 In this video we're going to be looking at an easy example for Newton's first and second law. What we have here is a box on wheels that we assume to be perfect so there is no friction. And then there is a force pulling on the box with 50 Newtons at an angle of 30 degrees. So the first step as always with this type of problem is to draw the free body diagram. So I'm going to be isolating my box and whenever I cut through something there will be a force. So I'm cutting here through the force. If I do my free body diagram means I will have a force F here at an angle of 30 degrees. Then I was cutting away the surface which means I will have a normal force on the friction. Well friction in this case is zero so no friction but we have a normal force always away from the surface towards the object. So my normal force goes up here. And then as we're on earth we have the weight or the force of gravity straight down. Fg is straight down. And that's my free body diagram. Now what I can add is I can assume which way I think it's going to be accelerating. I think this box will actually to the right. So I draw next to my free body diagram not connected on it my acceleration. So first step is done. Free body diagram is done. Second step choose a coordinate system. So here it looks like a regular coordinate system. We'll do the trick. First step the law is involved. So I have gravity so I will have the law of gravity Fg7g. Well the title kind of gives its way. I'm going to assume that there is in its first law involved. Some of the forces must be zero then the acceleration must be zero. And Newton's second law must be involved. So if the sum of all forces is not zero then the acceleration is proportional to the force with the length. So those are there. Step four is splitting the whole thing up in components. So I have x component and I have the y component. So in y direction whether my force in y direction I have the null force goes in y direction. I have gravity in my direction plus Fg. And I have the y component of my force here. So plus force y is well in y direction. Do I have an acceleration in y direction? No. My acceleration is in x direction. So the law for y should be that the sum of all components in y must be zero. I'm assuming that my box will not move up and down. It will only start sliding to the right. In my x direction however I assume I have an acceleration in the x direction. So some of all forces in x is not zero but in A. And then what are my forces in x direction? Well all in my pool. It's a different color. x component here is equal to mA. Next step I'm looking at the direction. It's my x component in my positive x direction. So again series yes so plus. And how can I calculate my component Fx based on my force? This is the adjacent of it. So I have force times cosine of the angle. So in this case 30 degrees is equal to... Now I have to check this my acceleration in positive direction. If yes I put a positive. If my acceleration will be the other way into negative direction I will have to put a negative here. In A. On my y direction I have my normal force going up. I have my Fg going down. I'm plugging my equation Fg is minus mg. Then I figured out my y component is positive going up. So plus F. And now how do I figure out this one? This must be sine of the angle. Like it's so. So I have 15 new tens. Or I want to solve for the acceleration. So let's solve it first by letters. That acceleration is force. Force times cosine of 30. And mass. So my acceleration will be 15 newtons. Times cosine of 30 over 10 newtons. Which gives me 4.3 meters per second squared. On the other side I will solve for my normal force. So my normal force is equal to mg minus my force from sine. Trusting feature here. Often people think that the force is always equal to gravity. It doesn't have to be because here I'm pulling a bit upwards. Therefore my normal force is less than the weight of the object. So in my case I had a mass of 10 kg. Times 9.8 newtons per kg. Minus 50 newtons times sine of 30. So what do I get? I get 98 newton minus 10.5 newton. Let's double check. 73 newtons. So components are checked. Then 5 actually are resulted. So that's done. 6 check it. Thus the answer makes sense. So I had forces in the range of 50 and 100 newtons of weight. 98 newtons of weight. So I think the magnitude makes about sense. I had a mass of 10, 50 newtons. So 50 by 10 would be 5. So also my situation seems to make sense. Another check I can do is I have cosine on the left side. That means I have sine on the right side. Yes that is true. I can also think what would happen if I would pull directly to the right. If I would pull directly to the right I would have force times cosine of 0. Cosine of 0 is 1. So all my force would then be translated directly to next iteration. And sine of 0 is 0. So then my force would play here. So if I would pull only horizontally my normal force of people to the width. Last thing I could check. I could check the units. Newton per kilogram is equal to meters per second square. So newton per kilogram. A newton is a kilogram tens of meters per second square divided by the kilogram. So yes I get each percent square. So I am very confident that my answer is correct.