 There we go. Homework from last day. Most common mistake I see, kids get the forces okay and they do MG like that and then what they do is they do perpendicular this far and they draw this parallel to the ground. So they call this MG perpendicular and they call this MG parallel. That's not the way we drew this. You want MG to be the hypotenuse of your triangle, not this long one here. You want the parallel to be parallel to the slope so you would want to draw this as that far and that far. Otherwise your trig is going to be wrong and everything after that is garbage. Having said that then, can we all go back and look at example eight right here? I didn't get a chance to finish it and since I like this question, I like this question, I like this question, I like this question, I really should finish it. So here was our analysis, here was our situation. Nicole we said we had the mass sliding up a rough ramp so it's going to slow down really really quickly. In fact we said if it's sliding up friction and MG parallel are in the same direction. Your equation was winner plus winner equals MA. Oh we said friction is mu times a normal force which is mu MG, it was mu, I don't know the normal force, it was mu MG parallel perpendicular, sorry MG perpendicular which was MG coasts we said. We found that the acceleration was 6.02. Then let's continue on. Part C says find the time to stop. If I want it to stop I know v final is zero. Now I called the acceleration negative. The reason I called the acceleration negative now Brett is we're slowing down, we're slowing down so it's negative acceleration from last unit and if I crunch the numbers I'll get this zero minus 20 divided by negative 6.02. I'll get the time it takes for this block to come to a stop as zero minus 20 divided by negative 6.02. 3.32 seconds that's part C. How far up the slope does the mass travel before stopping? What are they asking me to find for part D? Distance. So I'll go D equals question mark. Let's see I know that T equals 3.32. I know that Vi I believe was 20. I know that A is negative 6.02. Oh I can do this. D equals Vi T plus a half A T squared. 20 times 3.32 plus 0.5 slowing down so negative 6.02 3.32 squared. And I think I said before I'll say it again I like this question I like this question I like this question. 20 times 3.32 plus 0.5 times negative 6.02 times 3.32 squared. And I get 33.2 meters. Is that right? You get the same thing? Anyone? Yeah. So having said that questions from the homework now is your chance to ask. Today is a continuation of the last lesson. Today is just called advanced inclines. We're gonna look at yuckier inclines but it's gonna be the same concept. Any questions you would like me to go over? Yeah. Eight. Eight is very similar to the one we just did. Were you writing down what we just did? No? Then I'm gonna say later on tonight or later on in class if you copy this down because number eight, say on your example eight, in number eight do you not have an initial velocity in your look at yours? Yes? Is it sliding up the hill? Is there a coefficient of friction? Is it asking how far it goes until it stops or how long until it stops? I think that's what we just did here. So I would normally absolutely do that but I thought I was preventing your question by doing this one right now so if you were zoning out I'd say that's too bad. Watch the video or look at the notes later. Any others? Love to do number eleven. Okay. A five kilogram mass is at rest on a 15 degree slope. Find the value of the coefficient of static friction. We did say there was two types and the author is gonna be accurate on that but we said we'll just pretend there's one. I'm gonna dull. So here's my ramp. What's the angle? 15 degrees so my ramp is way out of scale because that's probably near the 45 degree angle. Who cares? Here's the mass. What are the forces acting on the mass? Get the obvious ones. Good old MG down. Normal force. Now it says that it's stationary. So if it's stationary which way is friction acting? Upwards because the gravity wants to pull it downwards so I know friction this way. Oh but I need to break gravity up Brett. I'm gonna break gravity up into MG perpendicular and MG parallel. Is that okay so far? Okay. Who's winning? Well it's a trick question. I think it's a tie because it's stationary but when in doubt if I'm not sure I usually like gravity win because I know usually things want to fall down so I'm gonna go like this. I'm gonna call MG parallel because that's what's pulling it down the hill minus friction and Brett why does that equal zero because the mass is not it's excel it's not accelerating stationary so I know that the net force is zero and I heard you say because they're the same in fact Brett if you just did that right away I'd be okay with that step if you said look these two here have to be the same size. In fact I'd say excellent for spotting that without having to write the winner minus loser. Friction is what times what? I don't know the normal force. Oh