hello and welcome to the video today I'm going to teach you everything in a level mechanics in under an hour this video is absolutely every topic needed it covers all the examples so you don't need to worry about it and it will include some examples as well I'll drop the completed notes in the description below if you want to check them out please do if you're new around here I'd really appreciate that if this video helps you in any way that you share it with a friend and just subscribe and like the video it really helps the channel out let's get into it start with kinematics so there are some certain things we need to know what's vectors what uh scalar so distance is a scalar quantity hasn't got a direction uh displacement is the vector quantity of uh distance so obviously it has a Direction so if we take this example over here um so if we start to finish the total distance is 160 meters but the total displacement from the start is actually only 40 meters so that's the difference between distance and displacement position quantity uh distance from the fixed origin okay so the position at the Finish is 40 meters away from the start uh this is really important and people actually often neglect this and forget about it average speed is equal to total distance divided by total time average velocity which is this uh vector quantity as well it's displacement divided by time taken moving on to position time graphs slash velocity time graphs so uh if it's a flat line on a displacement time graph it's stationary uh if it's positive it's moving away from the starting point if it's a negative gradient is moving towards the starting point um now velocity time graphs the gradient means it is the acceleration so we can use rise over run to find those accelerations uh Flat Line means constant velocity um moving in the opposite direction means below the x-axis if it crosses the x-axis at that point velocity is equal to zero which means it's either stopped or it's changed Direction okay moving on equations for constant acceleration this is also known as suvat okay uh so s is and Suva the s-u-v-a-t stand for different things so s is displacement U is initial velocity so the the velocity that the particle starts out or the object starts at if there's a final velocity what it ends at and a is the acceleration these are the formulas uh and T is time obviously these are the formulas these are all given to you all you need to know is how to apply them so acceleration if it's being dropped is 9.8 and that's given otherwise um and then negative acceleration means it is decelerating um so let's go through these two examples uh a car starts from rest so straight away that's telling me that U is equal to zero and reaches a speed of 15 meters per second so V is probably going to be 15. uh traveling 25 meters so s is 25. with a constant acceleration so we can only use these suvat equations when it's constant acceleration okay if if there's variable acceleration there's another piece of mass that we're going to learn later on but we can't use that um how much farther will car travel in the next four seconds okay so s u v a t and also let's just draw a little diagram so a car has gone from 0 to 25. and then we want to know how far it's going to go in the next four seconds okay so s we know is 25 U is zero fear is 15. we need to find the acceleration that's the point of this and we actually need to split it up into two so we need to find which of these equations has Suva in it so the one without T and that would be this one so we're going to do V squared equals U squared plus 2as so 15 squared equals zero squared plus two times a times 25 so 15 squared is two two five equals 50a so a is equal to 4.5 meters per second squared okay so I've not finished there that's kind of the first half of the journey we now need the second half of the journey now it says assuming it's got the same acceleration so we're now I'll rub this out in a minute because I'm moving over into this uh question area but we now only think about the second half of the journey in the next four seconds so we're trying to find the distance traveled in the next four seconds we know T is four we know acceleration is 4.5 now actually and this is where you need to use uh your problem-solving skills if I'm taking this point as my new start point then my new initial velocity is my old final velocity okay I hope that makes sense and then I don't care about final velocity so I need s-u-a-t so the one with outer V in it and that would be that one so s equals UT plus a half a t squared so s is equal to 15 times 4. plus a half times 4.5 times 4 squared and when you tell them the calculator you get 96 meters and that's how much it will travel in the next four seconds okay let me move this out of the way let me just put up here it doesn't really matter we're going to move on quickly I've always thrown vertically upwards with a speed of 12 meters per second from a height of 1.5 meters above the ground so let me draw the ground and that is 1.5 meters calculate the maximum height reached by the ball so this ball is going to go up and then it's going to come back down now at this point and this is something definitely worth knowing velocity is zero okay because it's changed Direction whenever it changes Direction velocity has to be zero so let's write out our suvat okay so s is what we're trying to find U is 12 meters per second V is zero because it's changing direction and acceleration is minus 9.8 so it's minus 9.8 because we're taking up to positive and gravity is acting downwards so I need an equation that doesn't have t in it again it's this one so V squared equals U squared plus 