First, Navier-Stokes governs the fluid in your setup. So, anything apart from the fluid will be an external force in N-S equation.
Body-force means an external force that applies in the bulk of the fluid, like gravity or a magnetic force.
Interaction with a "body", as a wing, which is external to the fluid domain, is done through boundary conditions : the integral of the total stress along the normal to the boundary with the wing will give you the force exerted by the wing on the fluid.
I agree with user3823992 that it was incorrect to neglect the pressure differential. With the steady sinusoidal body force that's given, it's basically a hydrostatics problem with the pressure differential balancing the body force. Consider the Navier-Stokes momentum equation:
$$
\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \nabla)\mathbf{v}=-\frac{\nabla p}{\rho}+\mathbf{f} +\nu \nabla^2 \mathbf{v}
$$
If we assume the velocity $\mathbf{v}$ is zero then it reduces to the hydrostatic case, where:
$$
\frac{\nabla p}{\rho}=\mathbf{f}
$$
$\mathbf{f}$ only has a component in the s-direction, therefore so will $\nabla p$:
$$
\begin{eqnarray*}
\frac{dp}{ds} \cdot \hat s&=&\rho q \sin(s) \cdot \hat s\\
\int_{p_0}^pdp&=&\rho q \int_0^s \sin(s) ds\\
p-p_0&=&\rho q (1- \cos (s))
\end{eqnarray*}
$$
($p_0$ is just an arbitrary reference pressure that may have been present before the force was applied).
$\mathbf{v}=0$ obviously satisfies the continuity equation, although I think any other solution with a constant $\mathbf{v}$ would also satisfy the equation. This would just be bulk fluid motion that was present before the force was applied, and would tend to zero in steady-state if viscous drag on the walls is included.
In the case where the tube is closed like a torus, the flow is still governed by the Navier-Stokes equations. The momentum equation in polar coordinates (r, $\theta$, z) can be reduced to:
$$
\begin{eqnarray*}
\theta: f_\theta&=&-\nu (\frac{1}{r} \frac{\partial}{\partial r}(r \frac{\partial V_\theta}{\partial r})+\frac{\partial^2 V_\theta}{\partial z^2}-\frac{V_\theta}{r^2})\\
r: \frac{\partial p}{\partial r}&=&\rho \frac{V^2_\theta}{r}
\end{eqnarray*}
$$
The only component of velocity is in the $\theta$ (circumferential) direction. The first line is the body force balanced by the wall friction and the $\frac{\partial p}{\partial r}$ in the second line is necessary to provide the centripetal force for the curved streamlines. However, this is now a 2-dimensional PDE and I think it's pretty unlikely that you'd be able to integrate or find a simple function to satisfy it - to find the velocity profile you would probably have to resort to CFD at this point.
The Poiseuille equation has a nice, straightforward solution because it is axisymmetric and effectively 1-dimensional.
Best Answer
Firstly, i want to set the coordinate system so that there is no confusion:
In such a coordinate system, the acceleration due to gravity $\boldsymbol{g}$ is expressed as:
$$\boldsymbol{g}=\left(g_x,g_y\right)=\left(g\sin\theta,g\cos\theta\right)$$
where $g$ is the gravitational acceleration constant and $\theta$ is the angle of the incline.
Secondly, these typical text-book questions on velocity profiles for inclined planes always have a number of implicit assumptions:
With these assumptions the Navier-Stokes equations reduce to:
$$0=-\frac{1}{\rho}\partial_xp + \nu\partial_y^2u_x + g_x \qquad 0=-\frac{1}{\rho}\partial_yp + g_y$$
Note that the continuity equation is satisfied identically with the above assumptions.
Consider the ways in which a pressure gradient may form:
Neither of these cases apply here as no applied pressure is specified and the in/out-flow boundaries are open (otherwise no flow) such that a hydrostatic pressure cannot be generated. Conclusion the pressure along the incline wall must vanish.
Actually, since $g_y\neq0$, there exists a pressure gradient in the y-direction and it is purely hydrostatic:
$$\partial_yp = \rho g_y = \rho g \cos\theta$$
If this may seem unintuitive, consider the case where the incline is perfectly horizontal ($\theta=0$). In that case there is no flow ($g_x=0$) and the fluid is simply resting on the wall exerting a hydrostatic pressure due to gravity ($g_y=g$).