# Block on a wedge

1. Mar 11, 2015

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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2. Mar 11, 2015

### Staff: Mentor

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

3. Mar 11, 2015

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?