= Solution
Use locally bed-tangent <drag force>, consistent with interpreting $\tau$ as bed <shear stress>. Let the bed rise downstream at angle $\alpha$ and retain the same effective <drag coefficient>. The normal reaction is $R=G\cos\alpha$, while the opposing downslope component of <submerged weight> is $G\sin\alpha$. At impending upslope motion,
$$
F_D=G\sin\alpha+\mu_sG\cos\alpha.
$$
Thus the <inclined-bed sediment threshold> is
$$
\boxed{\Theta_{\mathrm{th},0}(\alpha)=\frac4{3C_D}(\mu_s\cos\alpha+\sin\alpha)
=\Theta_{\mathrm{th},0}\left(\cos\alpha+\frac{\sin\alpha}{\mu_s}\right)}.
$$
For $|\alpha|\ll1$,
$$
\boxed{\tau_{\mathrm{th}}(\alpha)=\tau_{\mathrm{th},0}\left(1+\frac\alpha{\mu_s}\right)+O(\alpha^2)}.
$$
An uphill slope increases the <sediment entrainment threshold>; a downhill slope lowers it, until spontaneous gravity-driven motion invalidates the resting-bed model.
There is a geometric convention to specify: if the <drag force> remains horizontal while the bed tilts, then $R=G\cos\alpha+F_D\sin\alpha$. The corresponding <force balance> gives instead
$$
\Theta_{\mathrm{th}}^{\mathrm{horizontal\ drag}}
=\frac4{3C_D}\frac{\mu_s\cos\alpha+\sin\alpha}{\cos\alpha-\mu_s\sin\alpha}
=\Theta_{\mathrm{th},0}\left[1+(\mu_s+\mu_s^{-1})\alpha+O(\alpha^2)\right].
$$
The subsequent bed <shear stress> analysis uses the first, locally tangent convention. The two interpretations should not be mixed.
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