Solve Block Wedge Problem: Find Speed of Triangle Block

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Homework Help Overview

The problem involves a block of mass m placed on a triangular block of mass M, which is on a horizontal surface. The objective is to find the speed of the triangular block when the smaller block reaches the bottom end, assuming frictionless surfaces.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for clearer definitions of variables and equations presented in the original post. There are attempts to rearrange equations and clarify the relationship between the blocks' accelerations. Some participants express confusion over the handwritten solution and suggest using LaTeX for clarity.

Discussion Status

The discussion is ongoing, with participants providing feedback on the clarity of the original post's equations and suggesting improvements. There is no explicit consensus on the correctness of the solution, but some participants indicate agreement with a proposed answer while others seek further clarification.

Contextual Notes

Participants note difficulties in interpreting handwritten equations and emphasize the importance of clear variable definitions. There is mention of a "book" answer, but the specifics of this answer are not detailed in the discussion.

AnwaarKhalid
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Homework Statement


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,

Homework Equations

The Attempt at a Solution


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

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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?
 
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?
 
haruspex said:
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?
Yes, that is acc. of the triangle as well as the block.
 
endoftimes said:
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?
 

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AnwaarKhalid said:
Yes, that is acc. of the triangle as well as the block.
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.
 
Common horizontal acc. of the blocks= (M+m)a= ma_x
 
haruspex said:
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.
I'm sorry I do not know how to type using latex. Hope , these two images are clear enough.
 

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AnwaarKhalid said:
I'm sorry I do not know how to type using latex. Hope , these two images are clear enough.
I said preferably with LaTeX. It is not mandatory. Typing using subscripts and superscripts (see X2 and X2 in the toolbar) is good enough.
 
  • #10
haruspex said:
I said preferably with LaTeX. It is not mandatory. Typing using subscripts and superscripts (see X2 and X2 in the toolbar) is good enough.
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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