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About this Lesson
 Type: Video Tutorial
 Length: 11:49
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 126 MB
 Posted: 07/01/2009
This lesson is part of the following series:
Physics (147 lessons, $198.00)
Physics: The Physics of Extended Objects (25 lessons, $35.64)
Physics: Statics (3 lessons, $5.94)
This lesson was selected from a broader, comprehensive course, Physics I. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/physics. The full course covers kinematics, dynamics, energy, momentum, the physics of extended objects, gravity, fluids, relativity, oscillatory motion, waves, and more. The course features two renowned professors: Steven Pollock, an associate professor of Physics at he University of Colorado at Boulder and Ephraim Fischbach, a professor of physics at Purdue University.
Steven Pollock earned a Bachelor of Science in physics from the Massachusetts Institute of Technology and a Ph.D. from Stanford University. Prof. Pollock wears two research hats: he studies theoretical nuclear physics, and does physics education research. Currently, his research activities focus on questions of replication and sustainability of reformed teaching techniques in (very) large introductory courses. He received an Alfred P. Sloan Research Fellowship in 1994 and a Boulder Faculty Assembly (CU campuswide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for NonPhysicists: a Tour of the Microcosmos” and “The Great Ideas of Classical Physics”. Prof. Pollock regularly gives public presentations in which he brings physics alive at conferences, seminars, colloquia, and for community audiences.
Ephraim Fischbach earned a B.A. in physics from Columbia University and a Ph.D. from the University of Pennsylvania. In Thinkwell Physics I, he delivers the "Physics in Action" video lectures and demonstrates numerous laboratory techniques and realworld applications. As part of his mission to encourage an interest in physics wherever he goes, Prof. Fischbach coordinates Physics on the Road, an Outreach/Funfest program. He is the author or coauthor of more than 180 publications including a recent book, “The Search for NonNewtonian Gravity”, and was made a Fellow of the American Physical Society in 2001. He also serves as a referee for a number of journals including “Physical Review” and “Physical Review Letters”.
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If the net force and the net torque on an object is zero, we say it's in equilibrium. And if it's also true that there's no motion, then we say that it's in static equilibrium. Static means it's sitting still and since there's no net force and no net torque, it will continue to sit still. So, that's what you mean by static equilibrium. And even then, there's this distinction: is it stable or unstable equilibrium? So, sometimes it's obvious, and sometimes you have to think a little bit about it.
Let me just show you an example of a complicatedshaped object and let me hang it and let's just ask what kind of a state is it in right not. It's not accelerating. It's not angularly accelerating. So, it must be, by observation, that it's in equilibrium and I claim its in static equilibrium; it's sitting still, and it's stable. Stable means that if I perturb it a little bit, if for a brief moment, it's not true that the sum of the forces or the sum of the torques is equal to zero, it will return back to its previous equilibrium position. There's some extra force here of friction, but in the end, right now, there's no net force, no net torque and you can just see that this is a stable configuration.
There is another configuration here, another way I can set it up. From the same pivot point, with the same object, which is also technically, static equilibrium, but it's not stable. It's very difficult for me to do: it's a balancing act. And you know, no matter how hard I try, it gets really close to being balanced and then it runs away. Technically speaking, there really is a balance point. I can, in principle, get that object balanced so that some of the forces are zero; sum of the torques are zero; it's not moving; it's staying up there; it's in static equilibrium. But the tiniest motion, the tiniest extra force for even a brief moment, moves it out of equilibrium and it runs away from that position. So, that's an unstable equilibrium.
How do you decide, how do you know whether something is going to be instable or unstable equilibrium? Clearly, important when you're designing things. You want to know if a little protobation is going to make them fall over or not. So, here's a picture of that funny shaped object. Here's the pivot point. And then there's an object, which I can have in any orientation that I want. And what I'm really am interested in is, "is this thing going to twist around and fall?" So, here is going to be my origin. And I'd like to figure out, "can I make the sum of the torques equal to zero?" That's certainly a necessity if we're going to have equilibrium? So, what's the sum of the torques on this object?
This object is made up of a bunch of chunks of mass, M which we can label M[i] and it's a little bit of a complicated problem. I have to figure out the torque, what's the force, which is causing this thing to rotate around? And it's gravity. I'm just talking about an object hanging. So, it's the gravitational force on this M[i] that I'm worried about and then it's resulting torque about this origin.
So I have to add up the torques of each and every little chuck in order to get the grand total, sum of torques, and then I can set that equal to zero. So, it looks like a nasty problem, but in fact we've done this problem before. You can just try it. When you set up the integral (which is required), you'll discover that the conditionsum of the torques equals to zero; its completely equivalent to pretending that all of the mass of this object was concentrated in one pointthe name for that point will be the center of gravity. And it acts as though it was just a mass of total mass m, its mass, located at that center of gravity. And I can just figure out the torque of gravity due that point object rather than a complicated object.
Better than that, because if you work it out, you'll see that the center of gravity in this case, is just exactly the center of mass, which we know how to find and evaluate. I make a distinction between center of gravity and center of mass because what I just said is only true when the force of gravity is mg m[i] g down: it's a uniform gravitational field. And then there's really no distinction, practically: center of gravity and center of mass. So, the basic idea here is that, instead of having to worry about this complicated shape, I can just think about it as a point object, (so here's my pivot) point object located right here at the center of mass, or the center of gravity.