2as 0 equals 12 squared plus 2 times negative 9.8 times s okay 2 times 1.8 to 19.6 S equals 144. so s is equal to 7.3469. now that's not the maximum height of the ball because it started 1.5 meters above the ground so I need to add 1.5 so max High is equal to 8.8 5 meters to three significant figures okay moving on to variable acceleration okay so this is where you're normally given like a function and it's either X or s in terms of T so that's kind of your displacement in terms of T and then you've got or you could get velocity in terms of time or you could get a in time uh acceleration time in terms of time now all you need to remember is this little map so if you've got x a function of X or s that sometimes they call it X sometimes they call it s um I don't know why they've got the different notation if you differentiate that function you'll get VT or the function of velocity in terms of time if you differentiate velocity you get acceleration and then of course this works backwards as well if you integrate acceleration you get velocity if you integrate velocity you get displacement so this is all you need to remember if you remember this your standard pure rules apply if you want to find certain points you might have to substitute in um points you get constants of uh integration and you just need to apply your pure knowledge so just as a reminder area under velocity time graph is displacement okay so think about it when I'm integrating um I get the area under the graph so if I integrate velocity I get displacement as this map shows gradient so if I'm differentiating gradient on position time graphs is velocity so if I differentiate gradient to Velocity then obviously gradient means great sorry differentiate means gradient so therefore when I differentiate displacement on the displacement time graph then I get the gradient which is velocity gradient on the velocity time graph is acceleration so this map if you remember this it helps you with displacement time graphs velocity time graphs and the actual variable acceleration questions okay let's do these two examples okay the acceleration of the particle at time T is given by a equals 12 minus 2T the power school has initial velocity of three meters per second when it starts at the origin find the velocity of the particle after T seconds so I need to integrate this so if I integrate it I'm going to get V equals 12 T minus t squared and you get your plus C don't forget your plus c now it says the power score has initial velocity of three meters per second so three an initial means times zero so 12 times 0 minus zero squared plus C so C is equal to three now velocity is equal to 12 T minus t squared plus three okay so that's part A done find the position of the particle after T seconds so I need to integrate this again so let's call it X so that's going to be t squared so it's going to be six it's going to be 2 cubed so it's going to have to be a third negative a third and then I'm going to get three T and then uh you've got plus C which is why it says when it starts at the origin so displacement at zero at time equals zero so C is equal to zero if I substitute zero into T they're all going to cancel out so that means X is equal to 6 t squared minus a third T cubed plus three T and that's it okay great moving on example two so I've just used this mine this kind of mapping system here to know whether I'm integrating or differentiating okay example two a train moves between two stations stopping at both of them it's speed at T seconds is modeled by V equals one over five thousand t 1200 minus t find the distance between two stations so let me just uh expand this out because uh I'm sure we can make it a lot nicer so if I expand this I get 6 over 25 T minus one over five thousand t squared okay so now to get the distance between it I need to differentiate sorry I need to integrate if I'm going from velocity to displacement we're going from velocity to displacement so I need to integrate it so I'm going to get t squared there 3 over 25. and then I'm gonna get T cubed here so what do I need to get there I need to divide that by three which is one over fifteen thousand okay plus c so train moves between two stations stopping at both of them so when tier zero X is zero so that means C is zero so I can get rid of that C now I need to find the time that uh it takes between the two stations and because I know it stops at both stations I can set V is equal to zero and solve so and actually this is where the factorized form will help me because I can choose that we know that t equals zero is an option but actually from this factorized form we can quickly see that t is equal to one thousand two hundred seconds as well so now that I know T is equal to 1 200 seconds that's how long it takes to for the train to get to the second station I can substitute that into X sorry into my function of x 40 okay and now I am just going to type that in a calculator and you should get 57 600 meters or 57.6 kilometers okay forces and assumptions so forces are obviously a massive part um of uh mechanics now key forces wait way to mass are different mass is the uh actual like number of particles the uh the atomic mass whereas weight is different for depending on the gravity so your weight on the Moon is different to your weight on Earth that's kind of the cliche example to to get the weight on Earth it's mass times 9.8 the acceleration uh gravity on Earth uh the reaction force so everyone every uh Force has an equal and opposite reaction so if you're standing on the floor the floor is pushing you up with a reaction force friction obviously if it's a rough surface then you've got a friction