 Alright, we're just about to wrap up particle kinetics. If you remember, we've got a couple of things going into it. Of course, the particle kinematics. And as you went through that last homework and any other problems, a lot of times to finish these up, you've got to bring in some of the kinematics to get enough equations for all the unknowns. The first of the kinetics that we worked with was the F equals MA, which generally works for general problems. If you need acceleration, or you need the forces and you have the acceleration, or you just have something rather straightforward about the kinematics that leads into the problem, or if the forces are constant, then that's generally a pretty good approach to these kinetics problems. We just finished up last week with the work energy equation, which works very well for any position-dependent part of the problem, even if it's just asking for how far did it go under a certain condition in life. But if the force is position-dependent, or there's a spring in the problem, then the work energy equation works particularly well, especially even though we're not doing it if you notice we're skipping chapter four, which is systems of particles. If you have systems of particles, the work energy equation still works very well because you just sum up whatever the particles are doing individually and just sum it up in each of the parts wherever it happens to be, and it just works real well. It's almost like an accounting problem when you have a lot of things going on, a lot of position-dependent things going on, and or a lot of particles in the problem. We'll get our last one here today, and then we will sort of go back over everything with rigid body motion, where we're no longer looking at everything as a particle but as an actual object. And so we'll pull in today the impulse momentum method, which works very well as we'll see for time-dependent problems, especially when the force is dependent upon time, which is a fairly useful way to look at things like rocket launches and the like. It depends on just what the setup is. Or if time is either given or requested, then it works particularly well. But as we'll see, as we develop it, it starts from f equals ma anyway. It's not that these are really different. These are just variations of a theme. Everything comes from f equals ma anyway. It's especially obvious with this method, the impulse momentum method, that it's just a different way to write the impulse momentum, a different way to write ma that allows us to take a different look at certain problems. So the ma side could also be the time rate of change of mv itself. We typically take m as constant. That comes out, then the derivative is just dv dt, which is the acceleration. But if we do it this way, then we can define the linear momentum as that quantity mv. Our book happens to use a capital G for momentum. I think it's just clearly obvious why you'd use a G there. So nothing more than what we call momentum before in Physics I. And then this becomes the time rate of change of the momentum vector. Remember momentum really is a vector. It's a full vector in its various incarnations. Students do sometimes forget though that it is really a complete vector and it's over me. Of course, the special case, if those forces happen to sum to zero, either there are no forces or what forces there are all sum to zero, then of course that was the situation of no acceleration. But when we look at it now, then it means that the momentum vector is constant. There'll be no change in momentum. So what forces tend to do then is change momentum and it allows us to see just how they do that when we look at the equations in this form. So let's start from here then and develop it a little bit farther because we're not quite where we want to be yet with this. So we have this time rate of change of the momentum vector. If we do a little bit of simple algebra, bring the dt over, then we have the sum of the forces dt equals dt. And of course that in itself is trying out for integration because that's the differential form. Only so much you can actually do with that. So we'll integrate the left-hand side from t1 to t2. Oftentimes t1 is zero, but it doesn't need to be. And we'll integrate this side from g1 to g2 and we get then that the change in the momentum vector will be the area under the force time graph which is the step there that makes this method the momentum method particularly useful for time-dependent problems. If we have force varying with time, then it's extremely useful to do it in this way to figure out what the changes in velocities are going to be as reflected in the change of momentum. So it's this form here that is the impulse momentum form because that left-hand side, the integral of the force time function is the impulse and then this other side is, well not really the momentum but the change in momentum but that's where we get the name for it as impulse momentum method. All right, pretty simple setup. Just comes right out of f equals ma directly and no great shakes here. I think the only thing that's new really is that our book happens to use g for momentum. I think in physics we use what, a small p if I remember and it's different in almost every book. So we'll leave that side up over there as we set up a couple problems. So imagine we're doing what we always do in these classes is move crates across the floor. So this time we'll do it with a force that's a function of time, mass of 10 kilograms and the force as a function of time in Newton's looks something like this. It goes up to a maximum of 100 Newtons, the time scale in seconds and we have a time-varying force applied to that that looks something like this. In 10 seconds it goes from 0 up to the 100. We'll make it easy to start with, make it linear and then