hey guys welcome back to bison maths this is the second in a series of videos that i'm doing to help you memorize everything that you need for your maths a level for excel the first one was on pure it's on my channel page if you haven't checked it out yet go and have a look for it this one's gonna be on mechanics and i'll be doing one soon on statistics if you haven't checked out my channel before you should go and see that i've got hundreds of videos of my real life lessons with students at my school and it's a different kind of way of learning it's like being in a classroom rather than just watching a video all the videos are organized into playlists and they all match up with the official pearson edxl textbooks and i've got nearly all of the chapters done by now there's also stuff there for further maths so if you're taking further maths go and check them out and of course if you like this video please do hit the thumbs up and do subscribe to my channel to get more of these as they get released so for this mechanics video people usually think that there are lots and lots of different topics and mechanics but what i'm going to really try and convince you of is that there are actually just two big topics there are kinematics which is about how things move and there are also forces which are all of the things like connected particles lifts and moments if we can break it down into those two types of topics and then discover what skills we need i think mechanics is going to be a really strong area for you so i hope you agree with the way that i've broken things down here let's get on with the video okay so you can see here that i've organized my memory page for mechanics into these two sections really there's one for kinematics which is about how things are moving and then there's one for forces so we're going to get started by having a look at the kinematics section that we have up here now when you have a question in mechanics one of the first things that you should really do is decide is it talking about something with constant speed does it have constant acceleration or does it have a variable acceleration because that's going to determine what kinds of skills that you need so we'll get started by having a look at something that has got constant speed in other words the acceleration is equal to zero so we know from gcse the distance is equal to speed multiplied by time speed times time and sometimes you need to convert between kilometers per hour in meters per second so to convert from the kilometers to meters you multiply by a thousand and then to go from the per hour to the per second well you would divide it by 60 to find out what it is per minute and then by 60 again per second so you multiply by a thousand divide by 60 divide by 60. and to go from meters per second to kilometers per hour we just do the opposite so we're going to divide by a thousand and we're going to multiply by 60 and then we're going to multiply by 60. now if you had constant acceleration you could actually use the suvat formula you know i've put these little f's with a heart next to it to say that it's in the formula book but you should memorize this off by heart so these kinematic equations for the constant acceleration you should know these off by heart that is that v equals u plus a t s equals u t plus a half a t squared s equals v t minus a half a t squared v squared equals u squared plus two a s and s equals a half u plus v brackets t and the thing to note about these equations is that each one of them has one of the s u v a or t missing you only ever need three variables and you then have a fourth one that you're trying to find so for variable acceleration you'll know in the question of its variable acceleration as it will be expressed in terms of t so in order to do variable acceleration i write them in this order i start off with the s because that comes first in suvat then v then a s u v a t so they go in that order and if you wanted to go from s or x because sometimes they might say that the displacement is in terms of x or is expressed as x equals to go from the displacement to the velocity all you do is you differentiate it with respect to time and to go from the velocity to the acceleration you differentiate it with respect to time so if you are going down you differentiate and that's how you remember that going down is differentiating you can also think of the units as well but this is this is my quick way of remembering it and then if you're going to go upwards you do the opposite you're going to integrate with respect to time obviously making sure that you add in your constants of integration as well so that's how you can tackle those first three kinds then we can have a look at the different graphs that we have here so we've got some speed time graphs now this first one is pretty obvious that the speed is going to just remain constant because we've said there's no acceleration the second one we've got some different stages and the third one we can tell that it's variable that we have here so um in general for these the distance traveled is going to be the area under the line and you can see in this first one here that if this is the speed that we have here and this is the time well the area underneath it is the distance it's a rectangle and we can see that that distance is going to be the speed multiplied by the time now for this one here you should split this into a few different shapes you could do a triangle a rectangle in a trapezium but i just see this as two trapeziums and you need to know that the area for trapezium is a half brackets a plus b h if you were going to do suv act for this graph you can only do suva in three different stages because there is not constant acceleration throughout the whole thing there's constant acceleration for part one part two and part three so you cannot apply these um subvat equations for the entire journey it would have to be piece by piece and then for the last one for the area under the line well we've got to be careful of this section down here because it goes into a negative area and negative areas will not give us the correct distance so if we were going to integrate this one we would integrate the velocity with respect