 Good day, everyone. My name is Brad Langdill, and I thought I'd give you a free ride on my plane, the incline plane that is. I thought we'd do an incline plane problem. It's a great application of the concept of net force. And when I do these problems with my class, I talk about two, they're new forces, but really they're not new. We're just thinking of them in different ways. The force parallel and the force perpendicular. And we're going to see how we use those forces to solve a problem here. So let's say we've got an incline plane. We've got a box sitting on the incline plane. The block has a mass of one kilogram, and there's a coefficient due to friction, which is the symbol mu of 0.1. And let's say we want to know the acceleration of the system, which is a pretty common thing to be asked. I always encourage my students, first of all, right off the bat to draw a free body diagram of all the forces acting in the system. So usually the easiest one to get is the force of gravity. We'll draw that in there. The force of gravity always points straight down near the surface of the earth, so that wasn't so bad. Then students usually say, well look, there's some force that's causing the block to move down the plane. We call that, in my class, the force parallel, because it's parallel to the surface of that incline plane. There's also a force of friction here. I'll draw that in red, and that's the resistive force moving up the incline plane. And there's a normal force as well. Let's pick purple for that. And it's really important that the students remember the normal force acts perpendicular to the incline plane surface. In a lot of other problems, it's just right up and down opposite of gravity. Here it's actually at an angle, and that's going to be important as we move forward. Now from there, it doesn't seem like there's a whole lot you can really do. I guess I forgot to put an incline in the plane too. Let's say this is up by like 30 degrees. It doesn't seem like there's a whole lot you can do, but we can adjust our diagram a little bit but we can start to do some trigonometry and work out what our acceleration would be. So I always tell students, look, you can just take that normal force and you can flip it around like so. And that's what we call our force perpendicular. There's a symbol for perpendicular. It's like a little upside down T. That's the same as the normal force. All I've done is I've just flipped it around. And then I can take my parallel force, and I'm just going to slide it, literally slide it down here, not change its direction, not change its magnitude, just move it down there so that at the bottom I've got a right angle triangle. You can go through a proof of similar triangle showing that this is 30 degrees. The top angle here is always the same as the angle of incline. And you've got yourself a little right angle triangle you can work with now. And that's going to help us when we go to find the net force. So when I'm finding net force, like in most every other situation, it's just the sum of the forces acting on the system. What do we have here to add up? Well, because the acceleration is going to be somehow down the ramp I'm betting, so my net force is going to be down the ramp or the incline plane. The forces I want to add together, there's just two of them. The parallel and the force of friction. Those are the only two that are co-linear to the net force. So those are the two I'll add together. Now, in place of the net force, I'm going to write mass times acceleration, just Newton's second law. But what about the force parallel? How can I figure that out? Well, take a look down here. You can see the parallel force is the opposite side of my triangle. So I can use a little bit of trigonometry to figure out what that would be. It's the opposite side and I'm going to use the sine ratio. So sine of 30 degrees times the hypotenuse, which in this case is the force of gravity. And I can calculate that force of gravity pretty easily. It's 1 kilogram times 9.81 meters per second squared. All we're doing there is just Newton's second law, mass times acceleration due to gravity. And I can run that through my calculator and see what we get. Sine of 30 times 1 times 9.81. So 4.905 Newtons. Okay, so that's going to be my force parallel. 4.905 Newtons. Now for my force of friction. Well, how am I going to find that? Well, most students will remember the force of friction equation. Let's do it in black. Force of friction is mu times the normal force. Well, okay, we have mu. That's 0.1. But what's the normal force in this situation? Well, take a look at your diagram. The normal force is the same as the force perpendicular. So I guess I'm going to have to find the force perpendicular as well. And in my right angle triangle, it's the adjacent side, so I'm going to use cosine. And you know, it's a good idea whenever you're figuring out one component of a vector. Really all we're doing is we're finding the x and y components of this force of gravity vector. Whenever you find one component, you usually can find the other one pretty easily just by switching the sine to cosine. And I find it's helpful to do both of them in almost every situation. I'm just going to change this around and we're going to do cosine. So 8.495, let's just call it 8.496 Newtons for a force perpendicular. Squeeze that in there. So my force of friction is 0.8496 Newtons roughly. And I can put that up into my substitution now. Now the trick is with this, you have to place that force of friction in as a negative number. And the reason it's negative comes from your diagram. Take a look at your diagram over there. The force of friction is acting in the opposite direction of what the force parallel was. So one of them has to be positive, one has to be negative. You have to subtract them to get the net force. It doesn't matter which one's positive and negative really. But most of the time in my class we say down the ramp is positive and up the ramp is negative. So that's why I put the force of friction as a negative. You know, I suppose at this point here I can probably change that mass to one kilogram. Now I can just go through and solve. So the acceleration is, let's see here, 4.905 minus, answer minus 8.4957. So I'm getting, I made a mistake there, didn't I? I've got to multiply that by mu first. So I'll just subtract the numbers I got on the board here. And of course if I divided that by one it wouldn't make a difference. So I'm getting 4.1 meters per second squared is my acceleration. Not too bad to do really. The big things to remember for a problem like this, the diagram. You've got to be able to make this diagram where you have your perpendicular and your parallel force, a little bit of trig, little step here with the force of friction equation and then a net force statement, of course, remembering to make sure we have our positive and negative signs correct. If you'd like to see more problems like this, please check out my website at www.ldindustries.ca. Good luck, happy physics.