 They make sense as we put the story together, whether the story's in words or in these equation forms that we're using here. And again, this can be a summation of any parts of the system that happen to be changing speeds. Remember the little warning that I gave you on Monday, even though somebody here will forget it, will take the bullet for the rest of the class and be the one person who doesn't remember the simple algebra of the difference of two squares is not equal to the square of the difference. Somebody's going to make that mistake. I know it happened every single year. So I'm just waiting to grade that first set of homework and see who does it. We may or may not be changing the speed. It's a problem dependent. We'll look at it as we go. But if all of the work is not used for changing the speed in some way, we can also raise or lower a part of the system or the system as a whole. And that's just the change in gravitational potential energy. And again, it depends on what pieces of the system and what height change they undergo. Remember this has nothing to do with any sideways movement. It's the movement parallel to, meaning other, only in the vertical direction of any of the masses in the system. So we might have several different masses moving up or down and each of them may have their own height changes in there. It doesn't matter. You can calculate them all separately and then just add them up and put that term in as the change in gravitational potential energy term. And we might have an elastic medium in this system or more. And again, it's nothing more than summing all of those little pieces up if we happen to have more than one. If we have two springs in the system and they have different spring constants, just calculate their change in potential energy separately and add them together. And it's dependent, of course, on the change in how much energy, how much potential energy is stored in the system at any one time. One of the nice parts about this right-hand side is all of these are point functions. Remember what I mean when I call them that? Remember what I called that? Define a point function on Monday? Sorry? Not path dependent. Yeah, not path dependent. It only depends upon wherever the system finishes and wherever it starts and the difference in those two things. Whatever happens in between is immaterial. Sometimes you can use that to your advantage and very much simplify a problem with that use. So these are all point functions on the right side. It's very much a path function on the left side, as obvious by the fact that it's an integral and it depends upon the position of the object at any one instant. So we have all those parts put together here. These two parts here also involve forces, but those are forces that we do not include over here because those are conservative forces. The easiest way to tell what we mean by that is to imagine looking, watching any of these systems undergoing its motion and we videotape it. If we run the videotape backwards, conservative forces are, it's always going to be obvious that the system's running backwards. You can tell from the film, watching the film, either forwards or backwards, you can watch that and tell that it's running backwards. For example, if we've got some object that's being pushed along the floor, a little bit later, it'll be somewhat farther down the floor and we might notice in the movie that the floor has been scratched because of friction. We might notice, even if in some way it's monitored, that the system's a little bit warmer because of that friction and if we watch that movie backwards, we would see an object here being pushed that way but it would move back to its original position and would actually repair the surface of the floor over what you just slid. You watch that and you know that's not possible, that there's no way that could have happened. That's the nature of non-conservative forces. If we had a rope in the problem pulling, then when we run it backwards, we see the ropes pushing and that's just, again, an impossibility. If we had a mass going up or down in the system, for example, we could have had a mass attached to this weight as it goes over there, if we run it backwards and look at just that, all we see is that there's a mass going up or down. Without knowing what's going on in the rest of the system, we can't look at that one part of the system and tell if it's running backwards or not. There's certainly systems that could lift masses and there's certainly systems that could lower masses. We'd have to see the whole rest of the problem to know what it's going, whether it's which way it happens to be running, but then we're looking at the non-conservative forces anyway. It doesn't make these non-conservative just because they're attached to non-conservative forces. When viewed all by themselves, we can't tell whether that system's running frontwards or backwards. The same is true if we happen to have a spring in the system somehow. As the weight, as the object moves, we could, if we looked only at the spring, all we see is a spring that's increasing in length or decreasing in length. There's nothing inherent in that movement that indicates which way the system's running. We'd have to again look at some other part of the system to tell. We couldn't look at the spring alone and tell whether it's running backwards or not. That's the nature of conservative forces. They can go either way that if system, in terms of those forces, the system can return completely to its original state. Non-conservative forces, we can't get back to the original state by just simply running the system backwards. If you have to push the box across the floor to run it backwards, you've got to push it back to a starting point and you're twice as tired. Not tired at all as you would when you started the problem. So that's the pieces we're looking at. And remember, this work is going to change each of these in the same direction as the sign on the work that's doing it. So we can use that to our advantage as we go through some of these