 Alright. As we work through our kinetics problems, remember those are problems that involve the kinematics we started with, which was the position, velocity, acceleration, and time business that we started the class with. Then the kinetics brings into it the fact that we have forces that enable us to do all this. Specifically the acceleration. In fact, that was our first link between the two where we brought in Newton's second law that allowed us to figure out how to get a particular acceleration if we needed it, what forces would we apply, or if we had a set of forces, what acceleration would we expect to result. So there's a type of things we're working on a couple weeks ago. Then last week we looked at the work energy equation as another way to solve kinetics problems. It's not separate from this, it's actually for the most part a different take on Newton's second law that casts it into a form that made it a little bit more useful for particular types of problems. Hopefully you got the point that those problems that are position dependent usually fit particularly well in the work energy equation. Work itself had to do with distance traveled or change of position if you will. Then several of the energy terms, particularly the potential gravitational potential energy term and the elastic potential energy term very much had to do with the position of the object in any one time. Now we'll get our third and final way to solve kinetics problems that using the impulse momentum equation. So if you read the thing I put online to replace Monday's class you would already know what that is but I'll spend a couple seconds developing it and then we'll jump right into some problems. If you remember from Physics 1 how we define momentum you may not remember that we call the capital G you probably remember something like a small t which is actually a little more common for some reason our author has chosen the capital G to represent momentum but it's defined as the mass of an object times the velocity it's moving and the typical way we get a feeling for that is thinking about a very big object moving real fast is a very different situation in terms of what forces need to be applied to it to change that speed to make it go slower because there's such a huge difference in the momentum of those. So we'll take that then and do a couple things with it. For example let's look not at the momentum but the time rate of change of the momentum a lot of what we've been doing in this class has to do with the rate at which things are changing. That of course will then be d and v dt or in this special case of a constant mass then that becomes mv dot or what we might more commonly know as ma which from Newton's second law we know to be the sum of the forces acting on objects so we have a new way to look at Newton's second law as not the unbalanced forces will cause an acceleration but they'll cause a time rate of change in the momentum. That's a slightly different idea. For constant mass it is exactly the same as this but it does allow us to do problems that don't have constant mass plus it's a little bit more powerful way to look at it when we look at a forces effect changes in momentum which is a little more complete than just saying the effect changes in acceleration. So we have a very similar but slightly more powerful tool. This becomes very powerful when you get into classes like fluid dynamics that you'll probably take in the next couple of years. Alright so we've got this thing here now. Slightly new thing that the sum of the forces if unbalanced will cause a time rate of change in the momentum. We'll do a little bit with this. First just take it out of the dot form so that I get that dt then multiply through by the dt. We've done this type of thing several times in the last couple of weeks. Now we have a differential form of the very same thing and then we can integrate from time 1 to time 2 and obviously g1 to g2 and we get the Newton's second law in either form really because they are so similar in a slightly different version in that this side we now have the ability to deal with forces that are time dependent. So whereas the work energy equation worked very well for position dependent problems where the force changed with position now we can very easily deal with problems where the forces change as time progresses. This side is called the impulse. Now g1 to g2. That'll make sense. Good thing that on the video you can't even see that so there's one or what the hell right up in there. Good catch Bobby glad you're awake. Nobody else seems to be. Go down g1 to g2. Why don't you help us up here? Alright so we've got what we call the impulse on this time the amount of force applied and the amount of time for which it's applied. That just makes common sense you push on something for a little bit of time or push on it with the very same force for a lot of time you're going to do different things to it even if you're just pushing something across the floor. The other side then integrates to then that's very easy to do that integrates to just the change in the linear momentum and so that then is let's change colors give it a pink box that then is the impulse momentum equation. The amount of force applied for however long it's applied will serve to change the momentum of an object. As you saw it comes directly from Newton's second law it's not separate from Newton's second law it's just a different version of it a different picture of it if you will. So let's see there's a couple things that gave you online that you all ignored so you can go look at them but we'll just do a problem I was going to do with it today anyway that uses exactly this. So let's imagine we're doing something very simple just pushing a crate across the floor with some force however what we haven't done before is allow that force to be time varying make this 10 kilograms and the variation of the force with time will let look something like this so this is in Newton's so we've got a couple of time steps here 1, 5, 10 and 15 5, 10, 15 seconds a couple of force things we can do with that say that's 20 or in it we'll call that 100 Newton's and we'll let our force vary something like this from 0 to 10 seconds it linearly increases to 100 Newton's from 0 so you come up to the thing you don't push on it real hard then you start seeing your muscle the veins are coming out and your biceps and stuff you get excited about that so you push harder and harder