A luggage with two wheels, ( but I don't care about them, it is just to say that touches the ground in two points) is being pulled by a force $F$ at an angle $\theta$ from the horizontal.
Now I understand that when if it moves at constant velocity, then I can use the conditions for equilibrium and put: \begin{equation} \sum_i{F_i} = 0 \,\,\,\,\text{and}\,\,\,\,\sum_i{r_i\times F_i} \end{equation} So the sum of all the forces and the torques is equal to zero. (Considering friction, resistance force and weight). Are the above equations correct?
The torque "describes " a rotation due to a force; it plays the same role of a force but in rotating case, in fact many equation for forces and torques are similar. So in this exercise, if we take the rear wheel as pole for calculating torque, $W$ would rotate the luggage downward (or we can say clockwise if we imagine an axis through the rear wheel perpendicular to the sheet) whereas $T$ upward (or counterclockwise). Hence if the torque of $T$ will be bigger than the torque of $W$ the luggage will rotate counterclockwise and it'll lift up.
If you choose the rear wheel as pole to calculate the torque, the front wheel will lift off the ground when the torque of $T$ will be greater than the torque of $W$:
\begin{equation} r_WW\sin\alpha\le r_TT\sin\beta \end{equation}
where $r_w,r_T$ are the rispective distances between the pole (rear wheel) and the points of application of $W$ and $T$; $\alpha$ is the angle between the vectors $r_W$ and $W$, $\beta$ is the angle between the vectors $r_T$ and $T$.