m/s^2 . The  force of interaction (in newton) between  the top link and the link immediately  below it will be :  Take g = 10 m/s^2 .
Achainconsistingof5linkseachofmass0.1kgisliftedverticallywithaconstantaccelerationof2m/s2.Theforceofinteraction(innewton)betweenthetoplinkandthelinkimmediatelybelowitwillbe:Takeg=10m/s2.
Commented by Tinkutara last updated on 20/Oct/17
Commented by Tinkutara last updated on 20/Oct/17
Book′s answer: 3. Is this right?
Booksanswer:3.Isthisright?
Answered by ajfour last updated on 20/Oct/17
let force between top link and  the one immediately below be N.  Then   N−4mg=4ma  ⇒  N=4m(g+a) =4.8 newtons
letforcebetweentoplinkandtheoneimmediatelybelowbeN.ThenN4mg=4maN=4m(g+a)=4.8newtons

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