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In-the-figure-shown-below-all-surfaces-are-smooth-strings-and-pulley-are-ideal-If-the-wedge-is-moving-with-acceleration-a-towards-the-right-then-the-acceleration-of-the-block-with-respect-to-the-w




Question Number 24301 by Tinkutara last updated on 15/Nov/17
In the figure shown below, all surfaces  are smooth, strings and pulley are ideal.  If the wedge is moving with acceleration  a towards the right, then the acceleration  of the block with respect to the wedge  at that instant is
Inthefigureshownbelow,allsurfacesaresmooth,stringsandpulleyareideal.Ifthewedgeismovingwithaccelerationatowardstheright,thentheaccelerationoftheblockwithrespecttothewedgeatthatinstantis
Commented by Tinkutara last updated on 15/Nov/17
Commented by ajfour last updated on 15/Nov/17
let length of string from top of  wedge to block m be y and from  wall pulley to wedge be x. Then  y+x+(√(h^2 +x^2 )) =l  ⇒   v_(rel) +v_M +((xv_M )/( (√(h^2 +x^2 )))) =0  or  a_(rel) +a+(((v_M ^2 +xa)(√(h^2 +x^2 ))−xv_M (((xv_M )/( (√(h^2 +x^2 ))))))/(h^2 +x^2 )) =0  so  a_(rel) =−a−(((v_M ^2 +xa)(h^2 +x^2 )−(xv_M )^2 )/((h^2 +x^2 )(√(h^2 +x^2 ))))        =−a−((h^2 (v_M ^2 +xa)+x^3 a)/((h^2 +x^2 )(√(h^2 +x^2 ))))       =−a−((h^2 v_M ^2 )/((h^2 +x^2 )^(3/2) ))−((xa)/( (√(h^2 +x^2 ))))  if only system has started now  from rest and initial relation  is sought for, we can have    v_M =0       a_(rel)  = −a(1+cos θ_0 )  the minus sign is there because   ((dx/dt)) is −ve here, and ((dy/dt)) is +ve  if wedge accelerates towards right.  so  ((d^2 x/dt^2 )) = a    (i have assumed)  but according to question it  is ((d^2 x/dt^2 )) =−a_(given in question)  .  given answer must be        a_(rel)  =a(1+cos θ) .
letlengthofstringfromtopofwedgetoblockmbeyandfromwallpulleytowedgebex.Theny+x+h2+x2=lvrel+vM+xvMh2+x2=0orarel+a+(vM2+xa)h2+x2xvM(xvMh2+x2)h2+x2=0soarel=a(vM2+xa)(h2+x2)(xvM)2(h2+x2)h2+x2=ah2(vM2+xa)+x3a(h2+x2)h2+x2=ah2vM2(h2+x2)3/2xah2+x2ifonlysystemhasstartednowfromrestandinitialrelationissoughtfor,wecanhavevM=0arel=a(1+cosθ0)theminussignistherebecause(dxdt)isvehere,and(dydt)is+veifwedgeacceleratestowardsright.so(d2xdt2)=a(ihaveassumed)butaccordingtoquestionitis(d2xdt2)=agiveninquestion.givenanswermustbearel=a(1+cosθ).
Commented by Tinkutara last updated on 15/Nov/17
Thank you very much Sir!  Your solution is right.
ThankyouverymuchSir!Yoursolutionisright.
Commented by Tinkutara last updated on 15/Nov/17

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