calculating the minimum force needed to lift a rigid awning - class 3 lever problem

Question

I am building a camper and one idea is to have an extra 4 solar panels as an awning bolted to a a rigid frame, hinged to the camper at the top. It hangs down the side of the camper. I will use one or two actuators to lift it to, say, a horizontal position. It will actually only hang down half way as it will be folded in half for travel, but the unfolding is not part of this problem, so it can be assumed to hang down unfolded as a rigid awning for this problem.

I want to figure out the size and power of the actuators I need. I could just guess and maybe get close or if not, just buy bigger ones through trial and error, but I'd like a sense of what I need here.

This is a class 3 lever problem, but with my limited physics background, I'm running into problems accounting for the change in angle against gravity as the awning is lifted. The work increases as the awning reaches 90 degrees (I think), so the most work is at the (corrected) END.

In addition, the location of the anchor point for the actuator is open. Though this may be determined by the distance the actuator can extend. For example, this is a 40" stroke, 400 lb capable actuator - https://amzn.to/3jSxIog. Would it work?

THE AWNING WEIGHS APPROXIMATELY 200 POUNDS (4 PANELS, 45 POUNDS EACH PLUS FRAME AND HARDWARE). IT IS 80 INCHES WIDE FROM THE HINGE (FULCRUM) TO THE OUTER EDGE.

BEFORE LIFTING - AT REST

|*! = hanging awning at rest (80 inches)
| !
| !
| !
| !
| !
|
|
| = side of camper


AFTER LIFTING 

|*_._._._._._. = hanging awning lifted (80 inches)
|    /
|   /
|  / = actuator that has lifted awning
| /
| 
|
|
| = side of camper
```

Answer 1

Your diagram is not clear enough. As it is the actuator would not work because it's deadlocked.

There are many configurations of levers, links, and brackets to do the job, each with its own pros and cons.

I show a very basic bracket welded to the owning at 35 degrees.

Depending on the length of this bracket and the actuators play you can roughly calculate how much force you need.

In this case, approximately, it is $ F_{vertical} = \frac{200*40}{L_{bracket}*cos35^\circ} $

So if $L>200*20/0.89*400 =12.5 inch, \ say \ 13 inches$ you are OK.

.

EDIT

Afer comments as to alternative geometry.

As an alternative Let's assume the actuator is a bit offset from the deadlocked center to center.

Then the force on the actuator is maximum at the position of the owning being horizontal and the angle between the wall and the actuator is $\theta$. assuming the actuator is connected to the owning at a distance of D, $$F_{actuator}*L*cos\theta = 200*40, \quad 400\geq 200*40/L*cos\theta \quad (1)$$

$$\theta =cos\frac{L}{D}$$

We plug in the $\theta$ and verify (1). If needed we adjust the D.

Answer 2

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.

Answer 3

OP here. I found this answer as well:

Calculate effort to raise hinged lever with distributed load -

So F = (200 lbs * 80 inches) / (2 * D inches)

If D is say, 20 inches then F = 400 lbs.

Solving this, I get 400 lbs of force needed if the actuator is connected at 20 inches from the camper (fulcrum). To have some wiggle room I will connect it slightly further out, say 24 inches, thus requiring 333 lbs of force. The Amazon products that have 400 lbs of force will work. Likewise, I may have two actuators, so half that force needed for each.

The calculation of where to connect the actuator on the camper side and what length to buy I will have to do some trigonometry.

Assuming I connected the actuator at 55 down the camper side, the actuator must extend to approx. 60 inches and retract to just over 24 inches (to keep it offset from vertical), so an extension of 36 inches.

I will match the distance calculation with the products available. (EDIT: ...I WILL UPDATE WHEN I HAVE A SOLUTION)