Find Constraint Equation for Block on Wedge

[PFuser1232]

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:
$$h - y = (X - x) \tan{\theta}$$

 

[Doc Al]

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.

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 $h - y = (x - X_{cm} + a) \tan{\theta}$.

 

[PFuser1232]

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?

 

[Doc Al]

We chose XX to be the distance to the edge of the wedge to simplify the trigonometry required, right?

Right.