Easy Resolving Vectors Trick - A-Level & GCSE Physics

Okay. So vectors are some of those things that confuse people real easy. So I'm just going to show you how to deal with them really easily. So you don't get confused with them ever again. Now, obviously this comes up mostly when you're doing mechanics. Um, that's your forces and your velocities and work done and all that kind of thing, but you can use them, uh, with pretty much any vector. Okay. So here we go. Let's say we have, uh, two vectors. Uh, let's say that we have a car that's right here and it's pushing forwards with a force of 50 Newtons. Uh, but there's a crosswind. And so that means that there's a force perpendicular to that at 90 degrees. And that's going to be a 10 Newtons. Now, if you want to find the resultant, the overall force, then you need to make a triangle. Obviously, uh, the resultant vector, if we add them up, it's going to look something like this. And that's because when we move this 10 Newtons over here, this is what we get a nice triangle. Don't forget that you can move vectors wherever you want. They are the same size and the same direction. It doesn't matter actually where they are. So now we have a right angle triangle, obviously to find out the resultant, all we have to do is use Pythagoras. But of course, what if we want to try and find the angle or the direction we're going to call that theta. Now, whenever you find an angle, you always have to do the inverse sine cos or tan of, uh, two of the sides of your triangle as a rule of thumb, use the sides that you've been given in the question, not one that you've just recently calculated. So in this case, I know I've got the opposite and I've got the adjacent. So I'm going to do the inverse tan of 10 divided by 50. So that's the inverse tan of 0.5. And that gives me 11.3 degrees. Now when you give your answer for a direction in a question, make sure that you're putting explicitly down which direction you're talking about. So this angle is here. It's 11.3 degrees from the horizontal. So I'm just going to add that in there as well from horizontal. Obviously, if I did this angle instead, let's call that phi. Then I would say that so many degrees from the vertical, either one of those would be acceptable. So it's nice and easy if you have two vectors and you want to find the resultant. However, what about if you have one of the vectors and you have, or rather you have the resultant vector there, let's say that that's, uh, 15 meters per second. And also we have an angle. How do we find out what the vertical and horizontal components of this velocity are? I'm going to call this one here V H. I'm going to call this one here V V. Okay. Velocities but one is horizontal. One is vertical. So there is a way that I can do this with the angle. Obviously I can turn this, I can make this into a right angle triangle, and then I can use my SOHCAHTOA sine, cos and tan to find out what these two sides are. However, there is a much, much simpler and quicker way to actually finding out what components are of vectors. Here's the thing is that components are always smaller than the resultant. Now that might seem a little bit obvious and a little bit Mickey Mouse, but there's a good reason why I say that. So we know that sine of any angle or cos of any angle, they're always going to be between zero and one. Now, okay, technically they're minus one and one, but we don't really deal with obtuse angles in physics. They're just a ratio sine theta, cos theta. They're just a ratio that just tell you how much bigger or smaller something should be than something else. So if we want to find out a smaller number, that is the component of our resultant, we obviously want to times by sine theta or cos theta. If we divide by sine theta or cos theta, we're going to get a bigger number. And that's how you get a resultant from a component. So how is this useful then? Well, what we can do is just say automatically that we're trying to find the components of 15 meters per second. So we either times by cos or sine. However, do we use cos or sine? Is there a quick way of actually knowing how to find that out? Well, yes, there is. Um, if you turn through the angle, we use cos theta. If you turn away from your angle, you use sine theta or rather than more general terms should be. If you turn through the angle, it's cos. If you don't turn through the angle, then it's going to be sine. The Bible says, turn away from your sin to be saved. I guess that's one way that I, that helps me remember which way round it is. If I turn away, then obviously I'm going to be using sine. So just to see how useful this is, let's just have a look at how quick it can be done here. I have a force of 12 Newtons, but whatever object is being pushed here is traveling in this direction of a distance of two meters. Now we know that work done equals F times D force times distance, but that's only true if they are parallel. Therefore I need to know how much of this force of 12 Newtons is actually going in the direction that the object is moving in. There's a real easy way of doing this. Therefore I can just say work done equals two. That's my distance times the force of 12. And I know that whatever force is going in the direction is going to be smaller than 12 Newtons because that's my resultant. But am I going to be a timesing by cos or sine? Well, to get to the distance from my force, I'm turning through the angle. So therefore I know I'm going to be timesing by cos theta. Let's say that this theta is 10 degrees. So I just punch into my calculator two times 12 cos 10 and that gives me an answer of 23.6 joules. Let's just have a go at one more, but doing the opposite. So let's say Bob here wants to go north by 100 meters, but the only direction he can walk is actually this direction here. If I wanted to find out how far he's gone, then obviously I'm going to need to use the angle in between them. Now let's say for instance, that in fact, this time I don't know the angle in between, I actually know this angle here. That's actually going to be 25 degrees. So let's just have a think about this. Here's my component of my resultant and I'm actually trying to find out the resultant. Therefore, I'm actually going to have to divide by cos 25 or sin 25 in order to find out a bigger number. That's my resultant question is, am I going to be using cos or sin? Well, am I turning through the angle? No, I'm not. I'm actually at the angles there. So I'm not turning through the angle. So that means I'm actually almost turning away from the angle. Therefore turn away from your sin, turn away from your sin. So there we go. That's how we find it out. All we have to do is find out, let's say this is D, D equals 100 divided by sin 25. We're dividing because we know we want a bigger number and we're using sin because we're not turning through the angle. We're turning away from the angle. Punch that into my calculator. That gives me a distance of 237 meters. And there you go. That's the quick and easy way to find out a component or a resultant vector from an angle that will save you a lot of time in lessons, in with your homework and in exams as well. And it will avoid all sorts of confusion. Let's just have a look at one more. Here is a brick on a ramp. Now we know that it's weight mg pulls downwards like so. Now this is the example that confuses the heck out of people. So this is one of the best uses of this quick rule. Now of course we know that this angle here is going to be theta. So first thing you do, whenever you come to a question that has anything on a slope and we have the weight going straight downwards, we want to resolve that into components that are parallel and perpendicular to the slope because we don't care about what's going on anywhere else. So there is my component that's parallel to the slope that will accelerate it down the slope. And then we've got the one that's perpendicular to the slope as well. Now if this is theta here, then you can automatically say, oh, this is theta here as well. And we know that both of these are going to be smaller than mg. So let's just have a think, have a go at it yourself, use the rule. We're going to be finding smaller components. So are we going to be timesing or dividing by cos or sin and which one is going to be cos and which one's going to be sin. Pause the video, give it a go yourself real quick. Okay, you're back. We know that this component is going to be smaller than mg and we're turning through the angle to get to it. So that's going to be cos mg cos theta. We're taking our mg and we're just making it smaller by timesing by cos theta. And of course, if this one is cos, this one obviously has to be sin, but let's just check it really quickly to get from here to here from the resultant to this vector parallel down the slope. We're turning away from the angle, turn away from your sin. So therefore that's going to be mg sin theta. So obviously you can see how quick and easy this method of dealing with vectors is. And I guarantee you that it will avoid any confusion that has caught you out before. And it will save you a heck of a lot of time in your exams and your work as well. Hope you found this useful. If you did, don't forget to leave a like, I'll see you next time.