in the next five seconds it comes down to 20, comes down to 20 Newtons and then is level from there. So that's our first take on it. If this was a constant force problem then of course the force would come out of the integral and we'd have just delta t, f delta t equals delta g and no great, great revelation for that form there. So let's find velocity after 10 seconds and then again after 15 seconds. Just to get warmed up, no great difficulty in this one. If we want to find the velocity we're going to have to find the change in momentum. Since it's a one-dimensional problem we don't need to use vector on this then and it's nothing more and since the mass is constant it comes out of the derivative we just have n delta v and that's going to be equal to the area under the force time graph. Just to make sure it works just to see it's a lot easier than if we were doing some other problems in this way. So v2 equals 1 over m well m is 10 so we'll put that in and then the area under the graph starting from rest so v1 is 0 and then the area under the graph is 1 half let's see for the first 10 seconds of course it's that area 1 half the base and the base is 10 seconds now these are of course not geometric triangles we're taking the area these are force time triangles and then the height is 100 degrees and that would be added onto v1 but in this case it starts from rest so v1 is 0 but we don't want to forget that so we get 50 we have Newton seconds per kilogram Newton is kilogram per meter second squared so one of the kilograms goes out one of the seconds goes out and we're left with just what we should be which is meters per second but Newton seconds this area here is the impulse the impulse added to the object that causes it to have a momentum change the momentum changes the other parts of it there are delta v and the m for the other part you can do it two ways you can do it again from start and figure out the entire area that will last you up to 15 seconds or we can just figure out how much additional area there is and that's going to be a lot easier so we'll call that v3 so v3 is going to be v2 plus whatever else is added on that's the delta v and then add on the extra area which comes in two parts the triangular upper part one half the base times the height base is 5 seconds on this upper triangle the height is 80 that's the upper triangle here but don't forget then we also need to add on this lower triangle that's 5 seconds wide by 20 Newtons high v2 we've got is the 50 seconds there sorry 50 meters per second very simple nothing you couldn't have done in physics one and I think that last 50 that comes out to be 50 this all comes out to be 30 so we get a final velocity of 80 meters per second last little bit of area there that whole area then of course is the change in moment and you can do it between any two times that are of interest so your change in p is just the magnitudes you're not doing change in p I don't know what change in p means change in the force you're just using the magnitudes you're not doing because the 80 was a downward change whereas the 100 was upward the force decreased but it was still in the same direction the force it's just a weaker positive force now but that kind of leads into the next part of the problem that we can look at that was just warm up a little bit of chance to get used to the units just to check to see make sure the units worked out and the idea was working but let's upgrade it a little bit let's upgrade it a little bit we're doing the same two things only now we're going to add to it that static coefficient of friction is 0.4 and the kinetic is 0.3 we had no friction in the earlier problem and now we do I think it'll serve to start another diagram and see what's going on so we have this same diagram so there's the 100 there's our 5, 10, and 15 and like a straight line it comes down to 20 stays level from there what's the difference now that we've got friction in the problem well remember when we start pushing on an object for a little while it doesn't go anywhere because static friction is big enough to oppose that force until some point when the force is actually big enough to overcome the static friction and then the box starts to move what does that look like on our graph this force here is the time dependent p maybe we should make this a little bit more general let's call it force because now we have two forces in the problem what does the friction force look like it's negative for a little while it's only in opposition to the applied force so for a little while we'll have a negative line going down at the same slope and that will be the friction force negative because it's in the opposite direction of what we've already been using as positive but at some time the friction force is big enough that the box actually starts moving then what happens to the friction force by up a little you mean like that remember the kinetic friction is a little bit less than the static friction once you get it moving it doesn't take as much force to keep it moving as it took to actually get it moving but then the static friction is essentially constant no matter what the velocity and would continue on like that at what's supposed to be level but still it's true that the area under the graph is the change in moment it's just for a while for this first section here what's the area zero because the area above and below are the same so after some particular time at some particular force then we start getting at least some net positive area because the area above is greater than the area below so we can redo this the question we had here so the area underneath well what we need to do first is figure out what these forces are and what that time is so that we can then take the area from then on so remember how to find these peak