and harder and it takes you 10 seconds to come up to full power then we'll let it in the next 5 seconds drop down to 20 you realize that you just can't keep it up for that long so the power drops back down and then you're comfortable with that and so it goes that way for a bit of time so if that object is starting from rest let's find out what the velocity is at some time later after you've applied a time varying force like this so let's say let's just to pick a time let's say how fast is it going 20 seconds later so we're looking for I guess what we could call v2 and that will come directly from the right hand side we'll assume it starts from rest and then the left hand side itself well you recognize that as the area under the FT graph so you can do that take a couple seconds to do that that's rather basic rather trivial even and then we'll increase our complexity with it shortly here everybody's trying to remember how to figure the area of a triangle just as I can handle this we're not that close to spring break yet it looks like we've got some coffee between this and that but the library slept a bit and still got up and made it back it must not look very far away one mile north? did you just move there recently? not at all better call a share this is inappropriate all right we'll do the same problem we'll come up with the exact same answer if we never make mistakes especially when we're integrating between g1 and g1 it's just too darn easy yeah how fast is it moving it? say 20 seconds can that be linear? huh? yeah no Jake I want you to integrate I think it's like changing area that's right there if you would place the rest of your we're already done took it at face value took it at the spirit with which it was intended maybe not face value spirit which is of course what you're looking for is the area under that graph up to 20 seconds that's the amount of momentum that was transferred to the block and that's what forces do for a lot of work what we're going to do as an engineer especially if you go into something like fluid mechanics it's going to be really important oh yeah it's transfer momentum got an answer Alex? check with someone don't give it to me check with someone be friendly there you go we don't agree just 90 oh come on 90 check around I imagine most of you broke it into a triangle there a triangle here and then a rectangle there I would think be easiest but it's not that's if you want you can come up with the equations here and actually integrate them but oh you're going to get the area of the triangle anyway but came up with what 90 was the general agreement yeah yeah if you got 85 that doesn't sound like the area of a square let's double check obviously well either it just worked because you guys were paying attention to it but what are the units here on impulse don't be shy do these seconds is that the same as the units on momentum which are meters per second those same units otherwise they wouldn't be equal barring a mistake at the board but we sure wouldn't count on that so alright so we're going to change this problem a little bit we're going to need some more board space so I'll move it over here same general deal we have that very same force we have that time set up there rewrite it so we have it here we expand this problem a little bit to change things a little bit this isn't the deal in real life this isn't what happens what really happens is that we have friction in these types of problems so let's throw that in I'm going to give you a static coefficient of 0.4 and a kinetic coefficient of 0.3 but everything else is the same so I caution you not to just jump into it too quickly without thinking of how this changes the situation naturally you expect that the velocity with which we'll end this problem will be less and we need to find out how much less but how does it change this problem we need a certain force to get it moving what is that force a certain force is needed to get it moving in fact we need to exert a force it won't move until P is at least equal to the static friction 39.2 newtons okay let's see that's that's about here maybe at 5 seconds somewhere in there obviously less than 5 seconds what happens before then in terms of our what what happens after that point I think I actually have that time when that occurs just yeah 3.9 just under 4 seconds so 4 back here somewhere just around 4 seconds a little less than 4 seconds you're finally going to have enough force applied and the box will then start to slide can we get that on to our our force time graph in such a way that maybe it could help us the same graph we had before when we didn't have friction it's a different situation now what do we do with this graph that might help us solve the problem in very much the same way we just did 20 in once it levels out so after this point what happens to friction do you remember if this is the force there's a little while where you're pushing on it but it doesn't move so you push on it harder and then finally it starts to move and what happens to the force then you need to keep it moving drops down a little bit and goes something like that Frank you just had this number here is the 39.2 news what was that number 29.4 so how does that know how we have to take into account this whole thing yeah after here it's essentially constant but how do we take into account this whole thing there's two different ways you could look at it one is you could say well p graphed like that is a positive force f is a negative force it will drop like that so for a little while there are one for one match however much we push on it friction is pushing back and growing at the same rate then what happens we finally reach the minus 39.2 that friction applies with our plus 39.2 and it starts to move the friction force drops so it goes back and then does something like that where that's the minus 29.4 we're getting from friction where we're only at that time applying a plus 20 with this impulse this area minus this area that's taking out momentum but it's still the area under the graph we just now have to take into account the concept of minus and plus areas as divided by the horizontal time axis isn't that times it was four where it just got scattered you figure it out it's about four, yeah but about me caused this thing to crash and everybody dies and you're responsible sit on the beaches in Bahama and think about that the courage you left behind I wouldn't so figure out figure out now how