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Question-217506




Question Number 217506 by Tawa11 last updated on 15/Mar/25
Commented by Tawa11 last updated on 15/Mar/25
2.5m/s^2  ?
2.5m/s2?
Commented by mr W last updated on 15/Mar/25
yes, if taken g=10 m/s^2 .  what are the accelerations of the  pulleys?
yes,iftakeng=10m/s2.whataretheaccelerationsofthepulleys?
Commented by Tawa11 last updated on 15/Mar/25
Sir, I only solve acceleration of the block.  Show me the acceleration of pulleys sir.
Sir,Ionlysolveaccelerationoftheblock.Showmetheaccelerationofpulleyssir.
Commented by Tawa11 last updated on 15/Mar/25
sir.  Q217504, am I correct?
sir.Q217504,amIcorrect?
Commented by mr W last updated on 15/Mar/25
i don′t understand the question.
idontunderstandthequestion.
Commented by Tawa11 last updated on 15/Mar/25
Hahahahahaha.  I think the question is missing a  parameter.
Hahahahahaha.Ithinkthequestionismissingaparameter.
Commented by Tawa11 last updated on 15/Mar/25
Thanks sir.
Thankssir.
Answered by mr W last updated on 15/Mar/25
Commented by mr W last updated on 15/Mar/25
2F_1 =F=100 ⇒F_1 =50 N  2F_2 =F_1 =50 ⇒F_2 =25 N  with g=10 m/s^2   F_1 −m_1 g=m_1 a_1   a_1 =((F_1 −m_1 g)/m_1 )  a_1 =((50−1×10)/1)=40 m/s^2  (↑)  similarly  a_2 =((25−2×10)/2)=2.5 m/s^2   (↑)  a_3 =((25−3×10)/3)=−(5/3) m/s^2   (↓)    A_2 −a_2 =a_3 −A_2    ⇒A_2 =((a_2 +a_3 )/2)=((2.5−(5/3))/2)=(5/(12)) m/s^2   (↑)    A_1 −a_1 =A_2 −A_1   ⇒A_1 =((a_1 +A_2 )/2)=((40+(5/(12)))/2)=20(5/(24)) m/s^2   (↑)
2F1=F=100F1=50N2F2=F1=50F2=25Nwithg=10m/s2F1m1g=m1a1a1=F1m1gm1a1=501×101=40m/s2()similarlya2=252×102=2.5m/s2()a3=253×103=53m/s2()A2a2=a3A2A2=a2+a32=2.5532=512m/s2()A1a1=A2A1A1=a1+A22=40+5122=20524m/s2()
Commented by Tawa11 last updated on 15/Mar/25
God bless you sir.  I really appreciate you sir.
Godblessyousir.Ireallyappreciateyousir.
Commented by Tawa11 last updated on 15/Mar/25
For mass 3kg  =  5ms^(−1)   For mas 2kg  =  2.5 ms^(−1)   for mass 1kg  =  3.75ms^(−1)
Formass3kg=5ms1Formas2kg=2.5ms1formass1kg=3.75ms1
Commented by mr W last updated on 15/Mar/25
wrong!
wrong!
Commented by Tawa11 last updated on 15/Mar/25
I will try more
Iwilltrymore
Commented by Tawa11 last updated on 16/Mar/25
1.8 ms^(− 2)   is this correct sir.  for the 3kg, 2kg and 1kg  pulleys
1.8ms2isthiscorrectsir.forthe3kg,2kgand1kgpulleys
Commented by mr W last updated on 16/Mar/25
no! how can they all have the same  acceleration? they are not fixed  with each other.
no!howcantheyallhavethesameacceleration?theyarenotfixedwitheachother.
Commented by Tawa11 last updated on 16/Mar/25
noted sir.
notedsir.
Commented by Tawa11 last updated on 16/Mar/25
Please solve sir.
Pleasesolvesir.
Commented by mr W last updated on 16/Mar/25
i have given an example above. if  you have understood this, i think   you should be able to solve it by  yourself. but i have opened a new  thread. you can follow the question  and solution there.
ihavegivenanexampleabove.ifyouhaveunderstoodthis,ithinkyoushouldbeabletosolveitbyyourself.butihaveopenedanewthread.youcanfollowthequestionandsolutionthere.
Commented by Tawa11 last updated on 16/Mar/25
you have not solve Q217565 sir
youhavenotsolveQ217565sir
Commented by mr W last updated on 16/Mar/25
i know. but it is solved! you can   compare with your own solution.
iknow.butitissolved!youcancomparewithyourownsolution.
Commented by Tawa11 last updated on 16/Mar/25
Commented by Tawa11 last updated on 16/Mar/25
Commented by Tawa11 last updated on 16/Mar/25
Is this correct now sir?
Isthiscorrectnowsir?
Commented by mr W last updated on 17/Mar/25
since you take F_1 =F/2, F_2 =F_1 /2,   the rest must be wrong. these are   such basic things! even when the   objects are in rest, we have that  F=2F_1 +M_1 g > 2F_1 . how can you   say F_1 =F/2?
sinceyoutakeF1=F/2,F2=F1/2,therestmustbewrong.thesearesuchbasicthings!evenwhentheobjectsareinrest,wehavethatF=2F1+M1g>2F1.howcanyousayF1=F/2?
Commented by mr W last updated on 17/Mar/25
it is assumed that there is no  friction between the strings and  the pulleys. in this case the pulley  don′t rotate. otherwise you can not  assume that the string on both sides  of a pulley has the same tension, but  have assumed F_(1, left) =F_(1, right) =F_1 ,  therefore no friction and no   rotation of pulleys!
itisassumedthatthereisnofrictionbetweenthestringsandthepulleys.inthiscasethepulleydontrotate.otherwiseyoucannotassumethatthestringonbothsidesofapulleyhasthesametension,buthaveassumedF1,left=F1,right=F1,thereforenofrictionandnorotationofpulleys!
Commented by Tawa11 last updated on 17/Mar/25
Noted sir.
Notedsir.

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