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Question Number 39970 by rahul 19 last updated on 14/Jul/18

Answered by MrW3 last updated on 14/Jul/18

let p_1 ,p_2 ,p_3 be the acceleration of string about the pulley 1,2,3 respectively (↷ positive). a_1 =p_1 (↑ positive for all accelerations) a_4 =−p_1 −p_2 a_2 =−p_1 +p_2 +p_3 a_3 =−p_1 +p_2 −p_3 a_2 −a_1 =−1 m/s^2 ⇒−p_1 +p_2 +p_3 −p_1 =−1 ⇒−2p_1 +p_2 +p_3 =−1 ...(i) a_3 −a_1 =−5 m/s^2 ⇒−p_1 +p_2 −p_3 −p_1 =−5 ⇒−2p_1 +p_2 −p_3 =−5 ...(ii) a_3 −a_4 =0 ⇒−p_1 +p_2 −p_3 −(−p_1 −p_2 )=0 ⇒2p_2 −p_3 =0 ...(iii) (i)−(ii): ⇒2p_3 =4 ⇒p_3 =2 from (iii): ⇒p_2 =1 from (i): ⇒−2p_1 +1+2=−1 ⇒p_1 =2 Q26: a_1 =p_1 =2 m/s^2 upwards ⇒Ans. a Q27: a_2 =−p_1 +p_2 +p_3 =−2+1+2=1 upwards ⇒Ans. b Q28: a_3 =−p_1 +p_2 −p_3 =−2+1−2=−3 downwards ⇒Ans. c

$${let}\:{p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} \:{be}\:{the}\:{acceleration}\:{of}\:{string} \\ $$$${about}\:{the}\:{pulley}\:\mathrm{1},\mathrm{2},\mathrm{3}\:{respectively}\:\left(\curvearrowright\:{positive}\right). \\ $$$$ \\ $$$${a}_{\mathrm{1}} ={p}_{\mathrm{1}} \:\left(\uparrow\:{positive}\:{for}\:{all}\:{accelerations}\right) \\ $$$${a}_{\mathrm{4}} =−{p}_{\mathrm{1}} −{p}_{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} +{p}_{\mathrm{3}} \\ $$$${a}_{\mathrm{3}} =−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} −{p}_{\mathrm{3}} \\ $$$$ \\ $$$${a}_{\mathrm{2}} −{a}_{\mathrm{1}} =−\mathrm{1}\:{m}/{s}^{\mathrm{2}} \\ $$$$\Rightarrow−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} +{p}_{\mathrm{3}} −{p}_{\mathrm{1}} =−\mathrm{1} \\ $$$$\Rightarrow−\mathrm{2}{p}_{\mathrm{1}} +{p}_{\mathrm{2}} +{p}_{\mathrm{3}} =−\mathrm{1}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${a}_{\mathrm{3}} −{a}_{\mathrm{1}} =−\mathrm{5}\:{m}/{s}^{\mathrm{2}} \\ $$$$\Rightarrow−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} −{p}_{\mathrm{3}} −{p}_{\mathrm{1}} =−\mathrm{5} \\ $$$$\Rightarrow−\mathrm{2}{p}_{\mathrm{1}} +{p}_{\mathrm{2}} −{p}_{\mathrm{3}} =−\mathrm{5}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${a}_{\mathrm{3}} −{a}_{\mathrm{4}} =\mathrm{0} \\ $$$$\Rightarrow−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} −{p}_{\mathrm{3}} −\left(−{p}_{\mathrm{1}} −{p}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{p}_{\mathrm{2}} −{p}_{\mathrm{3}} =\mathrm{0}\:\:\:...\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow\mathrm{2}{p}_{\mathrm{3}} =\mathrm{4} \\ $$$$\Rightarrow{p}_{\mathrm{3}} =\mathrm{2} \\ $$$${from}\:\left({iii}\right): \\ $$$$\Rightarrow{p}_{\mathrm{2}} =\mathrm{1} \\ $$$${from}\:\left({i}\right): \\ $$$$\Rightarrow−\mathrm{2}{p}_{\mathrm{1}} +\mathrm{1}+\mathrm{2}=−\mathrm{1} \\ $$$$\Rightarrow{p}_{\mathrm{1}} =\mathrm{2} \\ $$$$ \\ $$$${Q}\mathrm{26}: \\ $$$${a}_{\mathrm{1}} ={p}_{\mathrm{1}} =\mathrm{2}\:{m}/{s}^{\mathrm{2}} \:{upwards} \\ $$$$\Rightarrow{Ans}.\:{a} \\ $$$${Q}\mathrm{27}: \\ $$$${a}_{\mathrm{2}} =−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} +{p}_{\mathrm{3}} =−\mathrm{2}+\mathrm{1}+\mathrm{2}=\mathrm{1}\:{upwards} \\ $$$$\Rightarrow{Ans}.\:{b} \\ $$$${Q}\mathrm{28}: \\ $$$${a}_{\mathrm{3}} =−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} −{p}_{\mathrm{3}} =−\mathrm{2}+\mathrm{1}−\mathrm{2}=−\mathrm{3}\:{downwards} \\ $$$$\Rightarrow{Ans}.\:{c} \\ $$