problems. So let me put up a more complicated problem. And since it is a little more complicated, I've already printed it out for you so you don't have to draw it. You can if you want. Chris, you can correct the drawing or improve it with some of your artistic touches. What we have here is something like perhaps a garage door opener of some kind. Maybe the garage door is attached to this bracket and a motor pulls the bracket back as the garage door is open. There's a counterbalance attached to the system just to allow us to use a much smaller motor. Take some of the load off the motor, that's the point of the counterbalance. The very same way it works on an elevator system. And there's a spring-loaded attachment to it to allow the system to very easily return and drop the door back down to the bottom. So something like that is what we've got. As we begin this problem at point one, which is already moving with some velocity, what we want to do is design things such that it comes to a stop here at point two. We don't want it to keep going and smash into the structure of the garage, nor do we want it to stop short of that and perhaps it's not open enough for the driver to get into the garage. So with that, those values on the spring and there's some dimensions of the general hookup where the spring is, we want to find an appropriate mass for the counterbalance such that we bring it to a stop at point two. If the mass is a little too heavy, it'll go too far, if it's too light, it won't go far enough. There's a certain point where the motor can no longer pull against the spring and the system will come to a rest. So that's the setup we've got. We start with the work energy equation and see if there's any parts of it with which we are already finished because of the path or sorry, the point function nature of some of these that a lot of them could turn out to be zero right off the front and we could be done with them, have a smaller problem right from the start. So is everybody comfortable with what the setup is here? I'll give you just a second to make sure you get the numbers down before we start. And then of course a second to put the work energy equation on your paper because it's a great place to start. Yeah, man. Can you just read off the values and write it up? Oh, okay. This is the rest length of the spring, 0.25 meters. The spring constant, 0.32 kilonewtons per meter. The bracket is 14 kilograms. It's initial velocity to the right is 1.2 meters per second. The engine, the motor is pulling on the cable with a force of 60 newtons assuming that's constant. We want to come to a stop at 0.2, so that's zero meters per second. And then, I don't know if I mentioned that, there's a little bit of frictional drag as the counterbalance goes up and down its channel of about 180 newtons. So as the bracket goes to 0.2, the mass is going to drop and as it drops there's 180 newtons of friction as it slides down the channel. Okay, Joe? G1, is that going to the left? Yeah, because it's going from 0.1 to 0.2, so it's already going to the left. Sorry, going to the right. No, that's just a dimension error. Okay. Sorry, it's going to the right with a speed of 1.2 meters per second. So it's already on its way to close and we're just interested in getting the system right so that it comes to a stop at 0.2. Right? Everybody comfortable? Okay. Look at each one of these four pieces and decide if any of them are zero if so we're already doing a smaller problem. Are there any outside or non-conservative forces in the problem? If there are, this is not zero and we need to calculate those. Things like friction. Things like the motor being pulling like this. If we run things backwards the motor's not going to push with 60 Newton so we can clearly tell that that's a non-conservative force. So we have both the motor and friction that we need to account for in the non-conservative work term. Is anything changing speed? In fact, several things are changing speed. The two things of concern, the bracket is changing speed and the counterweight itself is also changing speed. So we're going to have to look at both of those. I'll put cb down for counterbalance, that massive point b. Is anything changing height in a gravitational field? Of course the counterbalance is. Nothing else is so that's the only term we're going to have in that one. And we have the spring on the system and it is changing in length so we're going to have to calculate that term. So none of them disappear. This is a pretty involved problem and then we've got several items in each of the four major categories. So it's as easy as anything I think to go back and just look at these one by one. That way you're doing a small problem. You're not going to mess up the units. In fact, just to hopefully give you confidence that it works nicely to do it this way, this is just a quick peek at what the work energy equation would look like if you put all of the values in. If you put everything in with all of our unknowns and all the knowns, you get this nice equation at the top there. The first thing you're going to do is leave off your units. So you're going to get down to this. It looks a little bit nicer, but don't forget you've got to solve this for the mass of that counterbalance b and you've got to do all of that without messing up one minus sign, not messing up one squared term. I just don't think that that might be a convenient way to start the problem. I don't think it's a convenient nor an efficient way to finish the problem. So my recommendation to you is to do these one at a time, one piece at a time, then we're doing small problems with just a few minus signs, a few squared and it's very, very hard to mess up any of those. It's hard. All right, we have work being done by two things in the problem. The motor, maybe if I put the m up here where the O was and we have work being done by friction. Now we're taking those all to be constant so the integral integrates nicely. It just becomes the force times the distance. So the work done by the motor is the 60 Newton force it