After idealizing the model a bit, we can assume that the set up is like on the picture. 
The panel $OC$ rotates around the point $O$. The linear actuator is $AB$, attached to the camper's wall at point $A$ and to the panel $OC$ at point $B$. One way to make sure this works is to set $OA = OB = a$, i.e. the triangle $ABO$ is isosceles. The actuator exerts force $F$ at point $B$, aligned with $AB$. That force exerts a torque on the panel $OC$ which is $\vec{T}_F = \vec{OB} \times \vec{F}$, pointing perpendicular to the picture. Then, the oriented magnitude of this torque $T_F$ is $|OB| = a$ times the projection of $\vec{F}$ onto the perpendicular line to $OC$ at point $B$. I drew $AH \, \perp \, OB$ as a parallel reference segment. If $\angle \, AOB = \theta$, then $\angle \, BAH = \frac{\theta}{2}$ so the projection of $\vec{F}$ is $F \, \cos\Big(\frac{\theta}{2}\Big)$ and the torque is
$$T_F = a \, F \, \cos\Big(\frac{\theta}{2}\Big)$$. The second torque that acts on $OC$ is the torque exerted by gravity, which is $\vec{T}_G = -\, \vec{OG} \times m\,g\, \vec{e}_y$, where $G$ is the center of mass of $OC$ (not drawn on the picture). If we denote by $L = |OC|$ and $L_G = |OG|$, then the formula for the oriented magnitude of the gravity torque is $T_G = -\, m\,g \, L_G \, \sin(\theta)$ and the differential equation of motion for this mechanism is derived from the general equation of torques:
$$I\frac{d^2\theta}{dt^2} \, \vec{e}_3 \, = \, \vec{T}_G + \vec{T}_F$$
which is equivalent to
$$I\frac{d^2\theta}{dt^2} \, \vec{e}_3 \, = \, {T}_G \, \vec{e}_3 + {T}_F \, \vec{e}_3 $$
and thus reduces to the differential equation:
$$\frac{mL^2}{3} \frac{d^2\theta}{dt^2} \, = \, -\, m\,g \, L_G \, \sin(\theta) \, + \, a \, F \, \cos\Big(\frac{\theta}{2}\Big)$$
From this equation, we can express the force magnitude
$$F \, = \, \frac{\left( m\,g \, L_G \, \sin(\theta) \, + \, \frac{mL^2}{3} \frac{d^2\theta}{dt^2}\right)}{a \, \cos\Big(\frac{\theta}{2}\Big)}$$
and observe that the demand for force depends not only on the position (i.e. the angle $\theta$) of the panel, which a static effect so to say, but also on the angular acceleration $\frac{d^2\theta}{dt^2}$, which is the dynamical effect.
For the mechanism to work, you would like that initially the panel, which is at rest and thus has angular velocity zero, should acquire angular acceleration $\frac{d^2\theta}{dt^2}$ that is positive, so that the angular velocity goes from zero to a positive counter-clockwise angular velocity and then $\frac{d^2\theta}{dt^2}$ it could become zero for a bit, so that the angular velocity is kept constant and the rotation steadily increases and at the end of the motion, the force $F$ can be turned off ($F=0$) so the panel is left with the negative angular acceleration produced by the gravity torque (i.e. decceleration) that stops the motion. All of this prompts us to ignore the first moments of the motion, when the angular acceleration is positive, set the angular acceleration to zero in order to obtain some sort of average minimum force required to maintain at least constant angular velocity that secures the upward motion.
Thus, you could express as a demand for force the formula
$$F \, = \, \frac{m\,g \, L_G \, \sin(\theta) }{a \, \cos\Big(\frac{\theta}{2}\Big)} \, = \, \frac{2\, m\,g \, L_G \, \sin\Big(\frac{\theta}{2}\Big) \cos\Big(\frac{\theta}{2}\Big) }{a \, \cos\Big(\frac{\theta}{2}\Big)} \, = \, \frac{2\, m\,g \, L_G }{a} \, \sin\Big(\frac{\theta}{2}\Big)$$ Since $\theta$ is between $0$ and $\frac{\pi}{2}$ the sine is $0 \leq \sin\Big(\frac{\theta}{2}\Big) \leq \sin\Big(\frac{\pi}{4}\Big) = \frac{\sqrt{2}}{2}$. Hence, it would probably be good to have have a linear actuator that could produce force satisfying the formula
$$F = \frac{\sqrt{2}\, m\,g \, L_G }{a} $$ where $L_G$ is the distance between the point of rotation $O$ and the center of gravity of the panel and $a$ is the distance between the point $O$ and the point $B$ where the actuator is attached to the panel, as well as to the side of the camper. Observe that the larger $a$ is, the smaller the necessaryforce $F$ would be.