if it's moving in that direction and tension from strings pulleys Etc okay assumptions objects are modeled as masses concentrated at a single point so there's no rotational forces strings are inextensible which basically means the tension is the same throughout strings and rods are light I.E you can ignore the mass of the string slash rod pulleys are smooth so no friction force and the pulleys need to be considered so you don't need to over complicate it or you make these assumptions so it's easier to solve okay moving on to resolving forces so if a system is in equilibrium then the resultant force is equal to zero that is the biggest thing okay that's the most important thing take care with objects on a slope because you're going to have to resolve forces in a different angle you might have to I would always resolve it along the slope and it just makes it a little bit easier uh but you're gonna have to do a bit of trick okay let's do this example calculate the resultant force acting on the particle uh given your answer in the form a i plus b j Okay so we've got 10 uh going down we've got 12 going left and we've got 35 acting at an angle of 45 degrees so I need to split this 35 into horizontal and vertical components now if you look at this right angle triangle I basically just need to find the other two sides so I can do that using trigonometry socatoa so this is the opposite so that's going to be sine so if I do uh I mean yeah so let me just make it crystal clear so I want the opposite so I'm going to do sine of the angle so sine 45 tons by so sine 45 times by the hypotenuse that's going to give me that side and let me make sure your calculator is in the right units and that gives you 35 root 2 over 2 and then so then I'm going to look at the adjacent because this is the opposite this is the adjacent I'm going to do cos 45 codes 45 times 35 and that that's broken that 35 down into vertical and vertical and horizontal components okay and actually in this case um of course because it's a isosceles triangle because this is also 45 this is the same so 35 root 2 over 2. okay so now I need to resolve the horizontal forces and that's going to give me the a so it's going to be 12 minus 35 root 2 over 2. so the difference between two forces because they're acting in different um different ways so it's going 12 to the right when I get this I get negative 12.748 I took the left to be positive here you can take whichever way you like to be positive so left positive which means it's the a is 12.75 I and then if I do the vertical components let's take up to positive so that's 35 root 2 over 2 minus 10. typing that in the calculator that gives you 14.75 so B is 14.75 J okay B the force is shown in the diagram uh act on a particle a of math 0.8 kilograms calculate the magnitude of the acceleration of the particle so this is our resultant force over here so our resultant force is 12.75 I Plus 14.75 J okay so I'm going to use f equals m a here which is Newton's second law so f equals m a and they want the magnitude so I can write this as a column vector or you can find the magnitude straight away in fact we know the magnitude so let's use the magnitude so the magnitude of this Vector oh actually no no that's wrong it's not 35. so we can find it just doing Pythagoras but let's do it this way you can do it either way you can either find it before or afterwards okay so I've written that as a column Vector the mass is 0.8 and then I'm going to get and let's just call it um I don't know s and t um and that's going to give us the acceleration vector for which we can find the magnitude from so I need to divide both sides by 0.8 so that's going to give me the well let me be consistent s and t which is the acceleration vector um so the bottom is going to be 18.44 and then if I do 12.75 divided by 0.8 I get 15.94 okay so that means the acceleration vector is 15.94 I plus 18.44 J now if you want to find the magnitude of this you simply just do Pythagoras because if I could draw this um as a right angle triangle and this would be 15.94 this would be 18.44 and you just find the hypotenuse so you square both numbers and square root the whole thing add them together and when you do that you get 24 points three seven dollars dot so 24.4 meters per second squared and that's the magnitude of the acceleration okay example two a box of mass five kilogram rests on a slope inclined at 30 degrees to the horizontal calculate the normal reaction and the normal reaction force and the friction Okay so drawing a diagram always want to draw a diagram and I've got a box which is five kilograms so the weight component is going to act vertically down because that's how it works and that's going to be 5 times 9.8 because that is the force it's the mass times acceleration the reaction force is Gonna Act perpendicular to the slope because that's the reaction force and friction is going to work up the slope in this case because uh if there was no friction it would just slide down the slope so friction is going to oppose that okay so I can do that now if I zoom in here um and I draw a little dotted line This angle is 30 degrees it's always the same as this one it's always the same as this one in here so now and that's a right angle let me change that for a dotted line uh so that's a right angle so now I can break this weight component into verse going quotation marks and horizontal because I want to resolve perpendicular and parallel to the slope okay um I don't want to resolve actually um horizontal and actual vertical so this magnitude so let's work out what 5 times 9.8 is so the magnitude of this is 49. so this is 49 Newtons okay so the friction is going