but look look look look another force same size as a normal force what? So I'm gonna drop down MG parallel and this is mu MG perpendicular and now I'm gonna start to do a little bit of trig. We said that as long as we draw our triangle the same way this 15 degrees right here Brett ends up being this 15 degrees up here. Let's do parallel opposite adjacent to hypotenuse which trig function. Turns out MG parallel is gonna be MG sine of 15 when you do the cross-multiply thing. What about MG perpendicular adjacent and MG is what I always go back to is the one that I know Andrew so I have an A and an H Brett which trig function there. Turns out it's gonna be MG cosine of 15. Oh and I forgot the mu in front and I heard somebody murmur under their breath the G's cancel. As it turns out the M's cancel and the G's cancel and I end up with sine of 15 equals mu cos of 15. How would I get the coefficient of friction by itself Brett? No. What's happening between the mu and the cos mathematically? Times inks so how to move the cos over? Just divide. It turns out mu is sine 15 over cos 15 and Matt remembers that sine over cos sine from math 12 is what? It turns out it's a tangent. You guys don't know that yet those of you math 12 you'll eventually learn that sine divided by cosine is actually the same thing as tangent. Turns out it's the tangent of the angle. Engineers would use that shortcut if they were you know designing roads and the grade on a road and things like that. Is that okay? Brett yeah? I like that question. I like that question. We call that a static equilibrium question. Very nice. Any others? Going once? Going twice? Okay if you're done with that if you want to hand it in either today or the next couple of days and take a look at the lesson I just gave you then. Less than seven please. Advanced inclines. So your test right now I figure is going to be around the tenth. Ready? What kind of curve balls can we throw at you? And I think at least one of these would be fair game for a written but I'll let you know level of difficulty as we go through. Example one says how much force is required to pull this ramp this mass up the frictionless ramp at a constant speed? I would do two things here. I think Emily if I was a good student I would underline frictionless because that makes the question easier but there's a trigger phrase. You with me now? Okay there's a trigger phrase here and the trigger phrase there is constant speed. Emily what does constant speed mean? Ah that means that my acceleration is zero. Now that means something really important this unit it means when I go winner minus loser equals you know what my equals is going to be? Why zero because what's force? What times what? And if your acceleration is zero Emily what's mass times except what's any mass even if it's a bunch of masses times zero what's it going to be? Ah so that's going to change my winner minus loser equation. How are we going to start this off? How have we started off almost every force question we've been doing so far? I think we're going to now I call it a free body diagram. Since they gave me this lovely picture I'm just going to label the forces right on here I'm not going to go walking over here representing the mass as a dot. If I had to I would if my diagram was too cluttered but for now let's label the forces. What are the forces acting on this? Get the obvious ones. Gravity I'll draw it nice and big MG. What else? Normal force which is at a 90 degree angle to the surface. What else? I heard it. Since it's a rope we've traditionally called a rope tension so let's put a little arrow here and put a capital T for tension. Friction? No by the way if they wanted to make this tougher could they have added friction? Which way would friction be acting if I'm trying to pull up and not succeeding? Which way would friction be acting? Down. What extra force? Mute times a normal force? I don't know the normal force. Oh look I know another force. I can get it. Now this is my free body diagram. Now I'm going to break gravity up into components with a dotted line and this is MG perpendicular and this is MG parallel and this angle right here is 40 degrees. Right we said as long as you draw perpendicular then parallel that top angle will always be the 40. We proved it once so we said good enough. Who's winning? Ah what did we say? It's a tie. So can you see what my along the incline equations going to be? What two forces are the same size in this diagram? That's my equation. Because Emily noticed that constant speed means a equals zero. In fact you know what Brett this is almost this is mathematically the same as yours that was standing still because standing still means no acceleration as well. Einstein was the one who said you're standing still you're moving at a constant speed. You can't tell a difference and mathematically there is no difference. Oh and I think tension is what they want me to find when it says how much force is required to pull this mass up at a constant speed. Okay MG parallel let's do the trig opposite adjacent or hypotenuse opposite adjacent or hypotenuse. Which trig function? Yeah I think most of the time parallel ends up being sine except not always I can think of a few weird ones where it does not specifically if we were tugging upwards at an angle and had to break another force into a component. So I don't memorize like we did last unit while I here it's I draw the triangle out but then I realize that I know the hypotenuse so it's going to be cross multiplying. So the shortcut is once you know the trig function it's the hypotenuse times the trig function which is sine 40 and we're done. The mass is seven g is 9.8 sine 40 degrees. How big is tension? 