Find Constraint Equation for Block on Wedge

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Discussion Overview

The discussion focuses on deriving a constraint equation that relates the acceleration of a block on a wedge to the acceleration of the wedge itself. The conversation involves exploring different definitions of variables and their implications on the equation, with a particular emphasis on the geometry involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes the equation $$h - y = (x - X) \tan{\theta}$$ as a starting point for relating the block's and wedge's accelerations.
  • Another participant questions the validity of this equation if ##X## is defined as the distance to the center of mass of the wedge, suggesting it should be $$h - y = (X - x) \tan{\theta}$$ instead.
  • A subsequent reply challenges the redefinition of ##X##, arguing that the sides of the triangle must remain consistent, and suggests a modified equation if ##X_{cm} = X + a##.
  • One participant acknowledges a previous misunderstanding, stating that since ##X## and ##X_{cm}## differ by a constant, their second derivatives should be equal, indicating a simplification in the trigonometric approach.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulation of the constraint equation based on the definitions of ##X## and ##X_{cm}##. The discussion remains unresolved as participants explore these competing definitions and their implications.

Contextual Notes

The discussion highlights potential limitations in the definitions of variables and the assumptions made regarding the geometry of the problem. There is an emphasis on the need for consistent definitions to ensure the validity of the derived equations.

PFuser1232
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Suppose we want to find the constraint equation relating the acceleration of a block on a wedge to the acceleration of the wedge (see attached images). We can write:
$$h - y = (x - X) \tan{\theta}$$
What if ##X## was measured to the centre of mass of the wedge for instance? Wouldn't the correct equation be:
$$h - y = (X - x) \tan{\theta}$$
 

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MohammedRady97 said:
Suppose we want to find the constraint equation relating the acceleration of a block on a wedge to the acceleration of the wedge (see attached images). We can write:
$$h - y = (x - X) \tan{\theta}$$
I see how you got this and it makes sense.

MohammedRady97 said:
What if ##X## was measured to the centre of mass of the wedge for instance? Wouldn't the correct equation be:
$$h - y = (X - x) \tan{\theta}$$
I do not see how you got that. You still need the sides of the same triangle, you've just redefined what X is. So you need to rewrite your first equation using your new X. For example, if ## X_{cm} = X + a##, then your formula would become ##h - y = (x - X_{cm} + a) \tan{\theta}##.
 
Doc Al said:
I see how you got this and it makes sense.I do not see how you got that. You still need the sides of the same triangle, you've just redefined what X is. So you need to rewrite your first equation using your new X. For example, if ## X_{cm} = X + a##, then your formula would become ##h - y = (x - X_{cm} + a) \tan{\theta}##.

I just realized the error in my reasoning. This makes sense now. Since ##X## and ##X_{cm}## differ by a constant, their second derivatives should be equal. We chose ##X## to be the distance to the edge of the wedge to simplify the trigonometry required, right?
 
Last edited:
MohammedRady97 said:
We chose XX to be the distance to the edge of the wedge to simplify the trigonometry required, right?
Right.
 
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