 Okay, another example with forces and free body diagrams. Suppose I have... I'm going to take the same book that I had in the last example. How do I want to use the numbers? I don't, but I'm going to, I guess, because I want to make people happy and they always want an angle there, so that's that. And here's the one kilogram book right there. Now, this is on a plane of ice. Okay, so if you let it go, it would slide. How do you stop it from sliding? Well, in this case, I'm going to tie a string onto it right there. Okay. So, and just to make things cool, let's say the string goes over a pulley and that pulley goes down here and there's some mass hanging over the end right there. So, let's see if we can find two things. I want to know how much mass I have to put on the end of that pulley in order to make the block stay there. And I want to find the force the ice pushes on the book. So, what's the first thing to do? Draw a free body diagram for the book. You have to be careful. When you draw a free body diagram, it's the forces acting on one object. Don't mix your objects up. That's not good, okay. So, here's my book and let me just redraw it right there. What's touching the book? Well, I have the string is touching the book and the plane. So, I have those two forces and then I have the gravitational force. Let me first draw the gravitational force. Straight down vector. Now, the string. One of the things about strings is that they only pull in the direction that the string is. You can't pull at a different angle than the string is stretched. It's just the way they work. You can't push with it either. So, if the string's pulling that way, then the string force, I'll call this FT for tension, is that way. And then which way is the ice pushing? Remember, it's the normal force. Perpendicular. Perpendicular to this plane. So, that means that way, okay. And I haven't exactly drawn those to scale, even though they're not too far off. But they all should add up to zero since the block is stationary and at rest. So, that's my free body diagram. Now, if I want to add vectors and I want to do fun things with vectors, then I need to know the X and Y axis, which isn't real. But I need to pick that, okay. So, which direction is the X axis and which is the Y? There is no wrong answer here, okay. Probably most people would just generically say that's X. There is a wrong thing you could do. You could pick X and Y, not perpendicular to each other. Don't do that. But here's X and there's Y. That would be okay. But, and that would work just fine. But I'm not going to do that. You should do that on your own and see that you get the same answer. I'm going to pick this is my X axis and this is my Y axis. Why would I do that? I would do that because I have a lot of experience solving these kind of problems and I know that this is going to make things easier. Maybe I should have done it the hard way so you can see how you could still do it. But I want you to do it that way. But if you look, this is nice because now I only have one vector that I have to find an X or Y component. This is just in the Y direction. That's just in the X direction. So it makes things easier. Okay. So now one other thing is this angle right here. I'll let you play around with it. But if you do this and you do this, you can see that this angle is a complement of 90, which is that. And so that's also the same angle. Okay. It comes up enough you should just figure that out at least once. So vector-wise I know that Ft plus Fn plus mg equals zero. But in this case I have two directions that matter. I have the X and the Y direction. Let me go ahead and there's only one vector that has components. This is the Y component of gravity and this is the X component of gravity. And now that may look weird. Gravity is not in the X direction. If this is your X axis there is. Okay. Okay. So in the X direction I have what forces? Well I have a component for tension in the negative X direction. Negative Ft. And then I have mg in the positive X direction. How much? Well that's the opposite side of the triangle. So it's going to be plus mg sine 25. And that's it. Let me go ahead and solve for the tension. Because the tension is if I actually could draw a free bi diagram for this. Ft mg different m. And those have to add up to zero. So whatever the tension is I can use that to find this mass. So I'm going to go ahead and find this tension. Ft equals mg sine 25. And then if I had a mass of one kilogram and g is 9.8 times sine of 25. Let me just put that in my calculator right here real quick. It makes people happy. 4.14 Newtons. And then over here Ft let's call this m2. Ft also equals m2g. I skipped some steps there. So m2 equals Ft t over g. So that would just be 0.42 kilograms. Now we can go into the y direction. So what force do I have? I have the normal force in the y direction and then part of gravity. And that's going to be the cosine of the angle because it's the adjacent side. So I have Fn minus mth cosine 25 equals 0. So Fn equals mg cosine 25. And real quick I get 8.9 Newtons for the magnitude. Now here's where you've got to be careful. People fall into this and say oh, normal force is equal to mg. It's not in this case. A lot of times it turns out the normal force is equal to mg. But don't fall into the trap of saying that all the time. You will cause yourself some problems. Okay.