to be equal to the horizontal in quotation marks uh component of 49 and the reaction force is going to be equal to the vertical in quotation marks uh component of 49 so I do this using soccer tour again so this is the opposite this is the adjacent so cos so 49 times cos uh 30 so that and that's what the adjacent is which is 42.4352 dot Newtons and that's going to be equal to the reaction force because it's in equilibrium it is rest is that rest which means is that in equilibrium so R is going to be equal to that because if it wasn't equal to that then it wouldn't be in equilibrium so R is equal to 42.4 newtons to three significant figures okay so then the friction is going to be equal at the opposite of this triangle so it's going to be 49 times sine 30 which is just 24.5 Newtons okay great work I'm just going to do a little check to make sure I've got it right great yes I have okay moving on coefficient of friction um well we've just done a little bit of friction now it's important friction is a limiting Force now what that means is um basically it will equal and opposite you so if you just like gently put your hand on the table and try and slide it across it's not going to move and that's because friction is equaling your Force okay but there is a maximum friction that the the surface that can give you and it depends on what surface like if you put your hand on ice it would slip much easier than if you put your hand on gravel for example um so if and then if you push harder then eventually your hand will go and that's because you've overcome the the maximum uh friction that that surface can give now so it's a limiting Force which means they will equal it until you go above the maximum so F Max which is what I've just been talking about is equal to Mu which is the coefficient of friction which depends on the surfaces involved and then R is the reaction force Okay so F Max is equal to Mu times R if a force is acting on the object but the object remains at rest then f is less than mu R when the object is moving uh the frictional force is constant and it will always oppose motion for questions looking at minimum maximum force needed to move a block on a rough slope look at the magnitude of the force p okay Let's uh have a look here this is what we're talking about in terms of the force P so basically what this is saying and this is all it is is think about which way it's going to move okay which way is it going to move then friction is going the opposite direction so in the top example when the block is on the verge of sliding down the slope okay then friction is going to be going up when the block is on the verge of sliding down the slope sorry when the block is on the verge of sliding up the slope the friction is going to be acting down the slope so friction is always opposing the motion okay now whether this block is on the verge of sliding down or up depends on how big p is okay so obviously if if p is small and we can work it out in a minute if p is small then it's likely to be going down the slope if p is really big then it's going to be going up slope okay Newton's Laws we've touched on some of these already Newton's first law um is basically it will remain at rest or constant velocity until a force is acting on it okay so everything will remain in equilibrium which means that rest will constant velocity uh until a force is acting on it an external force to be precise the second law is f equals m a but in words the resultant force acting on an object is equal to the acceleration of a body times its mass Third Law if an object a exerts a force on object B then object B must exert a force of equal magnitude and opposite direction back on the object so basically every reaction has an equal and opposite reaction okay example one okay calculate the acceleration of the object so we're going to be using f equals m a so we need to find the resultant Force so the resultant Force we're going to do 12 000 minus six thousand because this uh let's assume a car um is traveling to the right okay maybe it's a truck it's slightly bigger than a car um it's moving to the right and the driving force is bigger than the resistance so the resultant force is twelve thousand minus six thousand the mass of the car is eight thousand and the acceleration is what we're trying to find out so this gives me six thousand on this side 8 000 on this side so a is equal to three over four which is equal to 0.75 meters per second squared so that's example one done example two uh a man of mass 80 kilograms stands in a lift calculate the normal reaction uh of the left floor on the man if the lift is moving downwards with a constant velocity okay so lists are way more complicated than they need to be they're really not that difficult okay so we've got a little man and you don't need to over complicate it okay so the man has is giving a weight on the floor of 80 times 9.8 that is the force he is exerting on the floor uh on the left floor okay now because it's constant velocity the man and just think about the man in this scenario is in equilibrium so that means the reaction force from the left has got to be equal so this is just simply R is equal to 80 times 9.8 which is 784 Newtons okay so the left is moving upwards with an acceleration of two meters per second squared okay so a is equal to 2 upwards now this is where the and think about when you're actually in a lift when you're going up the floor feels like it's pushing you up and when you're going down you feel a little bit of lightness like uh so your tummy tons for the first bit just a little bit and you feel a bit weightlessness um that's because the reaction force is changing