A LEVEL PHYSICS REVISION: VECTOR MECHANICS EXPLAINED

hello and welcome back to another one of these a-level physics videos today we're going to be looking at the vector mechanics part of the mechanics um section so again this will also help with the a-level mechanics just as like an added little bonus so with vectors as we've said before they are things that have a size and a Direction so things like acceleration velocity Force Etc so we've normally up until now in GCSE for instance we've been dealing with vectors that go kind of horizontally vertically but the key Point here is that we always treat them in pairs right so they're either going up and down or they're going left and right now in actuality if you think about real life most of the time forces act at some angle Theta to a horizontal right they don't always act you know perpendicular to each other or just to the left just the right up or down so what we need to be able to do and this is an eight of marathon physics is we need to be able to break up these forces into components Okay so what we can do here is I can break this Force up into two perpendicular components a component that goes to the right and a component that goes up so just to explain this a bit in another way this Force here is acting it it has a mixture of components a component that makes it go to the right and a component that makes it go up which means the overall effect is that it goes up and to the right as a diagonal okay now the way we work this out is just using simple trigonometry right this is a right angle triangle so if we have the hypotenuse and we want the adjacent side we'll be using cos so we know that cos Theta equals F over the adjacent side which means this side here is just equal to F cos Theta okay so actually that's the other way around so a over f okay likewise here you can show that it would equal F sine Theta because again we would have sine theta equals the opposite side over F and therefore you rearrange and honestly that's all there is to it so you just need to keep in mind that when you're doing something like this this is a skill that comes up again and again again in physics and in maths if you are going to be finding the resultant force of kind of forces acting on an object you need to resolve them into perpendicular and parallel components so for example let's say we have a box right and there's one force that's acting directly to the right let's say 10 Newtons and then we have another Force which is acting up and to the left at 20 newtons an angle of 30 degrees if we were to try and resolve this um force and work out the resultant Force I can't just do 10 minus 20 right or 20 plus 10 or something like that whereas normally if I had or in the kind of old way of doing things my forces would always be acting parallel to each other so then work out the resultant force is a trivial matter of doing you know the difference between 5 and 10. and again that would be if the right is positive remember we took directions as positive in the super video as well so my result in this case be negative five but I can't do that in this case why because I have two components so I have to actually resolve this into parallel and vertical components so two perpendicular components so let's say if I take to the right is positive my actual resultant Force if I'm just taking horizontal components into account is going to be the 10 Newtons here minus this adjacent Point here because as you can see this is obviously going to be traveling to the left this is going to be traveling to the right and if I had a force that was going down then this component will be going down and this component will be going to the right okay so then that would be adjacent so that should be 20 cos 30 and again just to illustrate that we have that cos 30 should equal that adjacent side over the hypotenuse which is 20. okay so that's minus 20 cos 30. and then you can work out whatever that is so if we get our calculators out we have 10 minus 20 because 30. I also need to check if I'm in radians mode it looks like I am so I need to change the angle into degrees equals and it gives me a positive a negative number sorry so minus 7.32 Newtons so what does that mean well 7.3 Newtons since it's negative it's going to be traveling in this direction that's my resultant Force horizontally only and then obviously you'd have to include the vertical component as well so if we resolve this vertically well we only have one force right and that force and if we take upwards is positive because we only have one force is just 26 sine 30. which we can work out it's 10. I should have actually known that I didn't actually read the angle properly 10 Newtons so what does this mean well the overall resultant Force the actual magnitude of it if I were to work it out is a vector that has two components so to actually work out what the value is is all you'd need to do is take the magnitude of the vector which is adding them together and squaring them which again should make sense because remember we have a component of the force going this way and a component going this way this component is 10 this component is 7.3 and whenever you're working out the resultant force is you're just working out the hypotenuse of that triangle like that so we're doing is adding the vectors together this doesn't make sense to you don't worry too much because we have got a couple of questions that are going to help out with this vectors topic so don't worry too much but again with practice this will become a lot easier but in general just expect that in physics and maths we're going to be kind of resolving vectors into their perpendicular components okay so that means the overall resultant force in this system is 12.4 Newtons acting to the left and up okay so let's get started with some questions shall we so again we have some multiple choice questions let's get started which of the following is a vector quantity so easy way to work this out this is energy this is energy and that's time none of those are vectors so that means momentum must be and again uh one way to work that out if you're unsure is to always look at the equations so we have P equals MV V is velocity which is a vector Times by a scalar so that must be a vector okay same with work done is force times distance um well it's not always four times distance and then you'll have kinetic energy but energy is not a vector because it doesn't have a direction right I don't say I have 10 joules downwards do I but I do say with momentum I give a Direction next is an object is in equilibrium under the action of three coplanar forces so coplanar just means they act on the same 2D sheet F R and W the diagram below shows the forces F and W the angle between them is 90 degrees which row shows the correct magnitude of R where R is the resultant Force right so all we need to do is if we draw these forces together right and it also gives us the um the values here now for this object to be in equilibrium its resultant Force needs to be zero so basically we're going to join this line up here to go back because this kind of closed system here will have a net force of zero the other way to look at it is if I were to draw this object out its weight is going down by 4.5 and it's being pulled to the left by 6.2 right so that's W and that's f so it's been equilibrium we need to cancel out that as those forces so one component is going to be to the right by 6.2 and the other component will be going up by 4.5 right because for there to be no overall force there needs to be a force that counteracts W going down and F going to the left okay so those are my two components of R okay so the 4.5 and 6.2 so now all I need to do is I'm really hoping this grouped all together yeah it did drag those over and