The divergence is a vector operator. This simply means that it is a differential operator that acts only on vectors. In this particular case,
$$
{\rm div}\equiv\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}
$$
Which, assuming an implicit summation,
$$
{\rm div}\equiv\frac{\partial}{\partial x_i}\hat{x}_i\tag{1}
$$
Since the velocity field is a vector,
$$
\mathbf{u}:=(u,\,v,\,w)=u\hat{x}+v\hat{y}+w\hat{z}\tag{2}
$$
then the divergence of this is the dot product of the velocity and the vector operator:
\begin{align}
{\rm div}\mathbf u&=\left(\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}\right)\cdot\left(u\hat{x}+v\hat{y}+w\hat{z}\right)\\
&=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}\\
&\equiv\frac{\partial u_i}{\partial x_i}
\end{align}
where $i$ is an index (1, 2, 3) that is implicitly summed over.
In the case of the Navier-Stokes equations, we have
$$
\left(\frac{\partial}{\partial t}+\mathbf u\cdot{\rm div}\right)\mathbf u=\frac1\rho\nabla p+\nu\nabla^2\mathbf u + \frac1\rho\mathbf f
$$
where the second term on the left is the one you are concerned with. First, you must take the dot-product of the velocity with the divergence operator (e.g., (2) dotted with (1)):
$$
\mathbf u\cdot{\rm div}=u_j\frac{\partial}{\partial x_j}
$$
which is a new operator. Then you can apply the vector $\mathbf u=u_i$ to this operator to get
$$
\left(\mathbf u\cdot{\rm div}\right)\mathbf u=u_j\frac{\partial u_i}{\partial x_j}
$$
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
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.