Another place you can do with friction applications is in the journal bearing. So anytime you have any axle going through something like the bearing and it's spinning, we can ask the question, what torque does it take on this rod to keep it spinning or to start it spinning from zero? To do that or any other kind of equilibrium problem, you have to figure out what your private diagram looks like. What are the equations of equilibrium look like? So we know that we're going to have whatever load is pressing on this rod, so that'll be W. We know that we're going to need that torque, whatever that in is, that's actually keeping it spinning. That's what we're looking for. What are the reactions from the rod? So once I consider the actual bearing, I know that on the surface of this bearing, I must have a normal force and a friction force like any other sort of friction situation. So this and and this f are the same as they were before and it's normal to the circle, f is tangent to the circle. I can replace that and I'm going to with the reaction that's the sum of these two arc and it acts at some angle theta between n and r. So that's how that theta is defined. I know that that r cannot be co-linear with W because if it were, there would be nothing to balance the zem. So that's the same as we saw it before when we were dealing with tipping. So I'm going to, same as we did with tipping, I'm going to slide my reaction force over so that I have a distance to multiply it by a perpendicular distance to balance that m. We're going to talk about that in just a second. But I can also say that I only now have two forces. I have W coming down and r coming up. If you only have two forces, they're equal in opposite. Because some of the forces in x tells you that r does not have an x component. But some of the forces in y tells you that r equals w. So I know that r is actually going to be vertical and that it's going to be acting at a distance. But what is that distance and what do I call it? I'm actually going to define a new thing. So rw is the radius of the bearing or the wheel. You know how big around is it? r a is the radius of the rod. I'm going to define a friction circle where r f is the radius x or whatever this distance was. You can look at this little triangle in here made up of r a and x. And remember this theta was defined as the angle between r and the normal force to the circle. So it's between that line and the vertical. Well that line and the vertical means that this triangle up here has that same angle in it theta. So I can actually say that this distance, which is the radius of my new friction circle, is r a times sine theta. Here I've drawn the actual friction circle. The friction circle in here, so I took this little triangle and slanted up to make that little triangle the radius of my new circle. So the leg of this triangle is the radius of my new circle. What I can say then is that r is tangent to the friction circle because that's the perpendicular distance between r and the line of action of w. So rw is going to be tangent to the friction circle. Now we're going to see friction circles again and it's still going to be the case that r is going to be tangent to the friction circle. So this idea is sort of pervasive with round things like this. When I take the sum of the moments down here at the center of the circle, oh, I can tell you that r times that perpendicular distance is equal to m. m is then r times this raise. r i sine theta. What is theta? Well, this is where you end up with the friction part. To start the bearing moving, then theta is going to be the friction angle, phi s. That's the arc tan of mu s. That's to get it going. To keep it going, you do the same thing with phi k's. So to keep it going, theta looks like phi k or the inverse tangent of mu k. So if I plug that all in, I've got r m equals r times r a sine of the inverse tangent of mu k. So to keep it moving, that's what I add on my first one. To keep it moving, this is what my moment needs to be. mu k is something on the order of 0.2, 0.15, and these are little numbers. And if you look at any kind of trig function, the sine of the inverse tangent of something that's small is approximately itself. So you're dealing with a situation where your moment is about, almost, please use this thought as part. But in terms of understanding what's going on, the moment to keep this turning inside its bearing is approximately equal to r times the radius of the actual times mu k. And r, if you remember, was equal to w. So whatever weight you've got times whatever radius you've got times the mu k of your help, how smooth is your bearing? That's about what it takes to keep it bearing going.