 In this video, we're going to take a look at static friction problem-solving examples. If you haven't watched my other videos about the general concept of friction and static friction in particular, as well as my video that describes this equation in more detail, you probably want to go back and watch them. They're in my playlist here for forces. We're going to take a look at a few possible example problems, not everything, but a few examples. Now, let's say we're given a situation where we know the coefficient of static friction between two surfaces is 0.56. If those two surfaces, my solid object and my surface that the solid object is sitting on, happen to have a normal force between them of 15 newtons, we could then go ahead and solve for the frictional force. Now, we have to be careful because there's this less sign in here. So that means the static friction is less than or equal to 0.56, which has no units because it's just a coefficient, and the 15 newtons multiplied together. What this means is that my static friction value has got to be less than or equal to 8.4 newtons. So as we talk about in some of the conceptual videos, if I actually have an object sitting on a surface and I'm actually not trying to move it, there's not going to be any friction there. But if I try to move it with some applied force, friction is going to try to act against it to hold it still. For any applied force less than 0.8, excuse me, 8.4 newtons, it's going to stay still. If I happen to apply an applied force of 5 newtons, then friction says, yep, I can handle that. My static friction is going to be equal to 5 newtons, which is less than the maximum. No problem. But if I tried to put an applied force of 10 newtons on that particular thing, friction is going to say, oh, I've got to be less than or equal to 8.4, and it's going to move. Static friction can't hold it in place. So a lot of times what they end up having you find is this maximum value. So this is really that the static friction maximum value is equal to mu s f n. And in any particular problem, you'll have to look to figure out whether the static friction is more or less than that maximum. If it's less, it stays static. If it tries to be more, it's not going to stay still. Now, there's some other ways we can take a look at this problem. And so if we have different knowns, we can actually go through and try to solve for those. So like our kinetic friction case, we can actually go through and use this to solve for our mu s. But we can only do that if our knowns includes the normal force and the maximum static friction. In other words, how much frictional force can I get before the thing is going to start moving? So just before it starts moving, that gives us our maximum force. And again, our normal force is typically going to be larger than our frictional force. So let's say we have these values. Now, like in the kinetic case, we're going to have to do some algebra here. And so again, we're going to work with that maximum value equation and rearrange it to say that the coefficient of static friction is going to be that maximum frictional force divided by the normal force. And then you can easily plug those into your calculator, the 6.8 newtons divided by the larger normal force, to actually get a value of, say, 0.48, plugging in my calculator and looking at significant digits that would round to 0.486. So that's how they find that coefficient of static friction is they have to take that applied force all the way up to the boundary of the maximum just as it starts moving. And that value of the frictional force just before it starts moving is going to be our maximum 1. Now, we could rearrange this one more way. And again, it's a little bit more complicated with this. But let's say we have known values for the static friction and the coefficient of static friction. And let's say I've got a value of 0.72. So it's kind of really a rough surface there. And let's say I've got a value for the frictional force of 26.5. Well, in this case, we should be able to solve for the normal force. But we have to be careful because we've got this less than sign. If it doesn't tell us that that static friction force is the maximum just before it starts moving, we still have the less than sign in there. So when I do that rearrangement, and let me rewrite the equation here before I start doing the rearrangement, the static friction is less than or equal to mu s fn. To get fn by itself, I have to divide both sides by mu s. And that means my frictional force divided by mu s is less than or equal to the normal force. Or if I were to rearrange this, the normal force has to be greater than or equal to the static friction force in this mu s. Otherwise, we wouldn't actually have enough static friction. So if I take these values and put them in, my 26.5 divided by 0.72. And again, that static friction, so that has to have a unit of newtons. What I get for this case is that my normal force has to be at least 36.8 newtons. Otherwise, it would have slipped. Now, if I've got a normal force of 50 newtons, then I can easily supply that amount of static friction force with that coefficient of friction. But if I were to drop down to, say, a normal force of only 10, it's not going to stay static. So the normal force would have to be larger than that amount. And if it tells us that this was actually the maximum value, then it would be exactly equal to it. So these are just some other examples of working with this static friction formula, which is a little more complicated than some of our other ones because of that less than or equal sign. Keep working and asking questions as you need to.