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The-system-is-released-from-rest-All-surfaces-are-smooth-Find-the-angle-at-which-the-acceleration-of-wedge-is-maximum-given-M-m-1-2-




Question Number 23617 by Tinkutara last updated on 02/Nov/17
The system is released from rest. All  surfaces are smooth. Find the angle θ at  which the acceleration of wedge is  maximum. (given (M/m) = (1/2))
Thesystemisreleasedfromrest.Allsurfacesaresmooth.Findtheangleθatwhichtheaccelerationofwedgeismaximum.(givenMm=12)
Commented by Tinkutara last updated on 02/Nov/17
Commented by ajfour last updated on 02/Nov/17
mgcos θ−N=mAsin θ      ...(i)  (A being acc. of wedge)  and   Nsin θ=MA               ....(ii)  so,  A=((mgsin θcos θ)/(M+msin^2 θ))  given m=2M , therefore     A=((2gsin θcos θ)/(1+2sin^2 θ))  =((2gtan θ)/(1+3tan^2 θ))        =((2g)/(((1/(tan θ))+3tan θ)))  For maximum A,   (1/(tan θ))=3tan θ  ⇒   tan θ=(1/( (√3)))  or   𝛉=30° .
mgcosθN=mAsinθ(i)(Abeingacc.ofwedge)andNsinθ=MA.(ii)so,A=mgsinθcosθM+msin2θgivenm=2M,thereforeA=2gsinθcosθ1+2sin2θ=2gtanθ1+3tan2θ=2g(1tanθ+3tanθ)FormaximumA,1tanθ=3tanθtanθ=13orθ=30°.
Commented by Tinkutara last updated on 03/Nov/17
Thank you very much Sir!
ThankyouverymuchSir!
Answered by Tinkutara last updated on 03/Nov/17
Commented by Tinkutara last updated on 03/Nov/17
Commented by Tinkutara last updated on 03/Nov/17
Commented by Physics lover last updated on 03/Nov/17
How did Mr Ajfour find the  minimum value of that trigonometric  equation.?  [Cot θ + 3 tan θ]
HowdidMrAjfourfindtheminimumvalueofthattrigonometricequation.?[Cotθ+3tanθ]
Answered by Physics lover last updated on 02/Nov/17
What about this one :  a_(wedge ) = (((mg×Sin θ × Cos θ))/(M+m))   as there is no force in horizontal  direction ⇒ horizontal acceleratuon  of the com of system should be  zero.And we can use the above  formula.   [a_(wedge) ]_(max ) ⇒[ Sin θ Cos θ ]_(max)    ⇒ θ = 45°   ????  what is wrong in this one ?
Whataboutthisone:awedge=(mg×Sinθ×Cosθ)M+masthereisnoforceinhorizontaldirectionhorizontalacceleratuonofthecomofsystemshouldbezero.Andwecanusetheaboveformula.[awedge]max[SinθCosθ]maxθ=45°????whatiswronginthisone?
Commented by ajfour last updated on 02/Nov/17
how mgsin θcos θ ?
howmgsinθcosθ?
Commented by ajfour last updated on 02/Nov/17
what would you say for acc.  of block, what is it, at what  angle with horizontal?  acc. of block is not mgsin θ   down the incline direction  from ground,  if observer observes  motion of block and wedge ,  both from ground frame.
whatwouldyousayforacc.ofblock,whatisit,atwhatanglewithhorizontal?acc.ofblockisnotmgsinθdowntheinclinedirectionfromground,ifobserverobservesmotionofblockandwedge,bothfromgroundframe.

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