fast it was moving at 20 seconds but also now figure out when does it come to a stop some things that are a much I think it would be a more difficult problem to do as a F equals MA oh yeah, 20 I need spring break see I'm too busy calculating in my head how I'm going to get into professor's gone wild videos I haven't made it in the other years it's always a best seller and I always just miss it you know you might see oh there are but my back's in the camera even with my new tattoo all will cross my back but I want to be able to say yes that's me right there you guys can pop the video in for your parents look, Professor Manning right there I'm sure that's what you want to do now of course is the net area under the craft and you may realize that for the type of things we're doing those are essentially the same add up the whole area or add up the areas separately and then on which time we're talking about some of the first problems you're under I want you to look at the thing online though so you can have credit for Monday's blast so that we don't have to make it up I don't have the answer for the 20 I do have the answer when it comes to a stop do you have that? we'll go check in you tell me they're positive and negative but are the magnitudes the same? yes or no? they must be because remember that the static friction is in direct response to the applied force up to the moment when it finally breaks free it starts to slide so what time did you get for when that happens? I don't know it's 110 seconds it's not 20 in 5 seconds it's 110 seconds oh yeah the 20 goes over to here it doesn't start moving until around 4 seconds so how do you find that time? well you know that this slope is 10 newtons per second and you configure this do you know where the 39.2 newtons came from? alright that came from that's f max when it's finally equal to the limit of the static friction which is when us once it reaches that then we have the maximum static friction that starts to slide from there I don't know how to get the time when it's zero well check with Frank to see if you've got the same velocity because I don't have to have that number action do you agree on that? it comes back to a stop because it will you're not applying enough force to overcome the friction so sooner or later it's going to come back to a stop how do we figure out when? your answer to that is in the pink box in the magic pink box in the magic box is it green or would you give it a go? yeah what's the area? it's going to affect everybody else make sure you top this it's going to prevent the system all the way down oh that's right well that's too bad it's going to be hard to make everybody sick would you have to do sneeze at the buffet if you're looking for a body I know but that's what you do I'm going to do it what do you do? the area at the bottom one thing to do is the area is right here we have this break because after that both things are constant maybe it's a little easier to find out how much farther do we need to go well actually that will tell you if it happens before or after this but you can probably tell from the picture that it's got more momentum put in than taken out up until here and then once you find the difference you can figure out how long it takes as you go by with each second that goes by you're adding a little bit you're taking way more momentum than you're putting in until it's played out that's one way to do it I would guess you guys did you agree on it? no were you into arm wrestle? do we agree on or do you agree on what the speed was after 20 seconds you got what? 35 35 meters per second following? you got that too? let's then do it this side of the equation is the area under the graph so we'll do it up to the 10 seconds you've already done the upper part you did that before the picture confused me with that 39.2 and that 20 trying to find out that small square right there what small square? 5 and 10 what's the name of the small square? above the axis that's not a square find a way too hard Alex break it into a couple of good regular solids you can figure out the area that might be one area very easy to figure out that that might be another one that might be another one then that one those areas underneath the equation being negative you just add up those areas so the area of 1.5 the base times the height 1.5 10 by 100 agreed? that's the amount of positive momentum added in the first 10 seconds however we're taking out because it's under the axis 1.5 at what time does it break free? 3.92 seconds because the slope is 10 1 tenth of whatever that value is 3.92 seconds and it happens at 39.2 that's negative area here and then you just keep adding on these areas well we'll do them in order A3 is a negative area because it's due to the friction it's under the axis what? 29.4 times 15.8 15.18 15.08 we'll just call it 15.1 seconds that's the length of time from the time it started sliding to the time I told you to find the finish velocity and times the 29.4 and then you add on these last two areas triangle 4 is 1.5 less than times 5 and it's height is 8 which is the last one and it's 10 by 20 10 seconds fair enough that's the sort of the brute force integration of the left hand side the area under the graphs is 0 to 20 when I told you to find the velocity it goes all together the area under the graphs that's the amount of impulse added to the object what's that equal to? that's equal to the change in momentum which is m v at 20 seconds minus whatever velocity it initially had which is just 0 what do you guys get for this whole area do you have that there? how much? 3.7 I did it at a different time and you still agree? you have a number for this 0 to 20 the total area? we'll go with Frank then what was it 39 3047 347 0.2 Newton seconds that's the amount of impulse that was added to the object and then that equals the change in momentum you have m you can yourself repeat for the bottom one 0 to 20 for the bottom one the whole bottom area is this triangle I labeled 2 and this rectangle I labeled 3 20 minus 3.92 that's not 15.1 somebody gave that to me you guys set me up 16.1 for the bottom one do you mean for the area 2 plus 3? 