to time between zero and four for that first section and then we would add on the magnitude of the integral of the velocity between four and six between um between those limits this is what the distance traveled would be so you need to make sure a negative area that you take the magnitude of it which i've represented here with these modulus lines and then the last thing that we need to know is that the accelerant at the acceleration is the gradient of the line and so you can tell really in this first one uh that the acceleration is zero here you can see there's variable accelerate there's three different accelerations and this last one we can see that the acceleration is variable okay then what we've got down here is some vector motion stuff this is usually done as a separate topic but i think it makes sense here so if there is no acceleration you just use a variation that the position vector is equal to the initial position vector plus the speed by time time is not a vector so it doesn't get underlined and then the super equations get adapted so that you have v equals u plus a t r equals the initial position plus u t plus a half a t squared and you get the initial position plus v t minus a half a t squared the other ones don't work so well because um of the squaring in v squared equals u squared plus 2as and then this last one that we've got here it's the same as above you're going to write them in that same order to go downwards you differentiate and you just differentiate the i and j parts separately and to go upwards you integrate both of them with respect to time now for projectiles i've actually drawn these two arrows here because projectiles are a combination of this set where there is no acceleration and this set where there is constant acceleration and that is here when we have acceleration sorry the horizontal motion the acceleration is zero so we just use things like we've got above like distance equals speed times time and for vertical motion the acceleration is minus g and those two things will separate that motion into horizontal and vertical and the thing that connects them together is t these are connected by t time i mean um so if they wanted you to find the speed and direction of motion at this point that i've got here a what you would do is you would find the horizontal speed using uh distance equals speed times time you would find the vertical speed using suvat and then if you wanted to find the speed you would do pythagoras and if you wanted to find the direction you would use the inverse tan and that's to do with this diagram we've got here if we're trying to find the speed and direction we find the horizontal speed and the vertical speed you'd use inverse tan to find that angle and you use pythagoras to find the uh the length of the hypotenuse now for the vector motion i just wanted to add in a couple of things that are usually short parts of questions if it said that it is moving in the direction two three moving is to do with its velocity and it's not going in the exact velocity of 2 3 but it is going parallel to it so you put that k in front of it to show that it's parallel if it says it is north east of the origin that means that its position is northeast would be 1 1 of the origin and it would be some multiple of that but really what we can see here is just that the i component would just be equal to the j component and that's because it's north east like this the other place where the i and j components would be equal to each other is down here so we get that i is equal to j here and i is equal to j here so that's the south west and north if it was south east or north west of the origin we would get that the i component was equal to the negative j components for both of these or the j component was equal to the negative i component so i'm just going to add that in that we've got here this is where they are equal to each other and both plus and this is where they are equal to each other but one of them is negative okay we're going to now have a look at some other bits to do with forces we're going to be looking at some modeling assumptions that we've got here so for these modeling assumptions for a smooth pulley people often say okay that means that there is no friction but how that has affected the model or how we've used that in our calculations it means that the tension on either side of the pulley is equal and for a light string people say that has no mass of the string has no mass but how we actually use that in the model is it means that the tension is equal throughout the string if you imagine a really really heavy rope the top of the rope would be having to support the whole bottom part of the rope so the tension would be much bigger the bottom of the rope wouldn't even be having to support the top part so the tension at the bottom would be smaller that's because it has mass so we have light string we can just say that the tension is equal throughout if it's an inextensible string if we've got connected particles this means that both particles have the same acceleration it was extensible it'd be stretchy you could have one of them moving the string just stretching the other one not even moving when something is a particle that means that we can ignore air resistance that's a really common one so i'm going to say another one too we can ignore air resistance and we can ignore rotational effects and that's because it has no dimensions a rod means that something is rigid so it doesn't bend it also means that it has no thickness so when we model things as rods we're saying it has no thickness so they ask you to say what's the limitation of the model it's not very realistic that a plank or a beam is going to have no thickness and i've got some more obvious ones here like if something is smooth or rough but they're pretty obvious smooth no friction rough there's friction and stuff like that okay so that's everything to do with kinematics so far really what you need to do at the beginning of a question is you say to yourself is it constant speed is it constant acceleration or is it variable acceleration and it will then tell you what kinds of skills that you'll be using we're now going to have a look at forces which is probably the