In today's video I'm going to show you how to solve a simple Newton's law problem using some basic steps and by following these steps you will be able to also tackle the most complicated problems you could think of. So this problem here classical problem we have a box that is on a surface. The box has a mass of 10 kilograms. We are pulling on the box with 20 new turns but we're not calling anywhere. So no acceleration and no speed. The question is what is the normal force of the box and what must be the friction that is acting on the box at that moment? So the first step is always to do the free body diagram, FBD free body diagram. What do we do with the free body diagram? We are isolating what we're interested in and are cutting it out of the environment. And wherever I cut I will have some force acting. For example here I had a force. So I'm going to be doing a simplified object. My box, I don't really care that it's a box, I can simplify the dot. And here I had a force, I had the pull at that direction. There are two degrees, there is my pull. I go along my line where I cut it out so nothing happens here. And here nothing happens here. And I hear something is happening. I cut the way the surface. Whenever I have a surface I will get a normal force which points away from the surface. And I will get a friction which is perpendicular to the surface. Now which way does the friction go? The friction is always trying to oppose the sliding. The sliding, not the motion, please remember I start the right way. The friction is opposing the sliding. Here the possible sliding will be if there will be no friction the box will move to the right. Therefore the friction moves to the left. Here is my friction. Now here on the earth, or we assume that this box is on the earth, so we will have gravity which according to me, the new thing will always point straight down. So the first step of solving is probably the most complicated step. Because if you make a mistake in your free body diagram, whatever follows, you will not be able to solve it. But if your free body diagram is right, everything should work out for you. So the second step, the first step was the free body diagram. The second step is to select the coordinate system. Usually what the coordinate system where as many forces as possible are aligned. So in this case my regular coordinate system is to the right and right are to just work out fine. So after my free body diagram I have my coordinate system. Next one, what knots are involved? So here I have gravity, so I do know that FG is MG. And I know that there is no acceleration, so I do know that some of all forces must be zero. So this actually means first of all, and this is the last limitation. Now fourth step, we are splitting this up in components. So what I am doing, I am going to split my pension half, I am going to write my x stuff here, and my y things here. And then I am going to go force by force and write it down where it adds. So first year my push or my pull has x component and the y component. So writing here and here. My normal force is only in the y direction, so plus F. Normal. My friction is only in the x direction. And my gravity again is only in my y direction, so plus FG. And what knots apply, well, it is first of all motions. So some of all forces in x must be zero, and some of all forces in y must also be zero in this case. Therefore all of those is zero. So I simply translated the first law of motion. All forces added up as vectors must be zero. All forces added up as vector must be zero. And I have no signs in yet because I kept them as vectors. As long as I keep the vectors, the direction doesn't matter. If I look at the magnitudes of reach components, then I have to put the sign in front of it. So in this case the push is in positive direction. The friction is in negative direction, zero. The push in y direction is positive, y. The normal force goes in positive direction, so up. And my gravity goes in negative direction. So negative. Next step, I'm going to plug in what I know. I know that my fg is mg, so I can put that one in. So my g is zero. I also can do some trigonometric here. So I know my x component, one down here, will be my push times cosr of 30. Nice friction is zero. And on the other side, for my y component of the push I add sign, so p and sign 30 plus fm, minus mg is zero. But now I can solve. So friction on the other side, so friction must be push times cosr of 30. So friction is push times 20 newtons, 20 newtons, times cosr of 30, which gives me 2 times cosr of 30. This means 17 and 3 newtons. And similarly for this side here, I can solve for the normal force. So I have that the normal force is equal to the take for the flow of the gel and g, and minus p and sign 30. Actually quite interesting, often people think that the normal force is equal to gravity, it doesn't have to be the case. Here the pool is lifting the box a bit, so the normal force is less than what it would be, if there would be nobody lifting on it. So normal force is mass times gravity, so if I take 10 newtons, let's go 10 kilograms times around 9.8, newton per kilogram minus 20 newtons and sign 30 is 0.5. So I have 10 times 9.8, gives me 98 newtons minus 10 newtons, gives me 88 newtons. Okay, I have solved. Now before I go on, so 5 solve it, and then 6 check if it makes sense. Check it. What do I mean by check it? Do my answers lie in the order of magnitude of the forces I had? You had about 20 newtons, 1788, I think we'll dare. In front of there would be like thousands, I would have to check if something is wrong. Another thing I can check is when I split it up in the components, for each force I have a one side cosar on the other side of the same angle. It cannot happen that you have cosar 30 and cosar 30 for the same force in X and Y. So we can double check that if a cosar here must have side of the other one. Another thing I can check is, are my answers positive? I solve for the magnitude, see it's not vectors, I solve for the magnitude, and they should not be negative. If math would end me, if it's something negative, then it means I've got the free body guy in line. Actually, if I were to rewrite, if somebody asked for the friction as a vector, so the friction as a vector, would in this case, see my code, see a system by my a centimeter by three, in X and nothing in Y, in U terms, I have the normal force as a vector, would have nothing in X and 88, in Y, in U terms. And I have solved my first simple, interesting solve problem.