In Taylor's book Classical Mechanics, pg. 259, he works through the following example: Consider the following block and wedge system: The block ($m$) is free to slide on the wedge and the wedge (mass $M$) can slide on the horizontal table, both with negligible friction.
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I am simply trying to write down the lagrangian in these generalized coordinates, but I run into a difficulty when he states the following: 
What I don't get is why the $x$-component of velocity is $\dot{q_1} \cos(\alpha)+\dot{q_2}$. Shouldn't the answer be $\dot{q_1}\cos(\alpha)-\dot{q_2}$, since the $x$-components of $m$ will be in the positive $x$, and the $x$-component of velocity of $M$ will be in the negative $x$ direction?
Best Answer
Let's look at the two options:
Taylor's expression: $$\dot{q_1} cos(\alpha)+\dot{q_2}$$
Your expression: $$\dot{q_1} cos(\alpha)-\dot{q_2}$$
Since the wedge will be moving to the left, $\dot{q_2}$ will be negative. We would expect $\dot{q_1}$ to be smaller as a result. Which of those equations gives us that result?
If you plug a negative number into the first equation for $\dot{q_2}$, you are adding a negative number to the x-component, making it smaller. That matches what we expect.
If you plug a negative number into the second equation for $\dot{q_2}$, you are subtracting a negative number from the x-component, which would make it larger. That is the opposite of what we expect.
Taylor's expression is the right one.
I've found this sort of thing confusing before. If you add the components, then the components are negative, you get the "opposite" result that you are expecting.