bigger part of this topic so to start off with if something is in equilibrium these are the kinds of words that you might see it would say that it's static or at rest it's on the point of slipping or there's limiting equilibrium or that it has constant speed what this means for all of these is that the forces are balanced and more specifically it means that the left and right forces are equal to each other and the up forces are equal to the down forces and then i've got here a vector forces triangle and you can see when you draw these vectors from uh in from nose to tail so they kind of follow around in a flow diagram they go back to the beginning from where they started this is because if you add vector forces together i'm just going to be doing three but you could be doing it with more or less if they equal zero that means that it is in equilibrium let me move on to dynamics all that dynamics means is that you can use f equals ma and i've said that if none of the above are true it is either accelerating or decelerating so there's an imbalance of forces what you do is you resolve using f equals ma in the direction of motion that's always the best way to do f equals m a in the direction of motion with vectors you use the vector version which is f equals m a with the f and a underlined where f is the resultant force r and the resultant force r can be found like this by taking however many forces you've got and adding them together i'm just doing it with three forces you could have four you could have two forces you could even have one and that would be your resultant force okay let's talk about friction so friction always opposes the direction of motion or the direction it is about to move in because it may not be moving but it may want to move in a particular direction so if it's static friction is going to be less than or equal to mu r where r is the normal reaction if it's in limiting equilibrium meaning it's about to move or it's dynamic so it's accelerating then friction has reached its maximum and friction is equal to mu r now i've written here that mu is greater than or equal to zero that is true sometimes you see that mu has to be less than one but this is not true you may find some calculations where mu could be 1.6 or two but generally mu is going to be less than four less than four or five but there's no upper limit really so on slopes they love these kinds of questions where they say is it going to move so i'm just going to quickly add some diagrams on here i think this one is going to want to move down the slope which means that the friction would be going up the slope and when you would put the weight coming down we would resolve it into these forces and so the one that's bringing it down the slope is mg sine theta so if mg sine theta is bigger than the maximum value that the friction can take then it will move and it will move down because it's bigger than the front of the friction can take if mg sine theta is less than or equal to the maximum value of the friction it will remain at rest okay now we're going to start talking about some of the force diagrams we're going to go into some basics here to resolve forces we know that there are two ways you can split them along the bottom you have that is adjacent to the angle so it's f cos theta and then we have f sine theta because sine goes with opposites it's all to do with the angle now if you have tension we're concentrating on what these particles feel this first particle that i have here would feel that it is being pulled so that's why the tension goes in this direction this second particle feels that the tension is pulling away from it so tension the arrows are going inwards for thrust this is where they kind of feel like they're pushing in it would feel like the pole is pushing into this one and for this one it's going this way so it kind of fills the opposite direction but this is how they should be for tension and thrust so boxes on boxes and lifts here now i've got a diagram this is meant to be representing a lift there are two boxes inside this lift there's box a which has small m kilograms box b is capital m kilograms and the lift itself is l kilograms and it's being held up by a rope that we've got here we're going to try and think about all of the different forces that are acting on these so for a it experiences and we're always drawing it from the point of view of the particle it experiences its weight which is mg and it experiences a normal reaction which is from b pushing upwards so i'm just going to write this here this is b pushing on a that one's pretty simple though okay now b is going to be experiencing its weight which is mg but it doesn't experiences it doesn't experience a's weight it actually experiences the normal reaction over here but going in the opposite direction because of newton's law that says that each action has an equal and opposite reaction so it feels r coming downwards and it also feels from the bottom of the floor of the left it feels a different normal reaction which is n i'm just going to call it n and this is the floor pushing on b that's why people get really confused with this diagram now for the lift i'm ignoring the particles that are inside the lift i'm just concentrating on the lift here if you were that lift what would it feel like well you would certainly feel the tension pulling upwards you would feel your own mass of the lift and that mass of the lift is going to be l kilograms so it will be lg as weight and it doesn't feel the weight of a and b in fact it feels this normal reaction of the floor pushing on b it feels b pushing on the floor with that exact same amount so there's going to be an n pushing it downwards like this that's where people get confused because you'll notice in these three diagrams so far there's only the weight of a the weight of b the weight of the lift there are no other weights associated with it it is all different reactions okay and then for this last one that we've got here if you consider the whole system you can consider that there is the tension and then coming downwards for the weight we're going to have little m plus big m plus big l g the whole system is probably a good one for this but sometimes they ask for some