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Block Wedge problem

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  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown. Assume frictionless surfaces. Find the speed of triangular block when the small block reaches the bottom end,

    2. Relevant equations


    3. The attempt at a solution
    I've uploaded my attempt at a solution. Please, tell me where I did wrong.
     

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  2. jcsd
  3. Dec 1, 2015 #2
    I think you might actually have the right answer, but you need to rearrange algebraically to solve for whatever it is you are supposed to have. I'm sorry it is hard to make out the photograph, but do you know what it is supposed to be?
     
  4. Dec 1, 2015 #3

    haruspex

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    You've lost me with your very first equation. You need to define your variables if others are to follow your logic.
    Is that Acom? Does that stand for the acceleration of the common mass centre of triangle+block? As a vector or in a specific direction?
     
  5. Dec 1, 2015 #4
    Yes, that is acc. of the triangle as well as the block.
     
  6. Dec 1, 2015 #5
     

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  7. Dec 1, 2015 #6

    haruspex

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    That only answered one of my questions. Anyway, I agree with the 'book' answer.
    Reading an image of your handwritten working is too much of a struggle. Please take the trouble to type it in, preferably using LaTeX.
     
  8. Dec 1, 2015 #7
    Common horizontal acc. of the blocks= (M+m)a= ma_x
     
  9. Dec 1, 2015 #8
    I'm sorry I do not know how to type using latex. Hope , these two images are clear enough.
     

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  10. Dec 1, 2015 #9

    haruspex

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    I said preferably with LaTeX. It is not mandatory. Typing using subscripts and superscripts (see X2 and X2 in the toolbar) is good enough.
     
  11. Dec 1, 2015 #10
    Common acc. of the masses = (M+m)acom=max
    ax= gsinxcosx
    Therefore, acom=mgsinxcosx\(M+m)
    Now,
    Tanx= h\BC
    BC=h\tanx
    Let the triangular block be displaced by x when the smaller block reaches the bottom end
    Since there is no external force
    Therefore,
    Rcomi=Rcomf
    M(BC/2) = M(BC/2 - x) + m(BC -x)
    which gives,
    x= mhcosx/(M+m)sinx
    Now, using
    v2 - u2= 2as
    v={ (2m2ghcos2x)/(M+m)2 }1/2
     

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  12. Dec 2, 2015 #11

    jtbell

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