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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 .
$$\mathrm{A}\:\mathrm{chain}\:\mathrm{consisting}\:\mathrm{of}\:\mathrm{5}\:\mathrm{links}\:\mathrm{each}\:\mathrm{of} \\ $$$$\mathrm{mass}\:\mathrm{0}.\mathrm{1}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{lifted}\:\mathrm{vertically}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{The} \\ $$$$\mathrm{force}\:\mathrm{of}\:\mathrm{interaction}\:\left(\mathrm{in}\:\mathrm{newton}\right)\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{top}\:\mathrm{link}\:\mathrm{and}\:\mathrm{the}\:\mathrm{link}\:\mathrm{immediately} \\ $$$$\mathrm{below}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:: \\ $$$$\mathrm{Take}\:{g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} . \\ $$
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?
$$\mathrm{Book}'\mathrm{s}\:\mathrm{answer}:\:\mathrm{3}.\:\mathrm{Is}\:\mathrm{this}\:\mathrm{right}? \\ $$
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
$${let}\:{force}\:{between}\:{top}\:{link}\:{and} \\ $$$${the}\:{one}\:{immediately}\:{below}\:{be}\:\boldsymbol{{N}}. \\ $$$${Then}\:\:\:{N}−\mathrm{4}{mg}=\mathrm{4}{ma} \\ $$$$\Rightarrow\:\:{N}=\mathrm{4}{m}\left({g}+{a}\right)\:=\mathrm{4}.\mathrm{8}\:{newtons} \\ $$

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