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There is a square board with two nails in corner A and B. We apply a torque $M$ to the square board. Nail A can move freely on a track only in the $x$ direction and B only in the $y$ direction. The question is to calculate the normal forces from the tracks to the nails. My problem is that I can't seem to find a way to calculate mathematically the accelerations in points A and B. The solution just gives them as stated below like it is something obvious, (but it isn't for me):

$\textbf{a}_A = l \alpha \textbf{e}_x$

$\textbf{a}_B = l \alpha \textbf{e}_y$

where $\alpha$ is the angular acceleration of the board.

How can I see that from a formula?

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First you need to figure out where the center of rotation is. Because of the contraints on A and B, the only point at which the tangent vectors are $\hat{e}_x$ and $\hat{e}_y$ is the upper left corner of the board. Since the distance to the center of rotation is then $l$ for A and B, by definition of angular acceleration, A and B accelerate with $\vec{a}_A = l \alpha \hat{e}_x$ and $\vec{a}_B = l \alpha \hat{e}_x$.

Now of course you still need to figure out $\alpha$, but I take it from your question that this part is clear to you.

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  • $\begingroup$ What do you mean with "the only point at which the tangent vectors are $e_x$ and $e_y$ is the upper left corner of the board" $\endgroup$
    – DoubleOseven
    Oct 23, 2015 at 12:41
  • $\begingroup$ Because A and B are constrained to move only in the $x$- and $y$-direction, respectively, you need to find a point at which the tangent vectors of the rotation fulfill these constraints, otherwise the board can not move. $\hat{e}_x$ and $\hat{e}_y$ are unit-vectors in the $x$- and $y$-directions, sorry if that was not clear. $\endgroup$
    – Rich
    Oct 23, 2015 at 12:44

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