normal reactions of the bits that are inside and i've got a video um on all of these things to do with lifts which i'll be linking in the description so you can check that out if you find that confusing okay let's now have a look at some more things to do with force diagrams so we're going to do connected particles and then afterwards i'm going to talk us through what the standard method would be for this so maybe this one is like a car towing a caravan now to begin with i'm going to say if this was the car there may be some driving force which is d and it would also feel the caravan pulling it backwards and that would be with t that tension in the tow bar and there may be some kind of resistance some kind of friction on the road which i'm going to call r1 now if you imagine that you were the caravan you would feel that the car was pulling you forward with the force t and there may be some other resistance that you've got here which i'm going to call r2 for that resistance obviously there's going to be normal reactions and weights but i'm just keeping mine on a level plane here okay this next one that i've got is going to be two particles with a pulley so let's start off with this one over here i'm just going to give it a mass of m1 so there's going to be m1g and there's going to be the tension pulling it upwards and for this one if you imagine being this particle there's going to be its weight which is m2g and the tension pulling it upwards so each time i use a different color i'm imagining that i'm a different part of the diagram and so my last diagram that i'm the last thing i'm going to think about is what it feels like to be the pulley well the pulley feels tension pulling it down here and tension pulling it down here and it also feels a normal reaction pushing upwards like this so if i was just going to quickly draw that diagram again that we've got let's just quickly switch this i would have for the pulley these different forces that would be acting so there would be a normal reaction coming upwards and it feels the tension coming downwards in these two places so for the pulley the normal reaction is 2t each color is me imagining on a different part of the diagram and what it would feel like to be that part of the diagram sounds strange but it's the best way to think about mechanics okay this time i'm just gonna do one type of uh connected particles over a pulley um i'm gonna say it's on a rough slope so i'm gonna add some forces in first of all for a there is going to be its weight which is gonna be m1g there is going to be its normal reaction there is going to be the tension pulling it up the slope and there is going to be the friction which is opposing its motion i'm presuming that it's going in this direction but it may not have to go in this direction it could be going down the slope um obviously what you would need to do is resolve this force into its two components so you would have m one g cos theta because it's adjacent to the angle and m one g sine theta for that one running that's parallel to the slope now i'm going to do in red here for this b particle it's going to have its weights which is m2g and it's going to have the tension that is pulling it upwards now again i'm going to use this green color to think what would it be like if i was the pulley well the pulley feels the string pulling it in these two different directions and as a result there's a normal reaction like this i'm going to do this diagram in a little bit more detail because this is a always a two or three marker that people miss out on now this angle at the bottom is theta this one at the top would be 90 minus theta because obviously this is a right angle down here so i just come over to this diagram that i've got we said that this whole angle was 90 minus theta so each of these individual angles that we've got here would be 90 minus theta over two and it's experiencing these tensions in these two directions and there is a normal reaction that's going in that direction exactly halfway between them so if i were just to resolve these forces i'm trying to find out these two red arrows the red arrows are adjacent to that angle so they would be cos of 90 minus theta over 2 which means that r is going to be two lots of those red arrows so it's going to be 2t cos of 90 minus theta over 2. these ones that i've got here and here are going to cancel each other out which is why i've basically ignored them there and they're cosine because they're running adjacent to the angle for that so that's how you find the normal reaction um or the resultant force of these two uh strings resultant force and normal reaction are going to be the same for the pulley so what's the method that all of these follow well start starting off the connected particles you do two f equals ma and f equals ma in the questions they will sometimes refer to these as an equation of motion after you've done both of those what you'll do is some simultaneous equations that will usually be to find out what the acceleration and the tension are but it could be other variables as well then there's always that part where the string is going to break or a particle is going to hit the floor and there's usually a standard method that we do for this so you start off by doing some kind of suvat and you do that to find out v for one of the particles because later on it will become u after the string breaks or after the particle hits the floor after one of them hits the floor then what you do is you find out a new value of the acceleration and there's going to be a new acceleration because after the string breaks or one of them hits the floor the tension becomes zero and that's going to change the acceleration and then there's going to be some more through that after you've done that and you will use the new acceleration and your value of u will be the v from before and that you have to kind of think to yourself carefully about how you're going to do that bit so that's your standard method f equals ma i think some a simultaneous equations that's usually the part a and b then you can do some suvat new acceleration more through that that's usually going to be the last part of the question it's really a a good one to be thinking about