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About this Lesson
- Type: Video Tutorial
- Length: 12:37
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- Posted: 07/01/2009
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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 campus-wide) Teaching Excellence Award in 1998. He is the author of two Teaching Company video courses: “Particle Physics for Non-Physicists: 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 real-world 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 Non-Newtonian 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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Static equilibrium means the sum of all the external forces is equal to zero on some objects. There's no net torque on the object and the object is at rest. Static equilibrium. It's really practical question in a lot of physical situations to be able to decide whether you've got static equilibrium and what would be required in order to have everything stable. So let just show you some problems.
Supposing you're hanging a chandelier and you've got some artistic ideas, so you think, all right, I'm going to hang this chandelier from my roof, I'm going to use three ropes, tie them together in a know like this. I like the look of this. I want this rope to be horizontal. Question? How strong do these ropes have to be? And, you're looking and you're saying, "well this thing has mass m, so it has a weight mg, so I think my rope just has to be able to hold the weight of the object." That's a naïve and reasonable guess and it's wrong. And what you really have to do, is this is a static problem. It's a static equilibrium situation where there's no net forces and everything's just sitting still. Work out Newton's second law. I've got the sum of the forces that's supposed to zero. That mean's the sum of the forces in the x direction should equal to zero. And in the y direction, it's equal to zero. And, what's my object? When you're going to work Newton's second law, you can't just look at a picture, you have to draw force diagram. So, here's my object, I think. It's got mass m, it's got mg, that's the force of gravity acting down. And I see there's a little rope up above it, so there's a tension key holding it up. And boy, this looks like an easy problem. T - mg = 0. Tension equals mg. See, the rope just has to hold the weight mg. And that's absolutely correct for this little rope here. But, what about this one, and what about this one? These are different ropes. They're connected together at some juncture. You should think of this as a knot or if you like you can think of that connection point as a little fictitious object. The knot itself can be viewed as an object and you know, if we want to think about the forces or tensions in these ropes, they're acting on that point.
So, let's do Newton's second law again. Some of the forces equal to zero acting instead on this little point. Now, what's the mass of the point? Well, it's zero and it's acceleration is zero. It's a static problem. I don't really have to worry about that. The right hand side is going to be zero. And if I focus my attention on that spot, theory gets a little bit more complicated. So, here I have a horizontal rope and ropes can only pull. So, let me call that tension T. Then I had this one that was tipped up at an angle, theta. And, again it's pulling. So, here's my imaginary object, the knot, and remember there was a tension which we just worked out. T was equal to mg. That is what's pulling down. In a given individual rope, the tension is the same at the two ends.
So, this is my force diagram and now I'm all set to work Newton's Laws. And you can't just set T + T + mg = 0. You have to work components. So, let's look in the x direction first. The sum of forces in the x direction, I've got to pick a sine convention so let me call this plus and plus, the usual story. So, I've got plus T, and I've got nothing from this force. And I've got minus because this tension has an x component in the left direction. And it looks to me like it's T cosine theta. And that adds up to zero because there's no net force on this point. And, I'm not done. I'm after T and T, the tension in these two ropes and it's one equation and two unknowns. So, I've got another equation, the sum of the forces in the y direction. So, I look at the picture, I've got this one, that's T sine theta. And then this one is negative, it's -mg, and that's also equal to zero. And I say, "Ah, very good." I can solve for T, it's . And now I know T, I can plug it in to this one. So I immediately get the results for both T and T. T is and T is equal to this times cosine of theta. It's, cos
So, I start to think about these equations and say does this make sense to me. If I want theta to be thirty degrees, so sine of thirty degrees is a half, this gives me mg over half. It's 2mg. So, sure enough this rope needs to hold more than just the weight. It's got to hold twice the weight. Because there's this other rope tugging on me too. And this rope, all you've got to do is plug in the numbers. It's going to be a little bit less that T because it's T cos . And, so you can work it out and decide whether you're ropes are strong enough. I look at this equation and I say, "what would happen?" I've got new artistic idea. "what happens if instead on hanging this one on the roof, I hang it on the wall over here." So, I want to make theta go to zero. So, I'm basically going to have two horizontal ropes and the chandelier hanging from the middle.
So, what does my equation say now? If theta goes to zero, T goes to infinity. And so does T. So, I'm thinking, you know, is Newton's Law breaking down? Is there crazy going on? Well, there is something crazy going on, but it's not the problem with the math. It's real. It's physics. See, here's my scenario. Remember, I'm trying to hand this mass so that the rope is horizontal and the others I put an angle. I can feel already that this one, the one that I call T does have more tension in it. My finger over here is hurting more. And if I try to lower the angle, keeping this one horizontal, I can tell that there's something crazy on. I can't do it. I can't generate that infinite tension required and the formula's really are telling me the truth. The physical reality is you can't just have two exactly horizontal ropes. The tension required would be humongous. Ropes would break and all of the story wouldn't be static any more. All we needed here was Newton's law. Just F = ma.