depending on whether you're going down or up um so the the lift so the basic reaction force is going to change so I need so we're going to do f equals m a and we're just considering the man again so acceleration is 2 mass of the man is 80. so I need the resultant Force to be 160 Newtons that's the resultant Force now the resultant force is equal to R minus eight well we've already worked out what 80 times 9.8 is minus 784. so we need that to equal 160. so if I add that to 160 then R is equal to 944 Newtons so therefore there is an imbalance there the reaction force is greater than the weight which allows the man to accelerate upwards that's relative to space obviously he is Stood Still In inverted commas in the left so the left is pushing him up which is accelerating him through space okay example three two masters are connected by a light string passing over a smooth pulley as shown below calculate the acceleration of the four kilogram block when released from rest okay so we can consider these things together because they're connected by a string or we can consider them separately and that's going to help us so we want to find the acceleration of this block here so with the tension is the same for both uh four kilogram and five kilogram block because uh the tension in the string has to be the same throughout so I'm just going to consider the five kilogram block um by itself so the weight is 5 times 9.8 which we know is 49 uh Newtons and T is uh T now it's not in uh it doesn't say that it's in equilibrium so we we can't say that t is 49 but we know that uh the resultant Falls is T minus 49 okay um so that's and that's going to equal m a because the news um because of Newton's Second Law so T minus 49 is equal to 5 a I'm just going to get rid of then so it doesn't confuse anyone uh so that's just considering that block um so now let's consider the other block and the acceleration is going to be the same for it the acceleration is going to be the same for both because they're connected it wouldn't make sense if one was traveling for um or accelerating faster than the other so now let's consider the other one now the resultant horizontal force is T minus two um have I got these going the other way no that that's uh let's take it to go left oh it's so annoying when it does that um just to make sure everything's moving in the same direction so let's say T minus sorry 2 minus t is equal to m a so this is just f equals m a I'm just finding the resultant Force I've taken uh positive and left to be positive uh you can take whichever way you like to be positive as long as it's the same direction of the system so T minus uh and then the mass is four of that block now I've got simultaneous equations here so let's substitute um T into one another um so that's so this is T equals two minus four a so 2 minus four a uh minus 49 equals five a so minus 47 equals nine a so a is equal to [Music] negative 5.22 meters per second squared now I took out two positive which means it's going to accelerate down okay so it's going to accelerate down so hopefully that makes sense um so just so it's clear I took up to be positive which means I had to take left as positive um but if you took the other direction to be positive I.E down and right then obviously it would change the signs of all of the forces and you would end up with the same answer anyway okay example four uh so you've got three forces uh with a on a particle of mass 10 kilograms find the magnitude of the acceleration particle so first of all let's write these all in so that's F1 in column Vector form that's F2 and that's F3 now I'm going to add them all together because that's going to give me the resultant Force so 2 minus 3 is minus 1 plus 4 is 3 1 plus 4 is 5 minus six is minus one okay so this is the resultant Force on the particle so I can do three minus one equals m a now I know the mass is 10. so is equal to 10A so a is equal to three over ten and minus a tenth okay so that's uh acceleration as a vector I've just divided both sides by 10 there if it wasn't clear so now to find the magnitude of the acceleration vector I just Square it and square root it all so three over ten squared plus negative one over ten squared all square rooted uh if you type in the calculator you get you're a legend if you've made it this far in the video please like And subscribe it really helps the channel out uh root 10 over 10. which is equal to 0.316 meters per second squared okay moving on projectiles okay now projectiles is actually a really simple topic which is often over complicated uh it's basically resolving forces with suvat uh you need to break it all down into vertical and horizontal uh so vertical and horizontal okay so vertical horizontal and once you've done that you can just do Sue that horizontally and to that vertically uh and you're basically doing the same thing as Suva um now in terms of uh certain things you need to know so if you want to find the maximum height then when you're doing it vertically um you want V equals zero just like normal to that if you want to find the maximum range how far it can go you want to set the vertical height is equal to zero um and and then you just solve with Suva now the thing is and this is uh the interesting point if something is projected let's say at angle Theta the horizontal component will not change okay it will not change unless there's air resistance so the horizontal component will stay the same the vertical component will change because of gravity okay so then you can actually just do speed distance time for the horizontal component for a lot of the time okay now obviously you can find the horizontal and vertical components