# Torque, Force, Work question on “Moving a Door”

[diagram below]

Hi all, I'm an engineer, but it's been 30-35 years since I did any of this, so I could really appreciate some practical/mathematical input.

I have a "door panel" which weighs 330 lbs (or 1500 N) at most and is suspended on a pole/pivot/hinge/ball joint P, which needs to swivel as much as physically possible about the E-W axis by applying either a force of F1 (a force applied at a certain angle and distance from P to the N-S axis of the door) --or-- F2 (a force applied parallel and at a distance from P to the N-S axis of the door). F1 --or-- F2 cannot exceed 330 lbs and I can only use one but not both.

Question 1: Which (F1 or F2?) would likely use the least amount of Force to do so assuming both are applied/directed to the same spot along the N-S axis and distance from P?

Question 2: Assuming I had to use F1 at what ideal angle (less than or equal to 45 deg, or greater than 45 deg?) and distance from P (further or closer to P?) along the N-S axis of the door should I direct F1 to so that I use the smallest amount of Force necessary?

Question 3: Assuming I had to use F2 at what ideal distance from P (further or closer to P?) should I direct F2 to so that I use the smallest amount of Force necessary?

I need the options, so I can determine where to put/direct F1 or F2.

PS. By "door panel", I really mean a set of Solar Panels, but I like to think of it as a door, and F1 or F2 would be motorized linear actuators with a maximum travel distance (around 12 inches) and capacity of 330 lbs each. This is actually for a solar tracking project I want to try implementing.

Thanks in advance for any advice on the Physics of the matter. Will use your input as a starting point.

--eric g

In order to make the panel rotate about the $NS$ axis through $P$ we need to apply a net torque $\tau$, so with Newton:

$$\tau=I\alpha,$$

where $I$ is the inertial moment of the panel about that axis and $\alpha$ the angular acceleration.

In the simple case of a $F_2$ (force acting perpendicularly to the $EW$ axis):

\begin{align}\tau &=F_2|PO'|\\ \implies F_2 &=\frac{\tau}{|PO'|}\end{align}

Obviously the larger the lever $|PO'|$, the smaller the magnitude of $F_2$ needed.

In the case of $F_1$, the angle of application $\alpha$ (angle wrt vertical), so the torque becomes:

$$\tau=|PO|F_1\cos\alpha$$

Applying the force at an angle diminishes the useful torque it generates.

You stated you got what you needed from Gert's answer - I just wanted to point out a couple of things.

First - unless there is some restoring force that returns your panel to the equilibrium position, what you are really asking about is what impulse would be needed to move it.

Second - if you have a given impulse $F\Delta t$, the change in angular momentum it imparts depends on the distance from this impulse to the axis of rotation. In your example, an impulse along the EW (longer) axis has greater ability to change angular momentum. However...

Third - the moment of inertia of your panel about the EW and NS axes, respectively, scales with the square of the dimension - that is, if the length of the panel along the EW axis is twice that of the length along the NS direction, the moment of inertia is four times larger. At the same time, your ability to affect the torque is only twice as large (you can get twice as far from the axis of rotation).

In general, then, this means that it will be easier to rotate the panel along the long axis, with the force applied as far from the center of rotation as possible, and perpendicular to the panel.

But I urge you to consider a couple of factors:

1. what is the range of motion of your actuator
2. what is the torque you are trying to overcome (I doubt it is inertia, given that solar tracking is quite slow), and how is that torque a function of the position of the panel

The point of the second question is this: if there is some (restoring) torque term that depends on angle, and the force your actuator can supply is constant, then you will need to position the actuator such that it provide the maximum torque (i.e., acts perpendicular to the panel) when the restoring torque is greatest. The more balanced your design, the less torque is needed. Do consider what will happen when there is high wind, snow piling on to your panel, etc... "real world" factors are likely to dominate your design considerations.

• Floris, thanks for the input as well. I haven't really tried looking into the high wind problem nor the foundation. In the last 12 yrs, the eye of the storm has passed through or near my home twice, looking at 100-200 km/hr winds. I think, I'll have to make the setup collapsible to survive such. I'm looking at 3-4 solar trackers in my backyard each with 1KW worth of panels. I thought about a balance design, looking at adding weights or the option of sliding the panels. Thanks again for the input, much appreciated. --eric g – ericg Dec 4 '16 at 12:13