Let me do another example. Supposing that I've got a car cruising across a bridge. And there's only two posts on this bridge and it's a very simple bridge. I just lay a big old plank on these two supports. And I would like to know are my posts going to buckle? So, I'd like to know what are the forces here and here. And you say, "gee, how could it really matter? This car has a weight, mg, and so aren't they just going to share the weight. Half mg, half mg. Again, you know, don't just try to intuit these questions. Draw force diagram, use Newton's Laws, work it out. What's the object going to be in this diagram for my force diagram? I think the best object will be the plank or the roadway because it's acted on by the forces I'm interested in, the force of the posts, and it's acted on by the car. So if I write down this as my object, objects can be elongated, that's my conceptual object. So, let me write down the force diagram in that picture. Here's my object. What are the physical forces? There's a car somewhere, it's not necessarily in the center, and it weighs mg and that's going to be pushing down on the plank. There's a post over here and which way does the post push? If the plank is sitting on the post, the post is pushing the plank up. So, let me call this one F[L] for the force of the left-hand plank, and let me call this one, F[R] for the force of the right hand plank. And I want to know those two forces. Are they equal? What are they? So, I say, "Ah, easy. I've got my force diagram." Some of the forces in the y direction equal zero. So I get F[L] + F[R] - mg = 0. And one equation, two unknowns. So, you think, okay, I'll use some of the forces in the x direction, but no, that doesn't help me because there are no forces in the x direction. I'm not stuck. Remember, for equilibrium, you need also to have the sum of the torques equal to zero. Some of the torque's about any point you like. So, you look at your force diagram and you say, "let me pick a point to be anywhere, how about here, right here where the left hand post is pushing." It's a clever choice because then this force is running right through my axis, my pivot point. it's not really a pivot point, nothing's pivoting. It's just my choice of origin. But there's torque of this force about that point. If you want to write down some of the torques, there's sines with the torques just as there's sines here. I was obviously, implicitly calling up positive when I wrote this down. For torques, you have to pick a direction. Let me call clockwise as positive. So, what am I going to get?
This force is certainly supplying a torque which is clockwise. So, there will be a positive torque here, from this force. This force acting over somewhere else is got a counter clockwise torque. So, this one's going to come into my equation with a minus sine. So, let's just write it out. Some of the torques. I've got force, remember it's r x f. And we call this distance D and the sine of the angle between the radius vector and the force vector is just 90 degrees. So, I just get mgd[1 ]plus, and then this sine has a minus sine when I calculate the torque, so it's -F[R] times, what was the distance? It was D plus D. And that is supposed to equal to zero. It's the sum of the torques. so I can solve for F[R].
And I immediately get F[R] equals Mgd divided by D + D. So, that's the answer, and you know when you stop and think about it, it seems kind of reasonable. It says is D is equal to zero, if the car is sitting over this post, there's not going to be any force over on the right hand post. And as the car moves its way over, this post will feel more and more of the force. And this one, less and less. Remember, that the sum, F[L] plus F[R] is always equal to the total weight, mg. So, they're sharing the weight but not symmetrically, unless the car is right smack in the middle. If D equals to D, then indeed I get .
You know, once again, it's kind of nice to just do it, let's build a little model here. Here's my roadway. Notice that I'm just laying the roadway down on the supports and I put a car. You know, I don't have a scale, but when I look at it, I can just imagine that this one is holding up the car and as I move it over, now it's this one that's holding up the car. And then I look at my equation and I see something funny. Oh, what if D was negative. This equation says F[R], so D is over here. What if it's off on this side? The formula says that F[R] would have to be negative. The way I've set this thing up, F[R] can't be negative because I don't have any glue or nails or anything. The normal force can only be up. My formula says it's going to be negative but it can't be, so what's going to happen? Is Newton's Law going to explode?
Well, in a certain sense. I was assuming equilibrium when I wrote down this equation and clearly not equilibrium situation. So, the equation is really correct and if I did have a little glue here, then I could have a negative force and it would be given exactly by this formula.
Solving equilibrium, solving static's problems, is it really just a question of drawing a nice force diagram, some of the forces equals to zero, some of the torques equal to zero. And solve away.
The Physics of Extended Objects
Solving Static Equilibrium Problems Page [3 of 3]
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