forces there the maximum static friction I know that's not your car Helen your horn doesn't work remember how to find this maximum static friction that's going to make the video great how do we find the maximum static friction this time biggest amount when finally the box will break free from the surface static coefficient in fact this is how we come up with this static coefficient coefficient actually run this test see what the maximum force is but then it's u s times times what the normal which sometimes is equal to the weight sometimes it is not be real careful with that it's extremely easy to come up with a problem normal force is not equal to weight in this case you're that lucky but not always alright so we can figure out what that is then 0.3 0.4 times the mass and so we can figure out what that maximum force is and we can figure out what time that occurs so that's 392 39.2 newtons and if we look at the graph we notice our applied force goes up a newton every tenth of a second so it makes it pretty easy to figure out what the time is at 3.92 seconds so we know at 3.92 seconds we finally get enough force where we start having positive area negative area below that and we can start then figuring out the rest of the details from there so it's not until 3.92 seconds that it actually starts moving so our change in momentum the total area with positive area above and negative area below ft and that will be beyond 3.92 seconds and so we can figure out what that is so we're looking for v2 after 10 seconds that's 1 over the mass once we divide through by m solve for v2 v1 remember is 0 so it's not a big deal and we take the area in the graph remember you'd like to calculate it and break this into some kind of some kind of different areas I guess so we have an area there area there area there up to 10 seconds remember that area below the axis is negative area because that's an opposing force that's a negative impulse that actually takes momentum system so how are we going to add those up let's see one now the base is 10 minus 3.92 which is what 6.08 right and the height is 100 minus 39.2 which is 60.8 is that right Ellen is that okay Chris you're happy are you ever happy Joe I know that finally it breaks through when p equals 39.2 because that's the maximum static friction and notice that the force increases 1 Newton every 10th of a second so I'm going to get that many Newtons in that many seconds it's that or you figure out with a slope of 10 Newtons per second where that T is that is area 1 and then area 2 has the same base it goes up to the 39.2 same units 10 seconds and then there's a third area but that's negative area it's underneath so we'll subtract it it's got the same same base and its height is 29.4 which comes from where kinetic friction UK times N 29.4 remember the static friction maximum static friction is a little higher the static friction usually again units of Newton seconds and that's the area at 3 now if you'd rather you can figure out what the sum of the forces is and graph that you could take this graph and recast it so this would be the sum of the forces and then actually add these together so that the we don't actually have a net force until this 3.92 seconds and then we go up still at that same rate for however long and then notice that the friction force is actually greater than the applied force out in the constant section so you'd actually drop below the line from the net force diagram but that should give you the same area we had here to start with I think it's easier if you just calculate the area straight away but you might not Chris likes things more difficult I know alright so if we do that then I believe we get 24.4 meters per second what was it the first time after 10 seconds 50 meters per second now with friction taken into account we only get 24.4 but with the time-marrying force just simply taking the area under the force diagram to calculate the impulse is a much easier way to do it using F equals MA alright take a minute or two and figure out B3 after 15 seconds we can either do it by starting over again or just figuring out how much additional area there is which is probably the easier way to do it so that would be 4 and you can take this whole thing as 5 if you want however you choose to divide it up isn't this going to be 0 because the friction force is higher than the applied force it might be just because the applied force is less than the friction force doesn't mean when it gets to that point it's not moving still you have to figure that out so you just need to calculate a couple other areas you can split this one into two if you want remember that part of it's negative part of it's positive so they're probably not good to put them together you'll just remember those are negative but don't cut it already see it's not hard it's a lot easier than using F equals MA with a varying F and that's a varying A certainly couldn't use the constant acceleration equations on this Joey, alright? got it? all we're looking for is the additional area the additional net area we've got a couple good chunks of positive and a fairly small chunk of negative so it's the net area that gives us the net change in velocity don't forget that you have to divide it through by the mass because that gives us those areas give us the change in momentum not the change in velocity directly Billy, got it? nope be careful if you're only calculating the additional area if you've only calculated the additional velocity divide through by mass so that's what it is after 10 seconds 15 seconds now that the applied force is dropping the kinetic friction force is still the same and we have these additional areas 5 5 8 additional impulse additional change in momentum 3 of 5 5 seconds by 20 100 is what we're at 294 across here what's that? anybody have