= Solution
Let $\Omega$ be the Earth's <angular velocity> magnitude. Use the local east, north and upward unit <vectors> $\mathbf e_E,\mathbf e_N,\mathbf e_U$, with $\mathbf e_E\times\mathbf e_N=\mathbf e_U$. At latitude $\lambda$, the Earth's <angular velocity> and the train's ground-relative <velocity> are
$$
\boldsymbol\Omega=\Omega\cos\lambda\,\mathbf e_N+\Omega\sin\lambda\,\mathbf e_U,\qquad \mathbf v'=V\mathbf e_N.
$$
Thus the east component of the <Coriolis force> is
$$
-2m\boldsymbol\Omega\times\mathbf v'=2m\Omega V\sin\lambda\,\mathbf e_E.
$$
For a north–south track, the train has no eastward <acceleration> relative to Earth. Its path curvature lies in the local north–vertical plane. The <centrifugal acceleration> and <Newtonian gravity> also lie in that plane, and the Earth's constant <angular velocity> gives no <Euler acceleration>. Hence the sideways <force> from the track must cancel the eastward <Coriolis force>:
$$
\boxed{\mathbf F_{\rm track,side}=-2m\Omega V\sin\lambda\,\mathbf e_E.}
$$
\b[Its magnitude is $2m\Omega V\sin\lambda$, and its direction is west] in the Northern hemisphere. The eastward tendency is the apparent <Coriolis force>; the requested track <force> has the opposite direction.
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