I'm working on an unassessed course problem containing this diagram:
The solution booklet uses $g\sin(\pi/6)=g/2$ for the component of gravity acting in the $x$ direction (which I'll denote $g_x$). By my reasoning, \begin{align} & \begin{aligned}[t] \sin(\pi/6) & = \frac{\text{opposite}}{\text{hypotenuse}} \\ & = \frac{g}{g_x} \end{aligned} \\ \therefore \; & \begin{aligned}[t] g_x & = \frac{g}{\sin(\pi/6)} \\ & = \frac{g}{1/2} \\ & = 2g. \end{aligned} \end{align} That seems unphysical. Could someone point out what I'm doing wrong?
The coordinate system $(x,y)$ has w.r.t. the inertial system the basis unity vectors
$$ \cases{ \hat e_x = (\cos\left(\frac{\pi}{6}\right),-\sin\left(\frac{\pi}{6}\right))'\\ \hat e_y = (\sin\left(\frac{\pi}{6}\right),\cos\left(\frac{\pi}{6}\right))'.\\ } $$
Then in the new referential we have
$$ \vec g = |g|(0,-1)' = a_1\hat e_x+a_2\hat e_y. $$
To know $a_1$ it suffices to make \begin{align} \vec g\cdot\hat e_x & = |g|(0,-1)'\cdot\left(\cos\left(\frac{\pi}{6}\right),-\sin\left(\frac{\pi}{6}\right)\right)' \\ & = |g|\sin\left(\frac{\pi}{6}\right) \\ & = \frac 12 |g|.\end{align}