A-uniform-chain-of-length-L-and-mass-M-is-lying-on-a-smooth-table-and-one-third-of-its-length-is-hanging-vertically-down-over-the-edge-of-the-table-If-g-is-acceleration-due-to-gravity-calculate-work

Question Number 22624 by Tinkutara last updated on 21/Oct/17

$$\mathrm{A}\:\mathrm{uniform}\:\mathrm{chain}\:\mathrm{of}\:\mathrm{length}\:\mathrm{L}\:\mathrm{and}\:\mathrm{mass} \\ $$$$\mathrm{M}\:\mathrm{is}\:\mathrm{lying}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{table}\:\mathrm{and}\:\mathrm{one} \\ $$$$\mathrm{third}\:\mathrm{of}\:\mathrm{its}\:\mathrm{length}\:\mathrm{is}\:\mathrm{hanging}\:\mathrm{vertically} \\ $$$$\mathrm{down}\:\mathrm{over}\:\mathrm{the}\:\mathrm{edge}\:\mathrm{of}\:\mathrm{the}\:\mathrm{table}.\:\mathrm{If}\:\mathrm{g}\:\mathrm{is} \\ $$$$\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity},\:\mathrm{calculate} \\ $$$$\mathrm{work}\:\mathrm{required}\:\mathrm{to}\:\mathrm{pull}\:\mathrm{the}\:\mathrm{hanging}\:\mathrm{part} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{table}. \\ $$

Answered by ajfour last updated on 21/Oct/17

W=△U+△K For minimum work △K=0 W=△U =(1/3)(Mg)((l/6))=((Mgl)/(18)) .

$${W}=\bigtriangleup{U}+\bigtriangleup{K} \\ $$$${For}\:{minimum}\:{work}\:\bigtriangleup{K}=\mathrm{0} \\ $$$${W}=\bigtriangleup{U}\:=\frac{\mathrm{1}}{\mathrm{3}}\left({Mg}\right)\left(\frac{{l}}{\mathrm{6}}\right)=\frac{\boldsymbol{{Mgl}}}{\mathrm{18}}\:. \\ $$

Commented by math solver last updated on 21/Oct/17

taking height with respect to centre of mass i.e 1/2× l/3 = l/6

$${taking}\:{height}\:{with}\:{respect}\:{to}\:{centre}\: \\ $$$${of}\:{mass}\:{i}.{e}\:\mathrm{1}/\mathrm{2}×\:{l}/\mathrm{3}\:=\:{l}/\mathrm{6} \\ $$

Commented by Tinkutara last updated on 21/Oct/17