that? I don't know if I have that number separately 5 times 20 that's this area of 6 147 oh, by the way that's 147 147 we still have net positive area by what? about 53 divide that by the 10 kilograms so that's an additional 5.3 no, something my number is right there a 153 an additional 15.3 so we get 39.7 so that additional area that is delta G because that's the impulse we're going to submit the seconds divide by the 10 kilograms and add it all alright one remaining question then find out the time when this comes to a stop because now the applied force is less than the friction force so it's going to come to a stop sometime so find t when v equals 0 and what we're looking for here now is when the rest of the area which will be greater underneath than above because it's a greater force below than above, remember it's only 20 up here so when that area is enough to take away the total momentum we have by that point which is well you figure it out what we're looking for is when do we get to a certain time that the total area above is equal to the total area below and it will come back to a stop then because the total impulse then is negative sorry it is 0 so we're out here to 15 seconds you need to figure out how much farther do we need to go to get a total area above equal to the total area below or a net area below enough to take off the last of the momentum that we had at that point let's see why don't we give this to you e3 397 so we have a leftover momentum of 397 because it's that 39.7 meters per second times the 10 kilograms so once this leftover area which is greater below than it is above rx and axis once that area above plus the area below equals minus 39.7 Newton seconds then we know it will come to a stop Samantha okay the frown the weeks how come you guys don't talk to each other we'll have to clean you too but I can see the tears what yeah that's always the story I'll always account for all the stories okay got it no I didn't kind of estimate it however you want to do it 39.4 extra newtons in opposition so how much time to get rid of 397 Newton seconds you can do the total area if you want but we've already got all that we only need to do the additional area leftover with a certain amount of momentum and we need enough impulse to finish that momentum got something check with Chris Chris is right I think about 59.7 seconds however you look at it the total area above equals the total area below it's not real obvious in this graph I could have made this one a little deeper but once the total area above equals the total area below then the total changing momentum is zero and since it started at rest it will then finish at rest at that time any questions on that are we all right to go on Joe Alan you're okay would have been a much harder problem to do I think just using half people than me because of the time varying force so let's look at another one this is a good Adirondack one you can see a little bit of flavor the Adirondacks here all right so pulling a log that's a log all right got a cable wrapped around it and then that's going over to a pulley system anchor there and then the road goes around and comes to a motor something like that pretty simple picture all right some of the details 500 kilogram log coefficient of friction at the base static 50% static 40 motor in startup will do something like this in startup the motor is going to do something like this after 3 seconds it'll reach its full force of 1800 newtons but before then it's got a little bit of a curve ramping up to that something like that and then once it gets there it'll run a constant force takes a little bit for it to actually apply the full force it needs to tension in the line is a function of time is the 1800 maximum but it takes t over 3 squared seconds to get there so when t equals 0 there's no tension when t equals 3 it's at full tension that's the pieces there all right so fine then the velocity after 5 seconds not a whole lot different than the one we just had for a little while there's not enough force in the line to actually get it moving but after a short period of time there is and so you need to figure out where those times are and what the force is at that time all right so we have a little block and tackle thing running on the log don't forget to take the pulley into account the log friction of course pulling it back but a little while it's just the static friction we're trying to overcome then later it's the kinetic friction when it actually gets moving and then you figure out the static force there after if it would help you go ahead and draw the friction force on here like we just did in fact I wouldn't mind it all if you did that I'd like to see what you put once we get up to a certain force it'll then start moving and only from then on are we actually adding impulse net impulse to the log text we give it some net change momentum again you gotta figure out the net the point where the static friction is overcome and we're adding net impulse and that's the net momentum to the system and adding momentum to the system means it's speeding up since it's constant mass most everybody has that static friction point it's maximum static friction is 2450 2450 Newtons only one of the systems well the system doesn't need to come up to that force because of the pulley over here it only needs to come up to half of that and it'll start moving because this momentum is not t but 2t so as soon as the motor is up to 1225 then the system will start up so somewhere in here somewhere in there it takes almost 3 seconds to get it to move sound about right? once it equals 1225 we overcome static friction you can figure out what time that is just by solving for the time t about 2.48 seconds 2.48 seconds it'll finally break free after that now we have enough met impulse that