This is very helpful RoyalCat, thanks very much for your help! If possible, I'd like to go over a few things, if you don't mind, just to make 100% sure I know what I'm talking about!
First, to answer your questions:
The "net moment of angular acceleration" is precisely what I'm looking for and how I'd like to put it. What I'm curious is whether or not, with the data I have, I can calculate it and give at least a rough number as to when (what angle) that acceleration occurs.
A key thing to understand is that for an object in equilibrium, the net force is 0, and the net moment about any point is 0 as well.
If the angle is 0°, where do you expect the surface reaction force (The resultant force of the normal and the static friction, I'll just refer to it as the normal force, but keep in mind I'm talking about the friction force as well) to act from? For the net moment to be 0, it would have to act on the same line of action as the force of gravity, meaning, directly from the middle of the base of the cylinder. As you increase the incline, the normal will 'migrate' ever closer to the edge, so that the net moment remains at 0.
You should try and explain this concept so as to illustrate why the critical condition is when the normal starts to exceed the base. (Theoretically, if the normal could act from that far away, it could produce a moment to counter-balance the moment of the force of gravity, but physical restrictions mean that it can't)
My background is in design, and I'm about to enter graduate school for architecture. The class I'm taking, which is more of a discussion based, non-graded "intro to structures" class, is what I'm doing this for. The goal of this project is not like normal homework assignments in which one must figure it out on his/her own and receive a grade; rather, the goal is only that I do a little teaching myself. If the assignment were graded, I think it would be on the students ability to explain a physical architectural condition by employing some of the fundamental concepts and terminology of physics. I'm getting a little more comfortable with some of the terminology, but I'm not quite there yet!
Ok, so here's my specific response to what you've told me so far, thanks again!
"The problem concerning a cylinder on an inclined plane is even cooler. You want the line of action of the gravitational force, acting from the center of gravity, to exceed the base. That means that there is a net moment around the same axis of rotation I referred to above. The critical situation is when the line of action of the force of gravity acting from the center of mass, intersects the point of the axis of rotation."The "line of action" is the direction from which the force (gravity) is being applied, right? So in this case the "line of action" is always vertical and in the downward direction? Would that make the "line of action" the same as the "center of mass"?
The line of action is not the same as the center of mass, just to be precise with our definitions. The line of action is an imaginary line that runs parallel to the direction of the force and intersects the point from which the force is acting (Note that the line of action is in the direction OPPOSITE the force as well, it continues in both directions. I'm in a bit of a rush now, but I'll try and add a diagram when I get home.)
The center of mass is the point from which you can say that the force of gravity is acting on the mass. For a system of i objects, each with a mass m_i (Not all of them are the same, i is an index) and a distance from a certain origin \vec r_i, the center of mass is: \frac{\Sigma m_i r_i}{M} where M is the total mass. For a continuous system, an integral will replace the discrete summation.
What that means for symmetrical objects is that the center of mass is smack-down in the middle. So for our cylindrical case, we can just say that the center of mass is at the center of its rectangular 2d cross-section.
Thus, as the tower begins to lean, the line of action moves horizontally (remaining perpendicular) in the direction of the lean. When the line of action falls directly on the edge of the base of the building, the building has equal weight on either side of the line of action (half working to tip the tower, half holding it down). But at the minuscule moment the line of action moves past the base, the net moment occurs, and down goes the monument.
Yep, the line of action is always perpendicular to the floor, passing through the object's center of mass. Turn the object enough, and it will exceed its base. At that point, where the normal is restricted to act from the pivot point, what is the total moment? Can the normal balance the moment of the force of gravity?
If I'm correct on my summary above, it may be enough information for me to do fairly well on my presentation, however if I can be a little more specific about how one would go about calculating this net moment with the information I have on the tower, that would only strengthen the presentation, I think.
SO: let's say that the friction between the base of the cylinder and the ground is always static, thus there is no sort of "sliding" to account for. Also, the tower is leaning perfectly in only one direction (no other angles to account for).
Yes, that's a perfectly reasonable assumption to make, and a vital one too, since if the tower started sliding down the slope prior to its toppling, we'd be in a VERY different problem (We would have to introduce another force, too, if the tower had a net acceleration down the slope).
I wonder if you can tell me how I can mathematically represent the net moment. I'm wondering what the equation would be into which I could (for example) plug in an angle (for which the line of action is to the right of the base edge (say, 1 degree)), and my answer will be POSITIVE. Then I show the same equation with the angle at which the line of action is precisely at the edge of the base, and my answer would be (perhaps?) ZERO. And finally, I plug an angle into the equation that is very large (say 60 degrees) and my answer is NEGATIVE.
You will get a net moment of 0 for every situation where the normal does not exceed the base. It will change its grip appropriately to counteract the moment of the force of gravity.
To get a negative answer, you would need a negative sine for alpha, so the object would topple CW (Positive=CCW direction is merely convention, for what I said to be true, your alpha would have to be measured from the horizontal axis, pointing right)
Once the force of gravity exceeds the base, the normal cannot compensate for the moment of the force of gravity around the pivot point, and so the tower will topple.
I'm not sure if this makes ANY sense whatsoever or if this is the way you might go about it, but it would be great if I could sort of buttress my argument with some very basic calculations. Thus, I can explain the line of action in sort of an intuitive way, but also provide a little bit of supporting data that, well, explains my explanation.
I'm not sure this is exactly the approach I should take, but I think it might give you an idea as to what I mean by mathematically explaining the net moment. I'm not sure how much work this would be to explain, but I appreciate any and all feedback you have! Thanks again so much! This is a wonderful forum!