pulls for a distance of 0.4 meters. Does the fact this is positive make sense? We need to check the minus signs every single step. Should this be positive? The force is to the right. The motion is to the right. It should be positive. It should be also Newton meters. Makes sense. This is for work. Add to it the friction term. It's nice in that the friction is given. We don't have to actually find it. We know it to be 180 Newtons. As the counterbalance drops because the bracket goes from one to two, how far does the counterbalance move? Because that's the distance over which the friction is doing work. Sorry? Yeah, because of the cabling you do what we did a couple... I think last week when we were looking at constrained motion you'll see that it will drop half of that distance. As the bracket moves 0.4 meters the counterbalance will drop 0.2. Is that positive or negative? As the counterbalance slides down friction force is going to be up in the opposite direction of the motion so this is going to be negative. That's all the parts we have doing non-conservative work so we calculated them separately we just simply add them up. What's that come out to be? A minus 12 I think? What's the minus sign mean here? This is the total work done and we're left over the minus sign because the friction term was a little bit bigger to the system as a whole of what does the minus sign mean? Energy is being removed from the system. Positive work is the addition of energy to the system negative work is the subtraction so there's more energy coming out through the friction than is being put in by the motor and this system we're talking about being of course the two masses. Alright so there's the first term now we can look at the kinetic energy term there's two things so we have a change in speed of the bracket I labeled A and the counter balance I labeled B and those can be calculated separately there's no need to put them together look at each one separately make sure the units are right each one may have its own individual velocities so I put this A here to remind me that in the bracket I use the velocities of A because they're not necessarily going the same speed at any one time if you'd rather put the subscripts inside on the B's feel free to this just seems a little cleaner to me before we get going are any of those terms zero? if any are zero we're done with them we're doing a smaller problem Bill says no David you says yes what? I thought you meant as a whole no we're just looking now if any of the little parts can drop out if the velocities hadn't changed or if they changed but returned to their original value we would have been done with that back here just no reason to write it out any farther if we don't need to all we're looking for is is there any velocity change if there is then we go down to this step and see if any of those velocities are zero are any of them? remember we're trying to bring the system to a stop so v2 is zero when the bracket comes to a stop the counterbalance will also come to a stop when the bracket comes to a stop I think it's important if you do each term individually this way if you don't you're very likely to lose these minus signs here and I say that from the experience of having seen students do that hundreds of times over the years so the mass of A 14 kilograms its velocity is 1.2 meters don't forget there's a minus sign in front of that is that going to give us proper units we have Newton meters over here so I want every other term to also be Newton meters kilogram meters per second squared and another meter we have Newton meters so the units are okay there this will be a negative term does that make sense? that means its losing kinetic energy is bracket A losing kinetic energy? of course it is first it's moving then it's not note also though remember Joey's question was what was the direction v1 was moving first for the kinetic energy term that wouldn't make any difference since we square the velocity we don't care which way it's going we just care if it's going and the second term this mb is the big unknown for the problem so we've got to leave that in there as a variable but we do know the velocity what's the initial velocity when we start the problem what's the velocity of the counterbalance this is moving half as fast as A is moving half as far we already got that if it moves half as far at the same time it's moving half as fast so its velocity is 0.6 now the mass is in kilograms our units are going to be fine so this term happens to reduce to something a little bit easier I believe minus 10.1 minus 0.1A double check this I think that's right this is all in units of newton meters we checked that at each point and we know that mb must be in kilograms for the units to work what's nice is if I do this I can leave out the units because I've double checked them now and down to just a very simple term with our unknown in it as a single variable did that double check out these two coefficients double check ok great alright so second term we have the counterbalance drops a little bit nothing else is changing in height so we don't need to concern ourselves a little bit so it's mb g delta h b but we know parts of that because mb is still unknown so we have to leave it like that g is 9.8 meters per second squared is this minus that's obviously the si value for g but should that be minus 9.81 meters per second squared I've got a no I've got a no this is normally another no no it shouldn't in these classes we treat this essentially as a constant excuse me a non-negative constant the fact that the block is moving down we take care of with the delta h and it moves down a distance of .2 meters but that's minus .2 meters because it drops that accounts for the fact that there'll be a decrease in the potential gravitational potential energy we get what a minus 1.96 mb is that right now again I know mb the mass of the counterbalance must be in kilograms for these units to work I know it must be a minus because it's the only thing I have moving vertically and it goes to a lower height or decreases it