of a force um using your soccer Tower and then it's just super okay and then it's just through that so let's just get into the example um a shot putter releases uh the shot at a height of 2.5 meters with a speed of 10 meters per second an angle of 50 degrees to the horizontal calculate the horizontal distance from the thrower to where the shot lands okay great question so first of all it's being released and let's draw the ground at 2.5 meters above the ground it's being thrown at 10 meters per second at an angle of 50 degrees so if I just plot this kind of projectile that's what I want okay now let's break this down into let me use a dotted line if I break this into horizontal and vertical components we can do that quite easily so the opposite is 10 sine 50 which is 7.66 and the adjacent is 10 cos 50 which is 6.4 three okay and these are meters per second meters per second okay so now I can use suvat uh so let's do and let's say this is vertical Okay so suvat okay and there are certain things I know so the vertical displacement is going to be negative 2.5 I hope that makes sense because it's starting from here it's going to end here if I'm just talking about and the vertical component it's dropped by 2.5 meters the initial vertical velocity is 7.66 and then the final velocity you don't know the acceleration is minus 9.8 because we're taking up to be positive so that's going to be useful we might want to find the final velocity or time I'm going to suggest probably the time now let's look at the horizontal okay and set up our super okay so s is what we're trying to find out ultimately that's our end goal the initial velocity is 6.43 that's come from here uh final velocity we don't know acceleration is zero because there's no forces acting on it now after it's been released so it's going to remain at constant velocity the horizontal component and the T we will get from uh well we're going to do simultaneous equations to solve this okay so let's call this t um okay so now I need to use something that's got s u a t in it so s equals u t plus a half a t squared so if I type in what I know okay so that's going to give me a quadratic in t which I can solve let me just make it a bit nicer so it's easy for us to solve using quadratic formula okay so and then when we go in to solve that I'm just going to type in my calculator okay so you get T equals 1.84 and T equals minus 0.277 okay now you can't have a negative time but the model doesn't know that and basically with the constraints you put in it's basically if it went backwards and we ignored the shot putter and it continued on this projectile it's giving you that time there okay it's giving you that time there so at this point here T equals 1.84 which means I can use that t here 1.84 because the time is the same it's the same shot put same sort of thing so now I can use S equals u t plus a half a t squared now a0 so that's very handy so s is equal to 6.43 times 1.84 which when you type in the calculator is 11.83 meters okay moments again moments are are not difficult um it's just knowing that it's perpendicular distance uh Times by The Force okay and that gives you the moment now if it's an equilibrium not only do the forces need to be resolved but the moments need to be resolved as well and that is clockwise and anti-clockwise need to be equal otherwise it will rotate okay now sometimes uh the forces aren't always perpendicular to the rod so we need to resolve forces uh in a different uh on an angle again okay so the result moment is the difference between the sum of the clockwise moments and some of the answer clockwise moments just said that a uniform lamina is basically like a sheet metal uh it basically has the same density throughout Center of Mass is at the center of the object and that is where the weight of all of the object Acts uh uniform Rod basically means the mass and the weight acts at the center of the rod or the midpoint an equilibrium is where the resultant resultant force is zero and total moment uh total moments also equal zero um so always draw a diagram uh and then you take moments about certain points and you can eliminate some of the forces some of the unknowns and then you kind of work by process of elimination example one a uniform Bridge across the stream is five meters long and has a mass of 100 kilograms it's supported at the end A and B uh okay great so let's just draw that first of all so I've got a bridge it's five meters long is uniform which is going to be helpful as a hundred kilometers 100 kilograms so that means that the mass of this bridge is going to act on here and there's a hundred times 9.8 and that's perfectly 2.5 meters either side okay a child of mass 45 kilograms is standing on the bridge at Point C given that the magnitude of the force exerted by support a let's label this A and B uh out support is three quarters the magnitude of the force exerted by its support B calculate the magnitude of the force exerted at support a and the distance AC okay so there's a child we don't know where it is um 45 kilograms which is 45 times 9.8 there's going to be a reaction force here and a reaction force here now I know because it says here that um the support a is three quarters so R1 is equal to three quarters times that of R2 um is that what they're saying or is it the other way around given that the magnitude of the force exerted by support a is three quarters of the magnitude okay so now I've got this the wrong way around so it's three quarters R2 three quarters R2 so a is three quarters of the magnitude of R2 yes good definitely got that right so that means I can actually replace R2 or R1 um in terms of R2 okay so let's