 This is a classical example of a problem using Newton's laws. There is a box sitting on a surface and there is a push applied from the top left down on the box. What we're interested in is finding the reaction forces, namely the normal force and the frictional force. We're going to start by drawing the free body diagram or the isolation diagram. To do so, I'm cutting my box out of my drawing here and wherever I cut through something, I replace that with a force in the free body diagram. So here I have my push. Down here, I was cutting through a surface. So whenever I cut through a surface, I'm gonna end up with a normal force from the surface away towards the center of the object. So I have a normal force going upwards and I also get a friction that is in the plane of the surface. Now the question is which way does it go? Friction usually tries to avoid any slipping. So in this case, in order to keep the box in place, the friction has to be to the left. I'm pushing it to the left and whenever we are talking about problems on the surface of the planet, we also have our weight or the force tube to gravity. So this is my free body diagram of this object. Next, I should specify what coordinate system I'm using on to solve the problem. I'm gonna go here with a classical x to the right and y up coordinate system. Now before I start solving this problem, I'm gonna write down any additional equation that I might know. For example, I know that my FG is equal to mass times gravity and I know that my friction, if it's static friction, it's always smaller than my coefficient of friction times my normal force. Just in case I need it. Now I'm gonna split up my problem in two directions. I have an x direction and I have a y direction. In x direction, as the thing isn't moving, I have Newton's first law of applying some of all forces in x is zero and the same in this case is happening in the y direction. Some of all forces in y is zero. Now I'm gonna look what directions are my forces in. If I start with here my normal force, my normal force is in the y direction. So I put it on the y-list. Normal force is in the y direction. Then my friction is in the x direction. So friction goes here. My gravity goes in the y direction and finally my push has both x and y components. So the push x component goes in the x direction. And the push y component in the y direction. Now I'm gonna think about the directions. Let's start with the x is my frictional force in the plus or negative direction. It goes against my x-axis. So I'm gonna put it as minus magnitude of the friction and my push goes in the positive direction. So plus my push x component is zero. So I would know that my friction or it doesn't even have an x component. So I could erase this. It's all of my friction. My friction is equal to the push x component which is equal to the push times cosine of in this case 30 degrees which is 17.3 Newton. I'm going on for the same thing in the y direction. Y direction. I have my normal force which goes with my y direction. I have my gravity that goes against it. So minus and I can plug in my equation Mg and I have my push y direction which is minus and my push times sine of 30. So I have that my normal force equals mass times gravity plus the push times sine 30. So if I calculate this I get assuming let's take some mass because mass was actually not given. So there was an error in the problem here. Let's assume 5 kilograms. So I have a normal force of 5 times 10 meters per second square plus the push times sine 30, 10 newtons equals 60 newtons. Now I was not asked for it, but I could actually calculate my minimal coefficient of friction needed because I know that my static friction has to be smaller than my mu s and fn. So my mu s has to be bigger than my friction over my normal force which is 17.3 over 60.