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Current time:0:00Total duration:9:23

AP.PHYS:

INT‑3.B (EU)

, INT‑3.B.2 (EK)

, INT‑3.B.2.1 (LO)

let's say we have some type of hard flat frictionless surface right over here that's my drawing of a hard flat frictionless surface and on that I have a block and that block is not accelerating in any direction it is just sitting there and let's say we know the weight of that block it is a 10 Newton block so my question to you is pause this video and think about what are all of the forces that are acting on this block all right now let's work through this together and to do it I'm going to draw what's known as a free body diagram to think about all of the forces and the reason why it's called a free body diagram is that we just focus on this one body and we don't draw everything else around it and we just draw the forces acting on it and there's actually two typical ways of drawing a free body diagram I'll do them both so first I could draw it this way so this is my block here now I told you that it weighs ten Newtons the weight of an object that's the force of gravity acting on that object and it would be downwards so we have from this 10 Newtons right over here we know that there is a downward force the force of gravity acting on the mass of this object of 10 Newtons it has a magnitude of 10 Newtons and the force is acting and downwards we could say this is the magnitude of the force of gravity and when you draw a Freebody diagram it's typical to show your vectors originating out of the center of that of that object in your drawing now my question to you is is that the only force acting on this if you think it is what would happen to the object well it would start a core would be accelerating downwards but I just said that this is not accelerating in any direction so there must be something that counteracts this and there is there's the normal force of this surface acting on the block that surface is what's keeping the block from accelerating downwards and I will do that with this vector right over here so it's going to be going upwards and it's going to have the exact same magnitude just in the opposite direction so I could say the magnitude of the normal force the normal force is magnitude right over here is also going to be equal to 10 Newtons going upwards and we can see that these two are going to net out and so you have no net force acting in this vertical dimension and I have no forces I haven't thought about any or drawn any in the horizontal direction and so that's why you have no net force in total and this thing isn't going to be accelerated now I mentioned that there's other ways to draw Freebody diagram another way that you might see it is like this where you see the body and from the outside of the body you see the vectors being drawn so in this situation you have ten Newtons downward and you would have ten Newtons upward this is another way that you might see free body diagram free body diagrams drawn now what I want to do is do something interesting to this block let me redraw it so I have my surface here my hard frictionless surface and it's flat and I still have my block here it's my 10 Newton block but now I'm going to apply a force to it I am going to apply a force like that is in this direction it's in this direction and it's magnitude let's say it's magnitude is 20 Newtons and just so that we know the direction this angle right over here let's say that that is 60 degrees I'll say theta is equal to 60 degrees the magnitude of this force is equal to 20 Newtons so what would the Freebody diagram now look like well it might be tempting to just draw the force right on one of these Freebody diagrams something like that something like that and that would not be inaccurate but we would have to watch out because this force is acting at an angle so if we were to break it up into its horizontal and vertical components some part of that force is acting downward and so you're actually going to have a larger normal force to counteract that and some other component is going to be working horizontally and so what we want to do is actually break things up because if you leave it in this angle it gets very very confusing so what I want to do is I want to break up this new blue into its horizontal and vertical components and to do so we just have to remember a little bit of our basic trigonometry if I have a force like this if I have a force like this and it is acting at an angle theta right over here with the horizontal and I want to break it up into its horizontal and its vertical components and its vertical components if the magnitude of the hypotenuse is capital F then the magnitude of the adjacent side to this angle this comes straight out of sohcahtoa from our right triangle trigonometry it would be the magnitude of our hypotenuse times the cosine of this angle and the magnitude of the vertical component that would be the magnitude of our hypotenuse times the sine of that angle and you could think about it the other way as well if the force was like this where it's just going in the opposite direction but once again you have this angle theta and now the components would look like this where the vertical component would have the same magnitude but now would be pointing downwards and the horizontal component would have the same magnitude but now it is pointing to the left it is the same idea if this force has magnitude F that's represented by the hypotenuse of this triangle then the magnitude of our horizontal component is still going to be F cosine theta the vector is not going in the other direction and the magnitude of our vertical component here is going to be F sine theta and so what about this scenario over here well in this scenario our vertical component is going to look like this and our horizontal component is going to look like this and so what's the magnitude of our horizontal component well it's going to be the magnitude of our hypotenuse times the cosine of the 60 degrees so this is going to be 20 Newton's times the cosine of 60 degrees and its really helpful both trigonometry and physic classes to know the values of your sines or cosines and your tangents at angles like zero degrees 30 degrees 60 degrees 90 degrees 45 degrees you could use a calculator here but it's useful to know that the cosine of 60 degrees is 1/2 so 20 times 1/2 this is going to be equal to 10 Newtons and if we want to know the magnitude of our vertical component well this is going to be 20 Newtons times the sine of 60 degrees once again this is useful to know it is square root of 3 over 2 and so 20 divided by 2 is 10 so this is going to be 10 square roots of 3 Newtons and so we can use that information we've broken up our original vector into two component vectors that if you took their sum you would get your original one but what's useful now is that we've broken it up into vectors that are parallel or perpendicular to our surface which will allow us to think about what nets against these things that I already have in place so let me draw that so actually I'll first draw this type of Freebody diagram so there is my block and I have I have the force of gravity acting on it downwards I will draw it right over here so that's 10 Newtons that is the force of gravity acting downwards now is that the only thing that I have acting downwards no I also have the vertical component of this applied force and so this is going downwards 10 Squared's of 3 Newtons and these aren't drawn perfectly to scale but hopefully you get the idea so this is 10 square roots of 3 Newtons just like that and now what is our normal force assuming that our surface is able to not be compressed in any way that it is a hard frictionless surface well now our normal force is going to counteract both of these forces our normal force might look something like this once again I haven't drawn it completely to scale but this would be 10 plus 10 square roots of 3 Newtons and what about now in the horizontal direction well I have this blue vector right here and so that is going to the right with a magnitude of 10 Newtons 10 Newtons and so now you can hopefully appreciate my Freebody diagram is really really really useful just looking at this I can predict what's going to happen to my block I would say look this upward force is completely netted out by these downward forces or these downward forces are completely meted out by this upward force and the only net force that I have is ten Newtons to the right and so that lets me know that hey since I have a net force to the right this block is going to accelerate to the right