1. Given the tetrahedron $S-ABC$ with the projection $H$ of vertex $S$ on the base being the orthocenter of $\triangle ABC$, $BC=2$, $SB=SC$, and the dihedral angle between the side face $SBC$ and the base is $60^{\circ}$, find the volume of the tetrahedron.
1. Let $A H$ intersect $B C$ at $D$, then $A D \perp B C$, plane $S A D \perp B C$, $\angle A D S$ is the plane angle of the dihedral angle formed by the side face $S B C$ and the base, $\angle A D S=60^{\circ}, S H=\sqrt{3} \cdot D H$.

To find $S H \cdot A D$, i.e., $\sqrt{3} D H \cdot A D$, the problem is then reduced to a plane geometry problem. Noting the equilateral $\triangle A B C$, from $\angle B A D=\angle D A C=\angle D B H$ we get $D B^{2}=D H \cdot A D$. Therefore,
$$
V_{S A B C}=\frac{1}{6} \cdot \sqrt{3} D B^{2} \cdot B C=\frac{\sqrt{3}}{3} \text {. }
$$

Note: The $S H$ and $A D$ in the problem are not uniquely determined, but their product is a constant, which is indeed what we are looking for.