44.1 yeah 44.1 Newton's. So what kind of questions now I'm starting to think about those essay questions remember there was one on your last test those using principles of physics right to explain questions. So what kind of additional questions could I add to this? Well oh as theta increases as the angle increases what happens to the tension. Okay as the angle increases what's going to happen to your tension is it going to get bigger smaller or stay the same and convince me what if we got to here what would the tension be more specific wouldn't it be? In fact aren't we using the Blackabee theorem again? From last day we okay increases use the Blackabee theorem. I would write more than that if it was a test Emily I would say if it's level ground tension zero if it's vertical tension is mg so as theta increases tension increases I could do that. In fact we sort of answered this question with that skateboard one never mind we'll come back to that one. Next page. Two objects are connected as shown M1 is 12 kilograms and it sits on a horizontal surface has a coefficient of friction of 0.17. M2 is 9.8 kilograms and it sits on a frictionless surface at an angle of 56 degrees find the acceleration of the system okay but no relax you know I'm gonna do first I already got the dope in front of me you still know someone doped for me I were another dope neither will I gulp I will plunge right in it's a force question you know what I'm gonna do first darn right I'm gonna label the forces oh that's a great point what do you guys want to use let's use 42 degrees means I stole that diagram from somewhere else and changed it what are the forces acting on this mask at the obvious one what else and I'll call it normal force number one since the other masses also on a surface so I'm assuming there's gonna be two what else which way why I have to use my imagination a bit there's no way this can be moving to the left if it's moving at all this is pulling it down into the right and it might not be moving if I get a negative I'm going to assume it's moving to the right if I get a negative acceleration now what that's telling me Trevor is friction isn't enough this mass is not enough to overcome the friction of this mass and it's not going anywhere but let's assume it's moving to the right so friction force one is to the left what else tension that's what's pulling it to the right what are the forces acting on mass to get the obvious ones gravity straight down what else normal force number two what else tension this way what else I would normally say yeah but it looks like that mass over there is friction list for some strange reason otherwise I would have friction acting in this direction to Caitlyn but don't need it I think that's it ah I got a real issue though because I want to incline before I do anything else now I'm going to break gravity up into its components mg perpendicular and mg parallel where this angle here and this angle here are 42 degrees who's winning now this is tougher what we're going to do is we're going to walk along the rope which is level for the first part and angled for the second part sorry angled this way for you guys for the second part who's winning all right well I can't be M1 so we have to make the assumption if any force is winning what's the only force that could be a winner here which way is this whole thing moving to the right and down what's the force that's pulling it to the right and down don't say mg mg parallel is here's my winner here's my winner so anything that ends up pointing down the hill winner anything that ends up pointing up the hill loser winner now I'm going to walk her along the rope I run into tension up the hill loser oh I run into tension that'll end up down the hill Brianna winner oh I run into friction which is going to end up pointing up the hill once it gets over here loser have I walked down the entire length of rope then I've got them all and since I've been looking at more than one mass it's M1 plus M2 times a and you know what I should call this I think I called this M2 so over here I should have an M2g parallel shouldn't I otherwise I might get my M's confused and cancel when they don't tension cancels yay what is M2g parallel well let's look at the trig here's my angle opposite adjacent or hypotenuse opposite how about here which