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Question Number 23224 by Tinkutara last updated on 27/Oct/17
Consider the system shown in the  figure. Initially the system was in rest.  (i) Find the acceleration of block if man  climbs the rod with acceleration a (w.r.t.  rod)  (ii) If the man climb to the top of the  rod then find the distance moved by the  block.
Considerthesystemshowninthefigure.Initiallythesystemwasinrest.(i)Findtheaccelerationofblockifmanclimbstherodwithaccelerationa(w.r.t.rod)(ii)Ifthemanclimbtothetopoftherodthenfindthedistancemovedbytheblock.
Commented by mrW1 last updated on 27/Oct/17
(1)  let a_R =acceleration of rod (↑) and block (↓)  a_M =acceleration of man (↑)  a_M =a+a_R   F=force between rod and man  F−mg=ma_M =m(a+a_R )  ⇒F=m(g+a+a_R )  (M+m−M)g−F=(M+m+M)a_R   (M+m−M)g−m(g+a+a_R )=(M+m+M)a_R   −ma=2(M+m)a_R   ⇒a_R =−((ma)/(2(M+m)))  (2)  L=(1/2)at^2   ΔH=(1/2)a_R t^2   ⇒ΔH=(a_R /a)×L  ⇒ΔH=−(m/(2(M+m)))×L  (the block moves up a distance ((mL)/(2(M+m))))
(1)letaR=accelerationofrod()andblock()aM=accelerationofman()aM=a+aRF=forcebetweenrodandmanFmg=maM=m(a+aR)F=m(g+a+aR)(M+mM)gF=(M+m+M)aR(M+mM)gm(g+a+aR)=(M+m+M)aRma=2(M+m)aRaR=ma2(M+m)(2)L=12at2ΔH=12aRt2ΔH=aRa×LΔH=m2(M+m)×L(theblockmovesupadistancemL2(M+m))
Commented by Tinkutara last updated on 27/Oct/17
Commented by mrW1 last updated on 27/Oct/17
that depents on which direction is  defined as positive. in your book the  +ve direction for the block is upwards,  so there is no “−”. i defined +ve direction  downwards, so i have “−”. the result  is the same, i.e. the block moves a  distance ((mL)/(2(M+m))) upwards.
thatdepentsonwhichdirectionisdefinedaspositive.inyourbookthe+vedirectionfortheblockisupwards,sothereisno.idefined+vedirectiondownwards,soihave.theresultisthesame,i.e.theblockmovesadistancemL2(M+m)upwards.
Commented by Tinkutara last updated on 28/Oct/17
OK Thanks.
OKThanks.

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