loses some gravitational potential energy so I'm down to a couple very simple terms it's getting to be a real simple problem that we'll be able to put back together shortly so let's do the last term what I'll do is I'll move my work term over here just because I will need it that's what I want yeah it's minus .12 meters okay for the potential energy of the spring we only have one spring in the problem so we only have to worry about that one but I do have a couple little pieces I need to figure out k is given it's in kilonewtons per meter I'm going to make that newtons per meter because everything else is just in newtons but I've got that term we're okay I need though to figure out what del one and del two are do you remember how those are defined your book doesn't do this I think this is a much simpler way to do it the book I think uses an x in here defines it in the same way but uses an x in here which I find problematic because that looks like it'd be the same x that this is moving and it's going to cause trouble so this means it's in here to the spring itself no reference to x or y or any other part of the problem remember how we defined del? David? that should be the difference between the the difference from the point of the rest of the spring okay the working length if you will at whatever particular point minus the rest length of the spring that will tell us how much when a spring is either stretched or compressed it stores potential energy so we need to know what those are l one well these are two two right triangles we have going here the first one when it's at point one it's the triangle shown there of point three meters by point two meters and l one don't worry about the fact there's a lead to the spring just take it as the simple distance that it is you can make these too hard if you want and l one comes out then to be looks like it's going to be somewhere between point three and point four maybe point three six I think I had point three six one as a matter of fact and l zero is the rest length of the spring that was given that's the spring you take the spring out of the box lay it on the table that's how long it is so that term is so del one is point three six one minus point two five we don't need to worry as much about minus signs because this term is going to square anyway but don't get in the habit of getting sloppy with minus signs point one one one right and so we'll square that put it in there del two we find in the same way just now we have a slightly different triangle because now the spring is quite stretched out now it's going to be anchored from point three all the way out to point seven so that's across the top the vertical distance is still the same so that gives us l two and what is that term point seven to eight so that's l two and we subtract from that five units in there and so that equals point four eight and now we have our potential energy term is this going to be positive or negative or can you tell it's not always easy to tell on a spring it sort of depends on just where it starts and where it finishes so would you expect this in this problem for it to be positive or negative this potential energy term positive the spring is relatively relaxed it gets stretched a great deal so it's got a lot more potential energy stored in it remember we've got to make the k term good for newtons rather than kill a newtons so that our units will work and we can just fill in terms of switch del one and del two here but the minus sign would have told us that and we get again it's positive like we expect what's that come out to be thirty 34 yeah 30 so 34 we'll call it 35 and that's newton meters newtons per meter times meter squared that's newtons meter so the units match all of the other terms we have okay let's need a little board space now now we're all done we can put it back together piece by piece port v minus 12 newton meters we know the change in kinetic energy term loop's almost erased it there minus 10.1 minus 0.1 a we know already that's newton meters and I know that the mass must be in kilograms the delta vg term is right there minus 1.96 mb I know it's in newton meters and I know the mass is in kilograms so we're getting to a very simple equation now and the spring term is all newton meters because I've already checked all the units on the small part of the problem end up with a very simple equation to solve there's no squares in it it's a linear equation very easy to solve now without making any mistakes hopefully much easier to solve than if we had left it in that the full form and just put everything into it in this fashion this is a much simpler equation to solve than that one is some of you love a challenge I know so refuse to do it this way you get it right I don't care but if you get it wrong sheepishly appear at my door and plead your gaze but we're now left with a very very straightforward problem 7th grader to do that now so you did all the hard work you hired that out now to your little brother Phil got it Chris I have to ask you did you dump everything into the full equation and do it from there okay you better be right we're just going to suffer our abuse so sorry Philly you said you had it Tommy you got it I got it 7.7 it's not that easy to solve this I wrote it down right minus there those are two minuses that was plus David what do you have how about 17 kilograms follow me like 17 kilograms Chris is just going to say that's what he got no matter what anyway so we can trust him Joey 17 kilograms we already knew the units were kilograms we worked that out earlier that's the only way all the units are going to work and a fairly complicated problem broken into just a bunch of small problems we could do one at a time track the units track the minus signs make a lot fewer mistakes I don't know maybe we should ask Chris if he got it right trust his answer yeah Chris he's just saying that to be nice to him he knows how much it means to me to be right okay alright let's try another one I'll leave this one much more to you guys let's see I want to save that one for