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Block on wedge which is not fixed.Find minimum μ to keep the wedge at rest.

  1. Sep 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a wedge of mass m whose upper surface is a quarter circle and is smooth. The lower part is rough. A small block of same mass is kept at the topmost position and the initial position is shown in figure. The block slides down the wedge.

    Find the minimum value of coefficient of friction between floor and lower surface of wedge so that the wedge remains stationery during the block's motion on it.
    9f0d7815b5.4122fa5d51.Xk4w50.png

    2. Relevant equations



    3. The attempt at a solution
    I tried to find the angle 'theta' at which horizontal force on the wedge is maximum.
    9f0d7815b5.4122fa5d51.Xk4w50.png
    By conservation of energy
    ##mgr=mgr(1-sin\theta )+\frac { m{ v }^{ 2 } }{ 2 } ##

    So ##{ v }^{ 2 }=2grsin\theta ##

    Horizontal force on the wedge(Fx) is ##\frac { m{ v }^{ 2 } }{ r } cos\theta ##


    So ##Fx=\frac { m }{ r } 2grsin\theta cos\theta ##

    ##Fx=mgsin2\theta##

    For Fx to be maximum ##d(Fx)/d\theta=0##

    so
    ##mg\frac { d }{ d\theta } sin2\theta =0##

    ##\theta ##=45

    So ##{ v }^{ 2 }=\sqrt { 2 } \ gr ##
    9f0d7815b5.4122fa5d51.Xk4w50.png

    Normal force on the wedge = ##mg+mg+\frac { m{ v }^{ 2 } }{ 2 } sin\theta ##=3mg

    Horizontal force on wedge= ##mgsin2\theta##=mg

    As wedge should remain at rest so
    ## 3\mu mg=mg##
    So ## \mu =1/3##

    But this is not correct.Where did I go wrong?
     
  2. jcsd
  3. Sep 26, 2014 #2

    ehild

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    Are you sure? The force of friction depends on the normal force between the ground and the wedge, and it is not equal to the weight of the wedge.

    The horizontal force of the wedge is the horizontal component of the normal force between the small block and the wedge. The normal force is not equal to the centripetal force. Do not forget agravity.

    ehild
     
  4. Sep 26, 2014 #3
    Untitled.png

    From figure 2

    ##N=\frac { m{ v }^{ 2 } }{ R } +mgcos(90-\theta )##

    or ##N=\frac { m{ v }^{ 2 } }{ R } +mgsin\theta ##

    Horizontal force in wedge ##Fx=Ncos\theta =\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) cos\theta ##

    From figure 3

    ##R=mg+Nsin\theta ##

    ##R=mg+\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) sin\theta ##

    Are these equations correct?
    Is my expression of v2 in #post1 correct?
     
  5. Sep 26, 2014 #4

    ehild

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    And you got already that ##v^2=2gR\sin(\theta)##, so N= ??

    Horizontal force in wedge ##Fx=Ncos\theta =\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) cos\theta ##

    Certainly not, R is the radius, a length, it can not be force ...
     
  6. Sep 26, 2014 #5
    ##N=\frac { m2gR\sin (\theta ) }{ R } +mgsin\theta =3 mg\sin (\theta )##

    ## Fx=3 mg\sin (\theta )cos\theta ##......(1)

    The 'R' in the RHS is not radius.It is reactionary force on wedge.Sorry I should have choose other variable for it, let it be A.
    So
    ##A=mg+Nsin\theta ##

    ##A=mg+\left( \frac { m{ v }^{ 2 } }{ R } +mgsin\theta \right) sin\theta ##
    Please consider reactionary force on wedge in figure(3) in #post 3 be A.
     
  7. Sep 26, 2014 #6
    On differentiating eq(1) wrt to ##\theta## and equating it to 0 I got ##\theta=45## maximum value of ##Fx=3mg/2##
     
  8. Sep 26, 2014 #7

    ehild

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    You need maximum coefficient of friction, not the maximum horizontal force. The force of friction also depends on the normal force and on theta.

    And substitute for v2 also in the expression of A.
     
  9. Sep 26, 2014 #8
    ##A=mg+3mg{ sin }^{ 2 }\theta ##
    I thought by equating ## \mu A=Fx## I can solve for ##\mu##???
     
  10. Sep 26, 2014 #9

    ehild

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    Solve for mu and find the smallest value.
     
  11. Sep 26, 2014 #10
    Should ##\theta=45##?
     
  12. Sep 26, 2014 #11

    ehild

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    No, why? Solve for mu in terms of theta, and find the minimum value of mu.
     
  13. Sep 26, 2014 #12
    I got ##\mu=3/5##
    Oh! Sorry.
    Putting value in ## \mu A=Fx##

    I got ##μ(mg+3mg{ sin }^{ 2 }\theta )=3mgsin\theta cos\theta ##

    or ##μ=\frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta } ##

    In order to have ##\mu## maximum

    ##\frac { d }{ d\theta } \left( \frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta } \right) =0##

    or ##\frac { d }{ d\theta } \left( \frac { sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta } \right) =0##


    or ##\frac { (1+3{ sin }^{ 2 }\theta )\frac { d }{ d\theta } (sin\theta cos\theta )-(sin\theta cos\theta )\frac { d }{ d\theta } (1+3{ sin }^{ 2 }\theta ) }{ { (1+3{ sin }^{ 2 }\theta ) }^{ 2 } } =0##


    or ##(1+3{ sin }^{ 2 }\theta )\frac { d }{ d\theta } (sin\theta cos\theta )-(sin\theta cos\theta )\frac { d }{ d\theta } (1+3{ sin }^{ 2 }\theta )=0##

    or ##(1+3{ sin }^{ 2 }\theta )cos2\theta -(sin\theta cos\theta )(3sin2\theta )=0##

    or ##2(1+3{ sin }^{ 2 }\theta )cos2\theta =2(sin\theta cos\theta )(3sin2\theta )##

    or ##2(1+3{ sin }^{ 2 }\theta )cos2\theta =3{ sin }^{ 2 }2\theta ##

    or ##2(1+3{ sin }^{ 2 }\theta )(1-2{ sin }^{ 2 }\theta )=3{ sin }^{ 2 }2\theta ##

    How to solve for ##\theta##?
     
  14. Sep 26, 2014 #13

    ehild

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    ##\mu## need to be minimum.

    Change to double angles. You have turned ##\sin(\theta)\cos(\theta)## to ##0.5 \sin(2\theta)## already, but remember the very important formulas

    ##\cos^2(\theta)= \frac{1+\cos(2\theta )}{2}##

    and

    ##\sin^2(\theta)= \frac{1-\cos(2\theta) }{2}##

    Store them in your memory!

    ehild
     
  15. Sep 27, 2014 #14
    That was a typo.


    Is ##\theta ={ cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \times \frac { 1 }{ 2 } ##??
     
  16. Sep 27, 2014 #15
    So ##\theta =26.565##

    As ##μ=\frac { 3sin\theta cos\theta }{ 1+3{ sin }^{ 2 }\theta } ##

    or ##μ=\frac { 3sin2\theta }{ 2(1+3{ sin }^{ 2 }\theta ) } =\frac { 3\times 0.8 }{ 3.199 } =0.75##

    ##\mu=0.75##
    Got correct answer :D.
    Thank you ehild.
     
  17. Sep 27, 2014 #16

    ehild

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    Splendid !

    ehild
     
  18. Sep 27, 2014 #17
    Thank you ehild. I could't have solved it without your help.:)
     
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