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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

letp1,p2,p3betheaccelerationofstringaboutthepulley1,2,3respectively(↷positive).a1=p1(↑positiveforallaccelerations)a4=−p1−p2a2=−p1+p2+p3a3=−p1+p2−p3a2−a1=−1m/s2⇒−p1+p2+p3−p1=−1⇒−2p1+p2+p3=−1...(i)a3−a1=−5m/s2⇒−p1+p2−p3−p1=−5⇒−2p1+p2−p3=−5...(ii)a3−a4=0⇒−p1+p2−p3−(−p1−p2)=0⇒2p2−p3=0...(iii)(i)−(ii):⇒2p3=4⇒p3=2from(iii):⇒p2=1from(i):⇒−2p1+1+2=−1⇒p1=2Q26:a1=p1=2m/s2upwards⇒Ans.aQ27:a2=−p1+p2+p3=−2+1+2=1upwards⇒Ans.bQ28:a3=−p1+p2−p3=−2+1−2=−3downwards⇒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!

Sir,howtowritetheinitial3eq.sa4=−p1−p2a2=−p1+p2+p3a3=−p1+p2−p3?Needmoreexplaination!

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

Eachstringrollsaboutapulley.Let′ssaytheaccelerationofthestringsarep1,p2,p3respectively.PulleyIisfixed,thereforetheaccelerationofblock1isa1=p1(upwards,hence+ve)andtheaccelerationofpulleyIIisaII=−p1(downwards,hence−ve).Block4movesrelativelytopulleyIIwithp2,thereforea4=aII−p2=−p1−p2SimilarlyaIII=aII+p2=−p1+p2Block2and3moverelativelytopulleyIIIwithp3,hencea3=aIII−p3=−p1+p2−p3anda2=aIII+p3=−p1+p2+p3

Commented by MrW3 last updated on 15/Jul/18

I hope it′s clear now.

Ihopeit′sclearnow.

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

Thank you so much sir ! ��������

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