Commented by rahul 19 last updated on 15/Jul/18

Sir, how to write the initial 3 eq.s a_4 = −p_1 −p_2 a_2 =−p_1 +p_2 +p_3 a_3 =−p_1 +p_2 −p_3 ? Need more explaination!

$$\mathrm{Sir},\:\mathrm{how}\:\mathrm{to}\:\mathrm{write}\:\mathrm{the}\:\mathrm{initial}\:\mathrm{3}\:\mathrm{eq}.\mathrm{s} \\ $$$$\mathrm{a}_{\mathrm{4}} =\:−\mathrm{p}_{\mathrm{1}} −\mathrm{p}_{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{2}} =−\mathrm{p}_{\mathrm{1}} +\mathrm{p}_{\mathrm{2}} +\mathrm{p}_{\mathrm{3}} \\ $$$$\mathrm{a}_{\mathrm{3}} =−\mathrm{p}_{\mathrm{1}} +\mathrm{p}_{\mathrm{2}} −\mathrm{p}_{\mathrm{3}} \:? \\ $$$$\mathrm{Need}\:\mathrm{more}\:\mathrm{explaination}! \\ $$

Commented by MrW3 last updated on 15/Jul/18

Commented by MrW3 last updated on 15/Jul/18

Each string rolls about a pulley. Let′s say the acceleration of the strings are p_1 ,p_2 ,p_3 respectively. Pulley I is fixed, therefore the acceleration of block 1 is a_1 =p_1 (upwards, hence +ve) and the acceleration of pulley II is a_(II) =−p_1 (downwards, hence −ve). Block 4 moves relatively to pulley II with p_2 , therefore a_4 =a_(II) −p_2 =−p_1 −p_2 Similarly a_(III) =a_(II) +p_2 =−p_1 +p_2 Block 2 and 3 move relatively to pulley III with p_3 , hence a_3 =a_(III) −p_3 =−p_1 +p_2 −p_3 and a_2 =a_(III) +p_3 =−p_1 +p_2 +p_3

$${Each}\:{string}\:{rolls}\:{about}\:{a}\:{pulley}.\:{Let}'{s} \\ $$$${say}\:{the}\:{acceleration}\:{of}\:{the}\:{strings}\:{are} \\ $$$${p}_{\mathrm{1}} ,{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} \:{respectively}. \\ $$$${Pulley}\:{I}\:{is}\:{fixed},\:{therefore}\:{the} \\ $$$${acceleration}\:{of}\:{block}\:\mathrm{1}\:{is} \\ $$$${a}_{\mathrm{1}} ={p}_{\mathrm{1}} \:\left({upwards},\:{hence}\:+{ve}\right) \\ $$$${and}\:{the}\:{acceleration}\:{of}\:{pulley}\:{II}\:{is} \\ $$$${a}_{{II}} =−{p}_{\mathrm{1}} \:\left({downwards},\:{hence}\:−{ve}\right). \\ $$$${Block}\:\mathrm{4}\:{moves}\:{relatively}\:{to}\:{pulley}\:{II} \\ $$$${with}\:{p}_{\mathrm{2}} ,\:{therefore} \\ $$$${a}_{\mathrm{4}} ={a}_{{II}} −{p}_{\mathrm{2}} =−{p}_{\mathrm{1}} −{p}_{\mathrm{2}} \\ $$$${Similarly} \\ $$$${a}_{{III}} ={a}_{{II}} +{p}_{\mathrm{2}} =−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} \\ $$$${Block}\:\mathrm{2}\:{and}\:\mathrm{3}\:{move}\:{relatively}\:{to}\: \\ $$$${pulley}\:{III}\:{with}\:{p}_{\mathrm{3}} ,\:{hence} \\ $$$${a}_{\mathrm{3}} ={a}_{{III}} −{p}_{\mathrm{3}} \:=−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} −{p}_{\mathrm{3}} \:{and} \\ $$$${a}_{\mathrm{2}} ={a}_{{III}} +{p}_{\mathrm{3}} \:=−{p}_{\mathrm{1}} +{p}_{\mathrm{2}} +{p}_{\mathrm{3}} \\ $$

Commented by MrW3 last updated on 15/Jul/18

I hope it′s clear now.

$${I}\:{hope}\:{it}'{s}\:{clear}\:{now}. \\ $$

Commented by rahul 19 last updated on 15/Jul/18

Thank you so much sir ! ��������

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