a test this one alright this one will draw by hand alright we've got a system starts from rest at this point we've got a box that's going to run along the surface it's being pulled in that direction another weight that's hanging there we'll call that one A and B a fit of originality okay and it starts from A attached to a spring so that's it start position at rest without any of the details put in yet alright now some of the details this distance is 18 inches the spring constant is 72 pounds per foot rest length is 12 inches the box A is 4 pounds box B is 8 pounds total friction that's pretty nice I guess you can't fault me for being nice on some of these things and what I want you to find the velocity of the system when X is 4 inches let me double check I've got all the pieces then when you've got that find what the maximum X is sooner or later the spring is going to get stretched enough as box A moves away from the anchor that it's going to bring the system back to a stop anytime the gravity springs all of the things that work energy equation works pretty nice but this is your problem we'll see how Chris does this time I'll dump it all into the equation start with the work energy equation see if any parts of it drop out if they do a smaller problem right from the start this will be 0.1 and somewhere further down here we need to well the first part is to find out where that is or what its speed is at that point start with the work energy equation and see if any of the parts drop out if so you're doing a smaller problem two of the parts drop out look at them one by one don't look at the whole system don't look at the whole equation just look at it one part at a time see what's going on start from rest start from rest in that position let go of it there the weight starts to pull it down starts to stretch out the spring I guess it's possible when released if the spring already had enough stretch in it that it wouldn't even let it move I guess that's possible that's not the case here I test all of these in my basement so I know this one moves well that'll be 0 in the absence of any non-conservative forces they come from the outside pushes, pulls friction those kind of things are in the work term are there any of those non-conservative forces in this problem do we have any speed change do we have any height change do we have any spring length change any of those will cause those terms to go any of those could cause those to go away or stay start with the work energy equation very easy first step just write that down just do that right off the top and you've got a smaller problem to work on if they all drop out you're done I guess that could be the case if the system didn't move that's a pretty boring problem Joe, what do you think? I think so too remember I said the system's frictionless we don't have any outside pushes, pulls, we don't worry about gravity because that's taken care of in another term we don't want to count for it twice we don't worry about the spring and force it exerts because we also take care of that in another term so you can do each one of those individually watch your units watch your plus and minus signs the recommendation is do them one little piece at a time see you do that, it starts getting smaller and we're looking for that velocity so that's going to have to remain a variable but it's the same for both of them so that's a little bit easier remember the mass of the two boxes remember what I said a couple weeks ago to do with the mass in the English system for these kinetic energy terms we need the mass of each of the boxes not the weight which is given the system starts from rest remember what I said to do with the mass in the English system a couple weeks ago what units to use we'll have to do it for A we'll have to do it for both is W over G you're given W G is we use the English units 32.2 for the second square remember what I said to do with the units in the English system didn't want us to use slugs someone suggested just doing just leave the units like they are right there because once you put them back in here the units are going to correct themselves in the end and it won't matter what unit you have nobody's not going to stick with those why make a move to slugs or pounds mass or anything else so that's about 1.8 what is that anybody have that? I don't have that written down yeah I don't have that written down 4 divided by 32.2 1, 2, 4 and the units are pounds second squared just leave them like that put it in the equation just like that since these both have the oops I crossed out the wrong one there since they both have the same velocity in this case we can bring a lot of this together it gets a lot easier the mass of B is half of that we can automatically work out with velocity in feet per second 1, 8, 6 feet 2 squared does that sound right? in foot pounds we already have feet over here we haven't used any other terms so we might as well do it in foot pounds we do have a some distance we're going to move here we do have the distance there in inches so we're going to have to be careful with it did somebody confirm that? okay alright so there's the masses and then that's that's 1 half the mass of B so we don't have to really figure that okay then do the other two terms the velocity is involved in the gravitational terms so we should be able to figure that one out we'll also watch plus and minus signs this should be positive because we're going from rest to some speed it doesn't matter which way we're going because the speed is always squared we don't square the velocity there's no such idea of squaring the velocity only block B changes at night all you have to do is make sure it's in the right units we have foot pounds here so we might as well stay in foot pounds there now's the time to do it it's very easy to do it when you're working on these little little sections of the problems your unit's in here don't lose them Tom once you've checked them they're all uniform and known and paid attention to then you can leave them out just to put the final form of the equation