trig function is M2g parallel so we're gonna get M2g sine 42 minus counter friction is what times what I don't know the normal force oh but look look look look look look look I don't know the force same size as a normal force what which which one M M1g okay and this one's nice and level so we are back to our familiar from grade 11 where mu where normal force was always equal to mg it's not on the angle so mu M1g and Connor how to get the a by itself what would I do with that bracket let's do that on this line is that okay well not divided by a mr. do it the acceleration is going to be M2 9.8 g also 9.8 kind of convenient sine 42 minus mu 0.17 m 12 m1 g is 9.8 all divided by 12 plus 9.8 bracket 9.8 times 9.8 I guess gonna use my squared button sine 42 close off the sign minus 0.17 times 12 times 9.8 close off the top divided by open a bracket 12 plus 9.8 close off the bottom do you get 2.0 3079 3 am I right 9.8 9.8 sine 42 minus 0.17 12 9.8 divided by 12% am I right yeah acceleration equals 2.03 meters per second squared before you turn the page before you turn the page find the tension that'd be a nice part B to this question to find the tension how many masses are we gonna look at just one which mass I don't know do both masses have tension on them then I can use either one I'll try and use the one where tension is on the winner side because it's easier to get the t by itself then I think I'm gonna use mass one what's my equation going to be here if I just look at mass one who's winning tension who's losing oh okay tension minus friction equals m1 a tension equals m1 a plus friction force one Mitchell friction is what times what I don't know normal force oh but look look look look look look in fact it's gonna be m1 a minus mu m1 g right that's not too bad equation to generate tension equals m1 12 a 2.03 minus 0.17 times 12 times 9.8 shouldn't it be plus yes it should be plus I don't know why it became minus suddenly mr. do it thank you who caught that corner I was wondering because I was doing the math in my head and getting a negative answer and going what the heck is this yeah anyhow it's gonna be 12 times that number plus 0.17 times 12 times 9.8 and I get 44.4 newtons how else could I make this nasty so look at the question that I gave you what if instead I gave you tension and mass one could you use tension and mass one to find the acceleration and then once you knew the acceleration what if I didn't tell you mass two could you use tension and mg parallel to find and acceleration to find mass two yep that's an easy curveball to throw at you I think there's one of the homework or there's one in your review which is why I'm not doing one right now example two how would you determine which way this system moves well there's a problem here if this is heavy and this is light I'm pretty sure this will pull this up the slope but Joe if this is pretty light and this is really heavy Joel I think this could pull it down the slope and it also depends on what friction is right here if friction is really really big maybe they're an equal how would I determine the direction now I'll be honest in real life I would guess and I would probably guess this guy is the winner and I would solve for a you know how I would know I had guessed wrong I'd get a negative answer and all I would do very quickly is point my friction in the opposite direction in other words if I thought this was winner this was sliding up I would have had friction pointing that way no this is winner than frictions pointing that way I don't even need to redo the equation I change a plus sign to a minus sign I tweak it slightly and I can quickly resolve it get the right answer that's what I would do it's the fastest way but here is the mathematical way what are the forces acting on this get the obvious ones what else what are the forces acting like this guy get the obvious ones sorry MG big MG normal force and tension oh and big M mr. Duke capital MG perpendicular and capital MG parallel look at your rope two forces cancel first of all or what one force cancels tension does which two forces are in opposition to each other along the rope and one G and and parallel if parallel is bigger you know which way block M is gonna slide down the hill if parallel is smaller you know which way block M is gonna slide up the hill and that's the answer to this if M1 G is greater than MG parallel big M slides up if M1 M1 mr. Duke if M1 G is smaller than big MG parallel M slides down but to be completely honest rarely do I crunch the numbers like that usually I assume the one that gravity is pulling straight down on is the winner and if I get a negative answer it's a quick and easy fix and it usually just means changing a few things on my calculator for that matter like example three says find the acceleration of the system okay what are the forces acting on mass one get the obvious ones gravity what else tension what are the forces acting on mass two get the obvious ones gravity normal force tension who's winning because