we're adding it'll start to move same thing as if the net force is now positive but since the force isn't positive the acceleration isn't constant since the force isn't constant acceleration isn't constant now you don't forget that the that's the force just to get it moving but the force will actually be required to overcome kinetic friction will actually drop a little bit it's actually 1960 and the force required by the engine then is half of that a little bit about half of the start up this area here is the net impulse am I right Travis? Chris said something about right? so you can do the integral from then from that 2.48 seconds up to the 3 you can integrate that and then after that it's a constant force constant acceleration system after that but no sense going on acceleration equations just add up the area out to 5 seconds that's all we're looking for in those 5 seconds passing the speed sound that's not what I got check it Travis it did take me a couple of times to get a little bit of mix up with some of the numbers you have something? it's closer than Travis's but still not what I had delta G equal to m delta V and that's what's the area under the under the curve so for 2.48 seconds up to 3 we have the force applied by the winch which is actually the 1800 q over 3 squared that's the full power the full force it's adding but you have to subtract from that the not the half of that because of the 2t so minus 1960 N over 2 and that's d that's just the additional impulse during start up but then there's another 2 seconds after that where now it's running at full power some of that's just coming over friction some of that is actually adding speed to it but I think that should be should be the pieces we need or there's different ways you can arrange that to look at it you can do the total positive impulse minus the total negative impulse no Chris is that okay up there earlier that was pretty good that worked out well it felt a lot better when you had that on the paper didn't you yeah you bet when you sit for your phd qualification you end up trying something like that luckily you can take them again okay is that making sense any way out of the value yet is your higher or lower mine was a little bit higher a little bit higher than that I ended up with 7.65 let me see if I have this broken out the pieces and not quite don't forget you divide through by the mass this is the impulse added the impulse per unit mass is actually changing velocity much over integrating you're not happy with that Tom you got something sort of you want to share Bill how you doing yeah no you do going 274 meters per second now okay but it's there's a bunch of minus signs that get in here you've got some integrating to do there don't forget the squared is over the three as well on the bottom should come up with I came up with 7.65 as the velocity after five seconds seven I did check this I did check this not always confident but of course you got I just put it up on the board makes it easy did you remember that this is the four of them log is twice the force being applied by the engine thinks 10% push the motor to the max after a little while start smoking probably at details on the calculation is probably just a little minus sign or a squared or something the physics is there the physics is in place we have that force being applied here during the startup phase and then net force being applied after running for a bit and now a constant force do this for me if you haven't already draw the friction curve in response so that's still a T time remember we know that at a little bit before this point it actually breaks for 980 in the kinetic friction region draw the friction response is that what you were doing there so that is just where is the this is the friction no 26 no no look at your graph just go to 2 seconds and say what would be happening there with what I've drawn folks I want you to sketch do this for me I want you to sketch the friction on here you've got to go back tell me what's wrong with this I will just you saw my dotted line and you thought it was a graph line I can't wait to space you a something that line there's nothing to happen I just cancel all of it out I overlaid it what dotted line did you think I was looking at oh no no I wasn't looking at that I was looking at just what you have there come on Tom draw something come on Phil speed it up a little bit Joey come on draw that graph for me I've got a couple minutes left my last chance to drag something out of here yeah John's got the same thing you do Alan I guess I'm wrong but that's just a integration detail so do that oh I see what you're talking about I don't have a room on my feet for that there you go that's supposed to be right down the middle okay yeah you see you had a net a net negative force it's actually going to go backwards you turn on the motor and the thing would take off backwards you're doing that too it has zero force in here a very low force but a huge negative friction force so it's going to actually take off backwards when you turn the motor on according to John Orl you can't push with a rope though alright now yeah Phil what do you got remember how big is the static friction in response to an applied force before it gets moving exactly the same exactly the same which means if the applied force is a curve like that then the friction force is going to mimic that it won't be a straight line like we had on the other problem the friction was a straight line on that problem because the applied force was a straight line then once it gets moving drops down and then we get the constant static kinetic friction from then on we have the maximum static friction there so that's what the curve up until then those two areas are the same since the net area is zero then the net change in velocity is zero