together the mass of B is twice the mass of A 2 4 8 pounds second squared for foot just leave the mass units like they are they'll work out 2.2 feet per second squared that's G no minus sign on it G does not have a minus sign on it delta H when the upper box moves 4 inches block B is going to drop 4 inches so we're putting a minus sign for the drop divide by 12 and we get feet and then everything will be in out of foot so see the mass units just immediately fix themselves when we need them very easy way to do it and so we get 2.66 and we know the units foot pounds or pounds foot doesn't matter which way to write those and so the last term to work on is the potential energy term makes sense so far Tom see you almost had the 4 without the units didn't you okay use the potential energy term the spring potential energy it moves 4 inches and we needed it to move but it's much easier to look at the units in that one little tiny thing than it would be if you dumped everything into the equation very easy to lose minus sign units lose little parts that were put together so the only thing left to do then is figure out how much the spring is stretched at each of those 2 places first one is easy because it's a 12 inch long spring and we had to stretch it 6 inches just to hook it up so the Dell one is pretty easy Dell 2 is not that difficult you just have to do a again a little bit of a bag of air 4 inches down 4 inches over 18 inches down Dell 2 remember the length of the spring is not that Dell term it's the change in length from rest to Dell 2 you can figure it in that by subtracting 1 foot from it remember we'll need that in feet just to make everything else work out the spring constant is in pounds per foot so we'll want the 2 length terms in feet, respectively Dell 1 is 6 inches but that's half a foot 2 5 3 7 feet that's not right starting at an inches but we've got to get it in the feet so now we have those 2 terms 72 pounds per foot spring modulus Dell 2 0.5 3 7 feet squared minus half a foot it's going to be positive make sense Dell term is never since it's always squared those are never going to be negative but the difference between them could be negative depends on how we end up but in terms of Dell we don't even care that it's squished or stretched just as long as it's not as a stress length and we'll get the units on this again of foot pounds 4 pounds foot to go with everything else anything that makes sense don't drop and lose your units yet 6 inches and you need it in feet so that's what that is and don't forget to square it the units wouldn't work if we didn't square it and that term I think comes out to be 1.36 foot pounds that's not right now we've got all the pieces put it back in back into the work energy equation put it back into there left hand term is 0 Dell to T that's got our unknown in it we already know the units work so we're not concerned with those anymore minus 266 for the gravitational term and 1.36 for the elastic energy term and foot pounds and that will give us V2 in feet per second that's the only way it's going to work out because that's what we have here feet per second with those units a mass is the only way it's going to work out give us foot pounds so we've checked all the units checked all the minus signs we've got a very easy problem left don't you think that reduces them what you do do you like that are you going to be one of the disciples use that method I don't know what you think that's a I think that's a much easier way to do it I don't know Chris likes the big long equations because then you can show them to girls and bars and stuff look at the stuff I can do it all work out and so V2 equals 2.64 per second that's the only way the units would have worked out 2.64 feet per second okay we're all right about the end so let's just set up the last part of the problem remember I wanted to find X max sooner or later the spring is going to be stretched out enough that brings the whole shebang to a stop so that's a different situation you need a start at the work energy equation somewhere it goes a little bit farther to X max so we only have enough time really just to set up that equation oh maybe it doesn't change enough yeah I guess it would now which of those terms are zero if any and either zero or got we have a smaller problem David's got at least one an answer of none David? X max is when it comes to a stop because if it didn't come to a stop there would go farther X max would be greater so that defines X max when the system returns to stationary it may not stay there it might oscillate because of the spring it might oscillate because of the spring but there is some point when it'll meet its rightmost travel here at three when it comes to a stop at that time it's gone from still at the start to still at the finish for our delta T remember we don't care what happens in between any other term zero you have a term David? do you? yeah we still don't have any work there's still no friction I guess we could have done the problem where after four inches we have a rougher surface or something but that's not the case so now we have a problem that's half the size it was when we started and much simpler to look at now don't forget that Vg will be a function of X max and so will Ve so both of those terms will have our unknown in it and still when you put it back together I think you get a relatively simple quadratic that just finishes everything up and that brings us to the end of the week I'll give you X max so you can double check it when you're looking for something to do over the weekend I have that goes out to 28 feet and again that just comes from re-figuring those last two terms both of which will have the unknown in it I'm not sure how far out it goes any questions? Chris did it's pretty simple I think if we keep it in those simple smaller terms it sure works better for me I don't know about your brain my brain just works a lot better to do these small problems rather than big problems it's a lot easier to watch the units a lot easier to watch the minus sign and the squares