I need to know who's winning so I can figure out which way I'm gonna point friction well let's assume the five is winning and let's point friction that way let's assume it's getting pulled up that okay Brianna oh and components M2 G perpendicular and M2 G parallel what's the winning force here overall what force is causing this whole thing to accelerate when you walk along the rope itself what's Emily giggling over and smiling for questions that we don't know the answer to but we can solve the first question I asked which is the winning force we may not know why Emily was smiling enigmatically but we know this who's winning there's two of them on there I need more specific please someone said gravity I got gravity here and gravity here and I got more specific the exact name of the force you think these guys are winning if this was winning then friction would pointing up the hill I've decided to let this be the winner I think this is pulling this up the hill even though that mass is bigger I'm gambling that the friction and the slant isn't that big is that okay or if I lost you I can't read you guys right now yeah Kara I have no idea what you're saying I heard the word why and then your volume went back to the original again I didn't have I had to guess something you just guess you know how I'll know I guessed wrong if my acceleration ends up being what negative I'm gonna let that guy now as soon as I decided that I pointed friction down the hill if I had that this guy be the winner friction would have been up the hill which is the only if you guess wrong change you have to make to quickly redo the question oh and to do that all you do is make the friction positive if it was negative or negative if it was positive very very easy fix to do so let's let this be the winner for now if I get a positive acceleration then Emily I'll know I'm right winner then I run into tension which is a loser what about this tension here well and when I follow it all the way around it ends up being winner down the hill friction ends up being a loser followed all the way it ends up pointing up and slowing down and there's one more m2g parallel ends up being a loser as well and that equals m1 plus m2 times a so again all I did here was guess what my winner was and I'll see if I'm right oh tension cancels I get m1g minus friction is what times what friction is what times what you times a normal force I don't know the normal force oh but look look look look look another force same size is a normal force what this is going to be mu m2g perpendicular minus m2g parallel that equals m1 plus m2 times a let's go label our diagram and do the trig this is 30 right here which means that this is 30 up here Jacob m2g parallel opposite of Jason their hot news I agree what about m2g I thought news I agree what about perpendicular I agree so if m2g perpendicular is adjacent you know which trig function I'm going to drop into here coast and if m2g parallel is opposite you know which trig function I'm going to drop in here my next line is going to be this m1g minus mu m2g coast 30 minus m2g sine 30 dedication and Sean how to get the a by itself I want to do that on the same line can I do that right now if that's okay divided by m1 plus m2 times a let's plug in the numbers and see if we did guess right now look up look up if we guess wrong if we got a negative acceleration Brienne then that would be a loser put a negative there that's pointing in the wrong direction but a positive well we'll think about it it would look like this this would be your winning direction so winner make it positive this right now you have pointing in the wrong direction you have to point it up so you would still want to leave it a loser but you'd need to make that a loser be a very very easy fix to do if you redrew it really quickly and if I confused you don't worry I'll show you in a second let's crunch the numbers what's m1 5 what's mu 0.05 times 8 times 9.8 coast 30 minus 8 times 9.8 screen froze well while I'm getting the screen back from the dead can you substitute in the numbers and then try and get an answer please there's what I get when I plug things in 5 mass 1g minus co-direction mass 2g coast minus mass 2g sign it's a bit of typing but really top is all one fairly easy line to type divided by the bottom line I can do this all in one step and I'm going to bracket 5 times 9.8 minus 0.05 times 8 times 9.8 coast 30 close bracket minus 8 times 9.8 sign 30 close bracket closed off the top divided by 5 plus 8 you get 0.4927 or am I wrong let me double 0.49 0.493 if we go to three sig figs 0.493 meters per second squared character we guess right yes now put your pencils down for a second put your pencils down what if we had guessed wrong here's all I would need to do so I get a negative here go I go like this look up really quickly draw friction in that direction this is now the winner this is now a loser friction is still a loser so all I would change is that and that that's a loser that's a winner it's a really easy