So, if you want equilibrium, you need, at a very minimum, the sum of the torques about this point equal to zero. And it's certainly not. Because there's MG, that's a force and that makes a torque about this pivot point. There could be another force herethere will be in general, but that doesn't make any torque about the pivot point. So this system does have a net torque. It's not in equilibrium, and it sure enough, if the center of gravity is over here, its going to fall down; the thing will accelerate.
So, what would I need, what kind of a configuration would I need to be in equilibrium? There's two. The nice one is the one where you have your pivot and the center of gravity or the center of mass is down here. Right below it. So, here's this force of gravity, pulling down on the center of mass, or center of gravity, and then there's the pivot force, whatever that might be, acting at the pivot. There is no net force, there doesn't have to be any net force in this problem. And there's no net torque, you can see that because about this point, there is no perpendicular R; there's no torque at all, from either force around the pivot point. So, zero torque, zero force; we can certainly be in static equilibrium here, and it's stable as well.
Because if you move this center of gravity off to the side a little bit, if you perturb it with some extra force briefly, what will happen? Well, if the center of gravity is over here, there will be a torque bringing it back again. If you perturb it over this way, there'll be an external torque bringing it back down. Always, bringing it back down to this configuration. That's what you need to be in stable equilibrium. A little perturbation will maybe set you swinging back and forth around the equilibrium point, but it brings you back.
There is another equilibrium for this system. Namely, when the center of gravity is all the way up at the top. That's the one that I was trying to find, and it's tougher to do. In fact it's essentially, practically impossible to do, because even though it's true that in this particular moment, in this exact configuration, there's no net force and there's no net torque about the pivot point. So this is technically an equilibrium point, the tiniest perturbation of the center of gravity, moving it over here, makes a torque, which makes it want to get even further away. It will start to accelerate away from this equilibrium. So, that's an unstable equilibrium.
So, when you have a complicated object and a pivot point, condition for stability is really very simple. If the center of gravity is below the pivot point: stable; if it's above, even if it's in equilibrium, it won't be stable equilibrium. And the nice demonstration of this is the little bird toy here, that I've got. It's got wings, which are a little bit below its head. And there are some hidden masses in there. So the center of gravity of this complicated shape is located spatially, right below the beak. And so it's really like this situation. The center of gravity is below the support point and this is stable equilibrium. I can tip it, and jerk it a little bit and it restores itself. It always comes back to this equilibrium configuration. Technically, this is called metastable, not stable, because if I give it big perturbation, it will fall down. But for little perturbations, it's stable.
How about this scenario? When you have the center of gravity above the pivot point? Can you ever have stability? I'm standing here, my center of gravity is above my feet and I seem to be pretty stable. So, what's the story there? The difference I'm not standing on a pin or a tiny point. I'm standing on a base. So let's just draw a picture. What happens when you're standing on a base?
So, imagine that I have something like this little mock tower of Pisa here. And that seems to be stable, I can put it on the base, it's even really stable. I can give it little perturbations and it doesn't fall down. So, what's going on? Well, I'm going to draw a picture, and let me draw two pictures. Here's the configuration as I just showed it. Notice that it has a big, fat base. So, the center of gravity of this thing is pretty close to the bottom.
And now let me change the story just the slightest little bit by adding a little bit more stuff to the top. So, when you add more mass over here, you move the center of mass up and over a little bit. And now, the new configuration is like so. The center of mass, or the center of gravity is now just shifted up and over a little bit cause it has gotten a little bit taller.
So, what happens in these two situations? Remember, I'm just setting it on the table. So, what sort of forces can the table apply? Let's look at this one first. The pivot point you can choose to be anyway you want. It doesn't have to be a physical pivot point, it's just an origin. So, let me ask about the sum of the torques around this point. There's certainly a torque due to the force of gravity, which is just the mass of the object times G times this perpendicular distance: that's a clockwise torque. What other torques are there? Well, there's a base. And there can be various normal forces of the table pushing up and they'll somehow distribute themselves naturally. But no matter what, each one of these little forces also provides a clockwise torque. So that grand total torque about this point is not zero, this thing is not in equilibrium at all, and if I let go, it falls over. This is the scenario.
So, what happened here? How can it be stable when I took off the top? Well, now the center of mass was not past the base. You can see that the center of gravity is providing now a counter clockwise torque about this arbitrary point and these forces are somehow distributing themselves so that, yes, you can have a net torque equal to zero and a net force equal to zero.
So, that's really the qualitative condition for stable equilibrium is: it's okay to have the center of mass above the base, but it has to be above the base; it can't be past the edge of the base. So when you're standing in a subway or a bus, and you can't hold on to something. What do you do? You don't put your feet together. Because if you do that and there's a little jerk; all of a sudden, if the center of mass of your body finds itself past your base of support, you're going to topple over. You're not in equilibrium any more. If you've planted your legs nice and wide apart, your center of gravity can be sort of anywhere, as long as it's inside over the base, you'll be stable. You can have equilibrium situation. So, that's the qualitative requirement for stable equilibrium.
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Understanding Stable Equilibrium and the Center of Gravity Page [3 of 3]
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Great but looking for more equations. Still glad I bought it!