Updated spacing between @ latex expressions

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Centripetal acceleration associated with circular rotation:
$$a=\frac{v^2}{R}$$
From the Newton's second law:
−$$\frac{mv^2}{R}=2mg\sin\alpha-N_x\cos\alpha-N_y\sin\alpha$$
+$$\frac{mv^2}{R}=2mg\sin\alpha -N_x\cos\alpha -N_y\sin\alpha$$
Law of Conservation of Energy
$$mgR=mgh+\frac{mv^2}{2}$$
$$h=R\sin\alpha$$
−$$\frac{mv^2}{R}=2mg(1-\sin\alpha)$$
−$$N_y\sin\alpha+N_x\cos\alpha=(4\sin\alpha-2)mg$$
+$$\frac{mv^2}{R}=2mg(1-\sin\alpha )$$
+$$N_y\sin\alpha +N_x\cos\alpha =(4\sin\alpha -2)mg$$
Along the axis $OZ_\perp$ "The Ball System":
−$$N_y\cos\alpha=N_x\sin\alpha+mg\cos\alpha$$
+$$N_y\cos\alpha =N_x\sin\alpha +mg\cos\alpha$$
$$N_y=mg+N_x\tan\alpha$$
$N_y$ found from the axis $OZ_\perp$ we substitute into $N_y$ found from the axis $OZ_\parallel$
−$$mg\sin\alpha\cos\alpha+N_x\sin^2 \alpha+N_x\cos^2 \alpha=2mg(2\sin\alpha-2)\cos\alpha$$
−$$\frac{N_x}{mg}=4\sin\alpha\cos\alpha-2\cos\alpha-\sin\alpha\cos\alpha$$
−$$\boxed{N_x=(3\sin\alpha-2)\cos\alpha \cdot mg}$$
+$$mg\sin\alpha\cos\alpha +N_x\sin^2 \alpha +N_x\cos^2 \alpha =2mg(2\sin\alpha -2)\cos\alpha$$
+$$\frac{N_x}{mg}=4\sin\alpha\cos\alpha -2\cos\alpha -\sin\alpha\cos\alpha$$
+$$\boxed{N_x=(3\sin\alpha -2)\cos\alpha\cdot mg}$$
When $N_x \leq0$, the shoe will break away from the vertical wall and will no longer act on it:
$$N_x=0$$
Condition of detachment:
−$$(3\sin\alpha-2)\cos\alpha \leq0$$
−$$3\sin\alpha-2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$
+$$(3\sin\alpha -2)\cos\alpha\leq0$$
+$$3\sin\alpha -2\leq0 \Leftrightarrow \boxed{\sin\alpha\leqslant2/3}$$
</p>
<h4>Answer</h4>
<p>
− $$F=mg\cos\alpha(3\sin\alpha-2)\mathrm{~with~}\sin\alpha\geqslant2/3;$$
+ $$F=mg\cos\alpha (3\sin\alpha -2)\mathrm{~with~}\sin\alpha\geqslant2/3;$$
$$F=0\mathrm{~with~}\sin\alpha\leqslant2/3.$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
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