okay we're nearly there we're going to be thinking about rigid bodies and how the forces are and then we're going to discuss this with moments as well to start off with though if something is uniform this means that the weight acts at the center if it's non-uniform the weight doesn't have to act at the center if it's on the point of tilting about a well for this diagram here it's about to tilt about here that means it's just about to lift off from b it means that there is no normal reaction at b so the normal reaction at b is equal to zero if it's going to tilt about b then the normal reaction a would be equal to zero okay when you're doing force diagrams for rigid bodies you always think about the forces that are acting on the the beam okay you're imagining that you are that beam so it would feel the 2g weight pushing down it wouldn't feel the normal reaction of 2g because that's what 2g that's what the 2 kilogram weight would feel there would then be some kind of normal reaction a and some kind of normal reaction at b and i'm going to say this is the uniform rod so there's going to be the weight of the rod you'll notice i've drawn the normal reaction at a and b as different sizes to try and remind us that they're not necessarily going to be the same size as each other and they definitely wouldn't be in this situation that we've got here so this first diagram that i've got is a ladder scenario and a lady a ladder is going to have the weight that would be pushing it down the wall would be pushing it outwards with that normal reaction the floor would be pushing it upwards with a normal reaction as well i'll have to use two different letters there and this ladder if it was on a slippery floor and a slippery wall it's going to want to move in this direction which means that there would be some friction to the right it's going to want to slip downwards so there would also be some friction upwards as well the second one we've got this is meant to be a part a rod sorry that is being um suspended by a string so there would be some tension and it's in that direction because that's what the beam would feel there is going to be its weight which i'm going to call w and then over here there's definitely going to be the wall pushing backwards i like to use the letter x for this because it's horizontal and then there's either going to be an upwards um force on this hinge or there's going to be a downwards force you just make a choice put it either up or down if you get negative you put it the wrong way around but there's no harm in doing that at all so i'm just going to get rid of that second one i'm just going to keep it going upwards and then this last one it looks like i've got a beam or a rod resting against a peg i'm going to say it's a smooth peg so there's just going to be a normal reaction like this and then there's going to be the normal reaction from the floor i could use r i'm going to use y to make it seem similar to the previous one and there's also going to be some friction i could use friction or i could use x and there's going to be the weight as well now if you wanted to find out the overall reaction of this bit down here all you would do is pythagoras to find out this overall reaction like this and to find out the angle of that overall reaction you could do trigonometry you could do inverse tan and find out what that angle was it's also going to be true for this one that you've got over here you can do that same process okay so we're nearly there we're just going to be thinking about what moments actually are to calculate a moment you do the force and you multiply it by the perpendicular distance you can if you want to do the perpendicular force multiplied by the distance i don't teach it that way but there's no harm if you do it either way they both work and all i've written here is that the anti-clockwise moments will be equal to the clockwise moments if it's in equilibrium now there are only three things that you can do in these moments questions if it's horizontal one there's only two things but the three things that you can do is first of all you can resolve it up and down and when you resolve this up and down you can clearly see because it's an equilibrium that y is equal to w the second thing is that you can resolve it left and right and when you resolve this left and right you can clearly see that x is equal to p obviously you can't do number two if it's just a horizontal thing um but i'm going to get rid of those brackets and the last thing that you can do is you can take moments and usually the best place to take moments is probably going to be this bit that we've got down here when moments are about maybe you just pick one place to take moments now for these different forces that i've got here this one x is not going to have any moment because it's going straight through a nor is y so i'm just going to concentrate on doing this w force now this w force here its perpendicular distance is from here to here which is a cos theta so that first bit which is trying to make it go in the clockwise direction is going to be a cos theta multiplied by w and that is going to be equal to the other one which is going to be up here let's just quickly switch a color and for p you can extend that line of action of the force the distance between these well this is theta the hypotenuse is 2a so it's going to be a 2 a sine theta so the moments about a are going to be 2a sine theta multiplied by p then all you do is you take number one number two and number three you solve them in whichever way you can to get whichever answer that you're looking for so this is everything that you need to know for mechanics for year one and year two um lots of this actually is quite simple but lots of it looks tricky so practice practice practice and it's gonna make you do really well in this topic okay so that's it that's everything you need to know for mechanics for year one and year two if you found that useful please do like the video go and check out my channel subscribe to it if you aren't already um there's loads of videos that i think will be helpful for you statistics one will be coming soon and i'll also be doing some more for further math as well okay good luck with eurovision