2 plus 3 is the total negative impulse added by the friction or positive impulse taken out by the friction however you want to look at it 2 plus 3 those not right the base times the height that's right for triangle 2 and it's negative because it's below the axis and then 16.1 times the 29.4 which is now the kinetic friction what dig? 349 do you want to fight Frank? I didn't okay alright so then divide that by the 10 kilograms for the mass we're left with the velocity at 20 which is 34.9 meters per second how about the second part of the question when does it come back to a stop that's whenever the upper area equals the lower area we know the upper area is greater by 20 seconds because that's exactly what we have here so how much more time needs to go by for the bottom to take away 349 newton seconds as the top is adding at the same time left over some time here deficit because it's a little hard to see the bottom is bigger than the top we'll finally use up that 349 I don't know if you can do this or not but could we say that could we just like some numbers could we just say that can we say that the top stops moving and lets subtract the 29.4 and then find the bottom area how many goes to the 500 the last one there what's the 500 because the top is going to be 500 if it stops right there stops right where like right at the 20 that area is 500 900 so then could you say I think could you subtract the 29.4 and this is 549 up there that difference I hope was the 349 we're very convincing at 20 seconds we know that the top has a surplus of 349 if that number's correct are you okay with that 349 bottom's okay that's the difference between the top and the bottom at 20 seconds alright then let's take it from 20 seconds on applying a force of 20 newtons against a force of minus 29.2 newtons somewhere we're going to reach a point where the bottom surplus is greater than the tops amount by 349 this isn't funny this is serious what do you just do with that 349 not x t different delta t maybe that's exactly what you do but I don't know if everybody saw it so we need to find out where the area a plus the area b when that equals the surplus we had coming into there actually the negative that would be negative that's positive and that area is based upon a certain amount of time beyond 20 seconds you could start all the way from the front and do the whole area with the and only coming at some time t you don't know leaving it as a single variable add up all the area set that to zero and then you only have a single t to solve for jake a couple different ways to go through it what time did you get frank 57.3 that's what I had to do so 37.2 seconds after the 20 the top area and the bottom area differ by 349 which is the surplus we had going into that point so a couple different ways to look at it well we knew that at 20 seconds when I asked you to find the velocity at 20 seconds we knew that at 20 seconds there was an area 349 wait what was the answer at 20 seconds the velocity of 20 seconds not at the time when you got 37 37 57.3 it comes to a stop so I have to add it in that 20 seconds before it's probably one all that is part of the game too it's just we know up to that point the top is 349 bigger than the bottom so why redo that area we already have it but if you want you can start right from the front and do the entire area but it's when they're finally equal top and bottom that it comes back to a stop then what happens it pushes us backwards now what does happen then what happens to these graphs after the 57.3 seconds it's come to a stop because that's what I told you to find that was at the 57.3 seconds what happens after that so if this right there is the 57.3 seconds what happens after that let's say that this force remains because there's a robot doing that we can't change it what happens though with the friction it equals what the friction is going to it's at what was that that was 29.2 it's going to drop down to 20 because it's going to stay there now this is the point where the two areas equal they continue to be equal after that so there's going to be no further change in momentum it's going to stay 0 from then on so they both just bottom out at 20 and it stays there what I'll give you is the problem that was pretty easy you've got to do the hard one problem that I put on the thing for Monday I gave you two problems on there one's fairly straightforward changes and the force is being applied to a crater in opposite directions to each other just as if we had the friction so you need to find the final velocity with that but then the second problem is a little bit more involved so let's take a peek at that one forces on an object puck or something were something like this this was f2 it's moving this way with 3 meters per second the instant those are applied and the two forces vary in magnitude with time one second goes to 20 stays there until 3 seconds 10 newtons then stays there to the 4 seconds so there's the first one that's f1 which is a maximum at 3 seconds and then shuts off after one second so that's the two forces there anybody color blind? too late to check now wish you were which color is green? how does this differ from what we just had we clearly have forces that are time dependent so that's the clear indicator usually that the impulse momentum equation will be appropriate so I want you to find by the way the velocity after 4 seconds let's see the impulse momentum equation is that all I've done is put in the limits of 0 to 4 seconds but then it's just our regular impulse momentum equation for this now this is a vector equation so you can break it into its component direction direction impulse will affect the x direction momentum and same in the y direction for any time we move to full two dimensions we then have our vector equations become two equations and you'll have two unknowns of that which will be vx and vy at 4 seconds don't forget though there is some initial momentum in the problem so close to spring period now this is I had two that I wanted you to give me to get credit for Mondays so this is one of them the other one was a little bit simpler oh yeah you sorry I get so excited 5 kilograms didn't it look like 5 kilograms what do you think a 5 kilograms looks like more 3d on a 2d problem so if you wanted to get this to me sometime I'll give you credit for Mondays jenny promise what's your favorite thing after 4 or well I've worn out because I've bought more than lots I remember when the camera was here it had a really nice one and so many circles in it I can draw like a police officer if you go to this company and look at what they have on there they have hundreds of them they have some with trees and people and toilets all the way we used to do all the drafting now it's just quick