this is my actual diagram for r that's what it's going to be where 4.5 is actually um it's not super clear but it's the upwards part here okay so just to go over my method for every force that was acting on this object I drew a force in the exact opposite direction with the same magnitude to cancel it out because equilibrium means it's not accelerating there's no resultant Force there's no force acting on the system overall then all I did is I made a triangle out of them now if you notice that gives me the direction so it must be one of either this one or this one up into the right now all I need to do is work out the magnitude well the magnitude is just going to be Pythagoras okay so we have again a triangle those lengths and we're just trying to work out the third side and R if you remember is just going to be 6.2 squared plus 4.5 squared so again with questions like this it might seem fairly easy once you've practiced right it might not it's not going to be easy for those of you seeing this for the first time but for those of you that have seen this before you might think it's quite easy bear in mind a lot of students made a mistake on this question because they rushed okay it's very easy for them to think R is a resultant Force for example and they would have picked one of the ones that are going down if you notice the magnitude is the same for each pair right so you have one magnitude and two directions another magnitude in two directions they picked the wrong direction but got the right magnitude so in this case the right magnitude was 7.7 Newtons which which means the row is a the correct one this one here but they picked the bottom one as in b because they looked and they saw that the forces were going down and to the left so they thought that would be the resultant force that is the resolved force of those two but R isn't a resultant Force it's the one that's making it in equilibrium so they completely breezed over this word here equilibrium which means there's no result Force which means all forces are canceled out with each other okay so that's where the mistake was made we have a very similar question from Edexcel so our plane is traveling north at 42 meters per second into the region with wind blowing from the East at 7.5 meters per second so to keep traveling straight up north it turns towards the east without changing thrust okay so again if you imagine it if you're going straight ahead and something hits you from the side you're going to start moving this way so to counteract that you turn into that so you can keep going in one straight line he says which of the following can be used to calculate the angle through which the plane turns well if you notice the angle that it turns is just going to be this angle here so if I draw this a bit more simply we have the 7.5 acting this way I mean the 42 acting this way so the angle through which I had to turn remember the plane at first was looking upwards yeah now this is actually a really important thing I'm going to work out this angle here and that's going to give me my value for how far it has to turn okay now the two sides that I have are adjacent and hypotenuse so to work out Theta here it's just rudimentary trigonometry so Theta should equal inverse cos 7.5 over 42. right adjacent size over the hypothesis so the question is why is that not one of the answers here okay so if you think about it and this is where we have to be careful that angle will give me Theta over here however I'm actually looking for the direction that it turns which would be this angle how do I get that angle well I do 90 minus inverse cos 7.5 over 42. does that make sense so again with these one mark multiple choice questions they look easy but they're designed to trick you so working out this angle is rudimentary but working out the angle that it turns is a lot hot it's not a lot harder it's just that you need to realize that it was pointing North and it goes to this direction so working at this inside angle it's just you just need to do 90 minus it now as you may know the COS graph is just a translation of the sine graph so this actually is the exact same as just doing inverse sine 7.5 over 42. make sense because again the COS graph is just um 90 degrees ahead of the sine graph so if you take away 90 from that so it's the other way around the cosgraph is 90 degrees behind the sine graph so if we're adding 90 to it it's going to just become the sine graph so it just looks like this okay and lastly we have a man who's walking at a constant horizontal velocity of 1.2 meters per second so let's draw that in the rain to the man the rain appears to be falling vertically at a velocity of 1.8 meters per second so like that okay now here's the thing the reason why this question is a bit weird and it's worth five marks is it takes a little bit of thinking because when you're walking in the rain the rain doesn't actually appear to be falling straight down right it appears to be Falling Towards you if you think about rain if it's falling straight down your face should never get wet only your head but as we all know through walking in the rain the rain hits Us in the face that's because we're walking into it so it's as if we are standing still and the rain is moving towards us when in actuality we're walking into the rain this is a concept called relativity which means that it's kind of indistinguishable which things are moving so let's say if I was running at you at 20 meters per second physically speaking that is the same as you running towards me at 20 meters per second or both of us running towards each other at 10 meters per second it would appear the same to both of us now of course as human beings you might think that's really weird because we have other points of reference right because I can look like for example I know I'm moving because I can see other buildings and stuff moving around me and I can feel my arms running but if somehow you're in the cold vacuum of space there's nothing around you right and I pushed and you were pushed towards me at 20 meters per second it would be the same it would look the same as if I was moving towards you at 20 meters per second so the only reason why we know that we're moving towards each other which one of us is moving is because we can see body parts moving right but in actuality it doesn't matter so for it to appear that the rain is falling straight down he must be walking in the same direction as the rain that's why the question seems a bit weird but to be honest you don't actually need to know any of that so we're trying to basically work out this resolved velocity here and that's it um so yeah it's a five mark question but keep in mind you need to actually draw it to scale which means let's say you wrote that 0.6 centimeters equals uh what um sorry one centimeter is 0.6 meters per second then this top thing should be then this top side should be two centimeters long and this side should be three centimeters long and then you need to be able to measure this third side in order to work out your value I'm going to cheat quickly because I'm just going to um use a formula instead but it's the same concept so again even without actually knowing what I was talking about with in terms of Relativity and all that fun stuff you can still answer the question perfectly fine just use the maths and you should measure it at 2.16 so technically we should give it to two significant figures or one decimal place because everything else is given to one decimal place okay well they wanted you to really do is actually measure it afterwards but again that question kind of confused people because they put the arrows in the wrong directions and things like that but this is completely fine and the correct answer and I'll see you in the next one