write that as three quarters R2 okay and this is point C okay and that's what we're trying to find one of the things we're trying to find okay now we've got all the forces in play here it's in equilibrium because the bridge hasn't fallen over so we can just resolve forces first and foremost um and that's just the vertical components so three over four r two so these are all the ones going up plus R2 is equal to 45 times 9.8 plus a hundred times nine point eight so that's just resolving vertical forces and I can do that because the bridge is in equilibrium so all the forces need to be equal so this gives me so three over four plus one is seven over four R2 45 times 9.8 is 441 9.8 times 100 is equal to 980. add those two things together 1421 and find out what R2 is 812 Newtons so I found um R2 which means I can find R1 so R1 is now uh so that's times it by three quarters and that gives me 609 okay let me change this one to eight hundred and twelve just making sure it all adds up double checking yeah great it all adds up fine so now they want to find the distance let's call it X between C and A okay so this is where we're going to take moments okay perpendicular distance so X Times by 441 that is the force exerted by the child so we're doing all the anti-clockwise or no sorry we start with all the clockwise moments so this force is clockwise this force is clockwise and this force is anti-clockwise so because it's an equilibrium the clockwise need to equal the anti-clockwise so x times 441 uh plus 980 that is the force of the weight of the bridge and that's got to be equal or hang on nearly made a mistake there so that's the force but we need to times it by the distance perpendicular distance which is 2.5 this distance is 2.5 and this distance is 5 so that's going to equal 5 times 812 so we can obviously solve this for x plus 2450 equals 4060. so solving this so X is equal to 3.65 meters okay now what does that tell me now that tells me that c is actually over here and that's perfectly okay that I got it wrong because I just gave it a generic uh the fact that it was left of the midpoint or I assumed it was left on the midpoint didn't actually make a difference um but the the final answer is 3.65 and and that's the important thing okay example two a uniform ladder I've done so many of these questions I just already know what it's going to look like of length three meters so this is three meters long and 20 kilograms great uh leans against a smooth uh verse wall so this is smooth so therefore no friction and the angle between the horizontal ground and the ladder is 60 degrees find the magnitude of the friction and the normal reaction force that acts on the ladder so there is a refractional force going this way okay because the floor is rough and we can indicate that with some dashes okay find the magnitude of the friction and the normal reaction forces that act on the ladder if it's in equilibrium so there's a reaction force there let's call it R1 reaction force there R2 we've got the weight here which is 20 times 9.8 and those are all the forces acting on it okay so if we resolve uh vertically here we can find R2 straight away um because R2 is just going to be equal to the weight of the ladder so that is like that because it's in equilibrium the vertical Source saw uh the vertical forces have got to be equal so when you type in the calculator you get 196 Newtons so that's one force found um so now if we take moments about this point here at the base uh we can find the R1 uh and we've got to be careful now we need to find the perpendicular distance perpendicular distance is the important thing so taking moments here let's start with the clockwise so you're going to have R1 Times by and I'm trying to find this distance here because if I extended this this is the perpendicular distance to R1 okay now I can do that because it's three meters and I've got the angle this angle is 30 degrees so if the hypotenuse let me just make the stream meters clear so this is three meters I can find the perpendicular distance um using that or you can use the triangle on the bottom with the 60 degrees you end up getting the same answer um because it's actually the same as this distance of course um so at the moment I've got the adjacent so it's going to be 3 times cos 30 and that's going to give me uh that distance there which is let's keep it as uh exact as possible so that is that distance there okay so I need to do R1 times three root three over two and that's going to be equal to uh now if I want to do this here if I I need to find this distance here um so I'm going to use uh now this distance is 1.5 so it's going to be 1.5 Times by and I want the adjacent again cos 60 Times by 20 times 9.8 okay so that's just giving me the anti-clockwise moment and they've got to be equal um and you can change where where you take moments from I I chose it because I don't know the friction yet so I wanted to find R1 so I can solve R1 here so let me do 1.5 times cos 60. times 20 times 1.8 that gives you one four seven and three root three of two so you get R1 is equal to 56.6 Newtons okay let me try and tidy up my diagram I don't need this big R1 here don't need this don't need this great don't need this great um do need to know that the angle is 60 though okay so I've got R1 I've got R2 I need to find the friction okay so I can just resolve horizontally now okay it's as simple as that so the friction is going to be equal to R1 because they're the only horizontal uh forces in it so therefore the friction is equal to 56.6 Newtons okay that's a Whistle Stop tour of uh mechanics I hope it helped 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