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Question-217565




Question Number 217565 by mr W last updated on 16/Mar/25
Commented by mr W last updated on 16/Mar/25
the strings are massless and there  is no friction between strings and  pulleys.  find the accelerations of all objects.
thestringsaremasslessandthereisnofrictionbetweenstringsandpulleys.findtheaccelerationsofallobjects.
Answered by ajfour last updated on 16/Mar/25
Acc. of 5kg block be A.  relative to 4kg pulley let acc. of 3kg be a_1 ↓  and that acc. of 4kg block itself be a↓ relative  to 5kg block.  3g−T_1 =3(A+a+a_1 )    ...(i)  T_1 −2g=2(A+a−a_1 )  ...(ii)  T_2 −g=A+a      ....(iii)  2T_2 −T_1 =4(a−A)    ...(iv)  200−2T_1 =5A     ...(v)  ..................  4(A+a+g)−2T_1 =8(a−A)  200−2T_1 =5A  ⇒  200−4(A+a+g)=13A−8a  ..(1)  addingr (i)&(ii)  g=5A+5a+a_1    ....(2)  uding (ii) in (v)  200−4g−4(A+a−a_1 )=5A  ⇒ 200−4g=9A+4a−4a_1   now using (2)  200−4g=9A+4a−4(g−5A−5a)  ⇒ 200=29A+24a  from (1)  {200−4g=17A−4a}×6  1200−24g=102A−24a  200=29A+24a  adding above two  1400−24g=131A  A=((1400−24g)/(131))  rest can be found now.
Acc.of5kgblockbeA.relativeto4kgpulleyletacc.of3kgbea1andthatacc.of4kgblockitselfbearelativeto5kgblock.3gT1=3(A+a+a1)(i)T12g=2(A+aa1)(ii)T2g=A+a.(iii)2T2T1=4(aA)(iv)2002T1=5A(v)4(A+a+g)2T1=8(aA)2002T1=5A2004(A+a+g)=13A8a..(1)addingr(i)&(ii)g=5A+5a+a1.(2)uding(ii)in(v)2004g4(A+aa1)=5A2004g=9A+4a4a1nowusing(2)2004g=9A+4a4(g5A5a)200=29A+24afrom(1){2004g=17A4a}×6120024g=102A24a200=29A+24aaddingabovetwo140024g=131AA=140024g131restcanbefoundnow.
Commented by mr W last updated on 17/Mar/25
thanks sir!  you got A=((1160)/(131))≈8.85 m/s^2   i got acc. of  5kg pulley   A_1 =((5590)/(421))≈13.28 m/s^2
thankssir!yougotA=11601318.85m/s2igotacc.of5kgpulleyA1=559042113.28m/s2
Commented by ajfour last updated on 18/Mar/25
I ll check sir. Thanks for solving the  projectile one too.
Illchecksir.Thanksforsolvingtheprojectileonetoo.