fix you can rewrite everything if you want to but you really don't need to okay and I thought I turned something on that was supposed to make all that stuff vanish and it didn't so we'll have to do this manually undo undo there we go oh now that you know that could you find the tension yeah I probably use this mask to find the tension even though tension is the loser there's way fewer forces here you have one two three forces here you only have two forces of the an easier equation to solve example four and I think this is the last one yep I like this question in that this question combines everything we've done so far this year we're going to use some kinematics and all sorts of stuff it says this at the top of the hill an 84 kilogram skier has an initial speed of 5.8 meters per second the ski hill has a coefficient of friction of point 08 after the skier skis through a vertical drop of 3.2 meters what's his final speed Nicole you know I'm gonna do here first once I wake up doll yeah I'm gonna draw a little picture well played Nicole well played I'm gonna have my skier skiing to the right and down so here's my hill and I'm trying to think here I think I need a theta and I think I forgot to give a theta in this question okay it's okay back there so I keep hearing little noises and it's irritating shush thank you um let's see I gave you a vertical drop of 3.2 yeah I need a theta assume theta equals 16 degrees 16 degrees here's my skier how can I find his final speed well that's v final what have they told me what have they told me vi that's nowhere near enough they told me the vertical distance but have they told me the distance down here can I figure it out though well yeah if I redraw this triangle with a 3.2 and a 16 and an X how far along the slope does this person go how can I solve that can I yeah opposite hypotenuse which trig function now this time I'm finding the hypotenuse which means I better write this out the sine of 16 equals 3.2 over X I think X is going to be 3.2 when I cross multiply divided by the sine of 16 what's the distance down the hill that this person travels you guys get 11.609 11.61 so he travels for a distance of 11.61 meters I have vf I have vi I have d what else can I find from this information well what are we been finding an awful lot so far during this unit the acceleration and Katie I think that's what I'm going to spend the most time on this question doing I'm going to solve for a here's my drawing but I'm going to put the skier right here what are the forces acting on my skier get the obvious ones what else what else is their friction yeah which way which way up the hill I need to break up MG into components so there's MG perpendicular and there is MG parallel by the way the only reason I didn't draw this on here is because to draw this on top of my skier I didn't think I could fit it in on because I put this on the left side of the margin so I quickly drew it here who's winning free so my equation is going to be MG parallel minus who's losing friction not very much friction we're on snow and that equals since it's only one mass ma this is going to be MG parallel minus friction is what times what I don't know the normal force oh but look look look look look I know the force in size is normal force what it's going to be mu MG perpendicular equals ma Connor how big is this angle how big is this angle 16 opposite adjacent or hypotenuse adjacent hypotenuse adjacent so for parallel the trig function because I know the hypotenuse this time it's going to be hypotenuse times sine MG sine 16 minus mu and perpendicular ends up being coasts this equals ma oh you know what I notice something cancels what John the masses cancelled turns out it means that by the way this is why an adult and a little kid can ski next to each other without having to constantly adjust their speeds otherwise wouldn't work parents could not ski with their kids oh a is gonna be 9.8 sine 16 minus 0.05 I think that was mu let me double-check 0.08 sorry times 9.8 times the cosine of 15 what's my acceleration down the hill cosine of 15 how about cosine of 16 mr. Duke I did that up here too you get an acceleration of 1.947616917 yeah how about 1.948 1.948 meters per second squared so I'm not done that's not what the question asked Nicole what did this question ask okay vf squared equals vi squared plus 2 ad vf squared equals what was vi 5.8 squared plus 2 times 1.948 times how far do we say you're sledding for 11.61 now 78.86 is way too high that's the it's you're going faster than on the freeway oh but that's cuz we didn't find vf what is this equation find us vf squared is 78.9 how do I find vf take the number and do what square root 8.88 meters per second about 32 kilometers an hour sure that seems okay okay what's your homework number one number two three four one two three four so far seven and then technically you can now do every question on the big unit review although I'm gonna look at some weird examples so you can work on those take home quiz next class