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Determine-the-speed-with-which-block-B-rises-in-figure-if-the-end-of-the-cord-at-A-is-pulled-down-with-a-speed-of-2-m-s-




Question Number 20344 by Tinkutara last updated on 25/Aug/17
Determine the speed with which block  B rises in figure if the end of the cord at  A is pulled down with a speed of 2 m/s.
$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{with}\:\mathrm{which}\:\mathrm{block} \\ $$$${B}\:\mathrm{rises}\:\mathrm{in}\:\mathrm{figure}\:\mathrm{if}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cord}\:\mathrm{at} \\ $$$${A}\:\mathrm{is}\:\mathrm{pulled}\:\mathrm{down}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}/\mathrm{s}. \\ $$
Commented by Tinkutara last updated on 25/Aug/17
Commented by ajfour last updated on 25/Aug/17
0.5m/s .  if block rises by y, then E goes  up by y, and C comes down by y.  lengrh of rope that available for  pull=4y =x (say)  now   4((dy/dt)) = (dx/dt)                4v_(block) =2m/s      v_(block) =0.5m/s .
$$\mathrm{0}.\mathrm{5}{m}/{s}\:. \\ $$$${if}\:{block}\:{rises}\:{by}\:{y},\:{then}\:{E}\:{goes} \\ $$$${up}\:{by}\:{y},\:{and}\:{C}\:{comes}\:{down}\:{by}\:{y}. \\ $$$${lengrh}\:{of}\:{rope}\:{that}\:{available}\:{for} \\ $$$${pull}=\mathrm{4}{y}\:={x}\:\left({say}\right) \\ $$$${now}\:\:\:\mathrm{4}\left(\frac{{dy}}{{dt}}\right)\:=\:\frac{{dx}}{{dt}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}{v}_{{block}} =\mathrm{2}{m}/{s} \\ $$$$\:\:\:\:{v}_{{block}} =\mathrm{0}.\mathrm{5}{m}/{s}\:. \\ $$
Commented by Tinkutara last updated on 25/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Answered by Tinkutara last updated on 27/Aug/17
Commented by Tinkutara last updated on 27/Aug/17
This is my try! Please do not delete  your own solution ajfour Sir.
$$\mathrm{This}\:\mathrm{is}\:\mathrm{my}\:\mathrm{try}!\:\mathrm{Please}\:\mathrm{do}\:\mathrm{not}\:\mathrm{delete} \\ $$$$\mathrm{your}\:\mathrm{own}\:\mathrm{solution}\:\mathrm{ajfour}\:\mathrm{Sir}. \\ $$
Commented by Tinkutara last updated on 27/Aug/17
Total length of string  = 3x + 4(y − x) = 4y − x = C  Differentiating w.r.t time,  4v_B  = v_A   ⇒ v_B  = 2 ÷ 4 = 0.5 m/s
$$\mathrm{Total}\:\mathrm{length}\:\mathrm{of}\:\mathrm{string} \\ $$$$=\:\mathrm{3}{x}\:+\:\mathrm{4}\left({y}\:−\:{x}\right)\:=\:\mathrm{4}{y}\:−\:{x}\:=\:{C} \\ $$$$\mathrm{Differentiating}\:\mathrm{w}.\mathrm{r}.\mathrm{t}\:\mathrm{time}, \\ $$$$\mathrm{4}{v}_{{B}} \:=\:{v}_{{A}} \\ $$$$\Rightarrow\:{v}_{{B}} \:=\:\mathrm{2}\:\boldsymbol{\div}\:\mathrm{4}\:=\:\mathrm{0}.\mathrm{5}\:\mathrm{m}/\mathrm{s} \\ $$

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