 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 in this problem here in 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 newtons but we're not going anywhere. So no acceleration and no speed and 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. What do we do with the free body diagram? We're isolating what we're interested in and we're cutting it out of the environment and wherever I cut I will have some force acting. For example here I had a force so what I'm going to be doing is I'm going to simplify my object my box I don't really care that it's a box I can simplify it a dot and here I had a force I had to pull that went in that direction 30 degrees that is my poach. I go along my line where I cut it out so nothing happens here nothing happens here now here something is happening. I cut away 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 memorize that the right way the friction is opposing the sliding here the possible sliding will be if there will be no friction the box would move to the right therefore the friction moves to the left it's my friction now we are on the earth or we're assuming that this box is on the earth so we will have gravity which according to Mr Newton will always point straight down so the first step of solving this problem is done and actually 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 fine so second step so first step was the free body diagram second step is select the coordinate system we usually want the coordinate system where as many forces as possible are aligned with it so in this case my regular coordinate system is to the right and y are to just work out fine so I have 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 the sum of all forces must be zero it's actually Newton's first love motion and this is the last gravitation now fourth step we're splitting this up in components so what I'm doing I'm going to split my page in half I'm going to write my x stuff here and my y things here and then I'm going to go force by force and write it down where it adds so first here my push or my pole has x component and the y component so writing it here and here a 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 not does apply well it's first stop motion 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 I didn't put any signs in yet because I kept them as vectors as long as I keep the vectors the direction doesn't matter yet if I look now at their magnitudes for each 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 that push in y direction is positive why 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 could put that one in so minus g is zero I also can do some trigonometrics here so I know my x component and now here will be my push times cosine of 30 minus friction is zero and on the other side for my y component of the push I add sine so p and sine 30 plus fn minus mg is zero but now I can solve so friction on the other side so friction must be pushing times cosine of 30 so friction is my push was 20 newtons 20 newtons times cosine of 30 which gives me times cosine this means 17.3 newton and similarly for this side here I can solve for the normal force so I have that the normal force is equal to the force of gravity or mg and minus p cosine 30 actually quite interesting often people think that the normal force is equal to gravity doesn't have to be the case here the pull 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 sine 30 0.5 so half 10 times 9.8 gives me 98 newtons minus 10 newtons gives me 88 newtons okay I have solved it now before I go on so five solve it and then six check if it makes sense check it what do I mean by check it like do my answers lie in the order of magnitude of the forces I had here I had about 20 newtons 17 88 I think we're there if one of them 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 cosine on the other side of the same angle it cannot happen that you have cosine 30 and cosine 30 for the same force in x and y so we can double check that if I've cosine here it must have sine on the other one another thing I can check is are my answers positive I solve for the magnitude see it's not vector so I solve for the magnitude and negative should not be negative if math would end me give me something negative then it means I put the freeway diagram right actually if I want to rewrite if somebody asked for the friction as a vector so the friction as a vector would in this case say my coordinate c system be minus 17.3 in x and nothing in y newtons the normal force as a vector would have nothing in x and 88 in y newtons and I have solved my first simple uh winter slug problem