Answered by mr W last updated on 17/Mar/25
Commented by mr W last updated on 17/Mar/25
δ_1  relative acc. of string to pulley 1  δ_2  relative acc. of string to pulley 2  a_1 =A_1 +δ_1   A_2 =A_1 −δ_1   a_2 =A_2 +δ_2 =A_1 −δ_1 +δ_2   a_3 =A_2 −δ_2 =A_1 −δ_1 −δ_2     F_2 =m_2 (g+a_2 )=m_2 (g+A_1 −δ_1 +δ_2 )  F_2 =m_3 (g+a_3 )=m_3 (g+A_1 −δ_1 −δ_2 )  F_1 =m_1 (g+a_1 )=m_1 (g+A_1 +δ_1 )  F_1 −2F_2 =M_2 (g+A_2 )=M_2 (g+A_1 −δ_1 )  F−2F_1 =M_1 (g+A_1 )    F−2m_1 (g+A_1 +δ_1 )=M_1 (g+A_1 )  (M_1 +2m_1 )A_1 +2m_1 δ_1 =F−(M_1 +2m_1 )g  7A_1 +2δ_1 =130   ...(i)    m_1 (g+A_1 +δ_1 )−2m_2 (g+A_1 −δ_1 +δ_2 )=M_2 (g+A_1 −δ_1 )  (m_1 −2m_2 −M_2 )A_1 +(M_2 +m_1 +2m_2 )δ_1 −2m_2 δ_2 =(M_2 −m_1 +2m_2 )g  −7A_1 +9δ_1 −4δ_2 =70   ...(ii)    m_2 (g+A_1 −δ_1 +δ_2 )=m_3 (g+A_1 −δ_1 −δ_2 )  (m_2 −m_3 )A_1 +(−m_2 +m_3 )δ_1 +(m_2 +m_3 )δ_2 =(m_3 −m_2 )g  −A_1 +δ_1 +5δ_2 =10   ...(iii)    δ_1 =((7800)/(421))  δ_2 =((400)/(421))  A_1 =((7800)/(421))+5×((400)/(421))−10=((5590)/(421))    A_2 =((5590)/(421))−((7800)/(421))=−((2210)/(421))  a_1 =((5590)/(421))+((7800)/(421))=((13390)/(421))  a_2 =−((2210)/(421))+((400)/(421))=−((1810)/(421))  a_3 =−((2210)/(421))−((400)/(421))=−((2610)/(421))
δ1relativeacc.ofstringtopulley1δ2relativeacc.ofstringtopulley2a1=A1+δ1A2=A1δ1a2=A2+δ2=A1δ1+δ2a3=A2δ2=A1δ1δ2F2=m2(g+a2)=m2(g+A1δ1+δ2)F2=m3(g+a3)=m3(g+A1δ1δ2)F1=m1(g+a1)=m1(g+A1+δ1)F12F2=M2(g+A2)=M2(g+A1δ1)F2F1=M1(g+A1)F2m1(g+A1+δ1)=M1(g+A1)(M1+2m1)A1+2m1δ1=F(M1+2m1)g7A1+2δ1=130(i)m1(g+A1+δ1)2m2(g+A1δ1+δ2)=M2(g+A1δ1)(m12m2M2)A1+(M2+m1+2m2)δ12m2δ2=(M2m1+2m2)g7A1+9δ14δ2=70(ii)m2(g+A1δ1+δ2)=m3(g+A1δ1δ2)(m2m3)A1+(m2+m3)δ1+(m2+m3)δ2=(m3m2)gA1+δ1+5δ2=10(iii)δ1=7800421δ2=400421A1=7800421+5×40042110=5590421A2=55904217800421=2210421a1=5590421+7800421=13390421a2=2210421+400421=1810421a3=2210421400421=2610421
Commented by Tawa11 last updated on 18/Mar/25
Sir, you assumed the cylinder are rotating here.  But Ajfour did not assume the cylinder is rotating.
Sir,youassumedthecylinderarerotatinghere.ButAjfourdidnotassumethecylinderisrotating.
Commented by mr W last updated on 18/Mar/25
The pulleys don′t rotate and i also  didn′t assume that they rotate.
Thepulleysdontrotateandialsodidntassumethattheyrotate.
Commented by Tawa11 last updated on 18/Mar/25
Thanks sir.
Thankssir.
Commented by mr W last updated on 18/Mar/25
if there is no friction between string  and pulley, the string can also   move relatively to the pulley without  that the pulley rotates.  in the workings above δ_1  and δ_2  are  the relative motions of the strings  along the pulleys. there are not the  rotation of the pulleys!
ifthereisnofrictionbetweenstringandpulley,thestringcanalsomoverelativelytothepulleywithoutthatthepulleyrotates.intheworkingsaboveδ1andδ2aretherelativemotionsofthestringsalongthepulleys.therearenottherotationofthepulleys!
Commented by mr W last updated on 18/Mar/25
Commented by Tawa11 last updated on 18/Mar/25
Ohh, thanks sir.  I appreciate your time.
Ohh,thankssir.Iappreciateyourtime.

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