 Oh, I just start recording and sure enough so plan today work through some examples I'll take questions from the review. I will be doing I will be around after school on Monday I'm not going to be doing like a formal formal tutorial. It's going to be hey I'm around come on in if you have questions, okay? And the reason is you guys are actually a couple of days ahead of my other blocks if I did a tutorial Monday after school My other blocks One of my blocks is going to be doing torques The latter question on Monday, so I wouldn't really have finished the unit. They would have just seen it all, okay? Uniform 15 kilogram pipe Of length blah blah blah has a hundred and sixty newtons of force applied four meters from its lower end as shown Using the point where the pipe touches the ground as a pivot Calculate the sum of the torques acting on the pipe So I'm looking at this one Justin. I don't think this one is in equilibrium I think what it's telling me is it's moving figure out what direction and what's the torque So I'm still gonna start out by saying perpendicular components. I'm gonna list all of my torques I think but this time I Don't think we're going to get equal Okay, a little bit of twist here What are the forces acting on this beam get the obvious ones? Okay, so there's going to be the center of mass right here and it's in screen frozen. I'm sorry There we go. Let's try that I hit that button. Thank you there's going to be the center of mass right here and Yeah, looking at my answers, I'm sure there's an unbalanced torque because they're saying what is the sum of the torques and it's not zero Mg and then we have Tension I think for this one what I'll do is I'll calculate Oh, there's more forces, but I think the rest are right there and since this isn't a written question I'm not going to freak out about getting the free body diagram right at multiple choice I want to cut corners be a bit of a rush what I'm going to do is I'll figure out the Clockwise torque from this I'll figure out the counterclockwise torque from this and I think I'll go bigger minus smaller Is the remaining difference? What's left over because I think clockwise and counterclockwise you can kind of think of them as opposites That's positive and negative numbers like we've done with other forces, but in this case. It's torques. We have to go components Mg perpendicular and Tension perpendicular Angles angles angles angles Jun Jun There's a Z. So this right here is 42 degrees No, I'm writing on top of the number that doesn't help you guys at all. That's right here is 42 degrees and Okay, the trick here for this 34 Imagine this going straight down 90 degrees right there 34 there so these two add to 90 56 and these two add to 90 You know the 34 ends up there and you probably noticed That when you have something on an angle the bottom angle ends up in the triangle right there You may have spotted that but I still walk through it because there's all sorts of weird diagrams out there Remember the one we did on the take-home quiz yesterday number three had the tension going in kind of a weird direction That was a tough one So I'm gonna say this The torques clockwise in that direction equal Mg perpendicular times its distance from the pivot how far oh Pipe is five meters long 2.5 and you'll notice again here Dylan here I didn't go some of the torques clockwise equal some of the torques counterclockwise because I looked at my answers And I said oh, they're telling me my net torque is not zero. It's not balanced. So I'm gonna find each one bigger minus smaller Which trig function relates mg perpendicular and mg You probably noticed most of the time when you're dealing with gravity it ends up being cosine The only time that wouldn't is if they didn't give you that angle if they gave you a different weird angle somewhere else and Could happen. So I haven't said to you memorize gravity as cosine, but it's been cosine almost all the time So I think this ends up being m g cosine of 34 Times 2.5 the clockwise torques ends up being math 12s make sure you're in degrees My in degrees I am The clockwise torques ends up being m 15 times g 9.8 times coasts 34 Times 2.5 and I get 304.67 What are the units for torque but can't be newtons newtons is force. This is not force Torque is what times what? force times perpendicular distance it is Newton meters, but that's not joules like it used to be um, oh Let's do Counterclockwise torques. I'll do that over here. So the torques Counterclockwise which is in that direction. I'm pretty sure that's tension perpendicular times. How far is it from the pivot? You have to do a bit of arithmetic Right in yeah for how our tension perpendicular and F Related and I think you've noticed that Most of the time not always but most of the time for tension which trig function has it been usually So this one ends up being tension sign of 42 Times 4 the torques counterclockwise ends up being Tension hundred and sixty sign 42 Times 4 And I get four hundred and twenty eight point two four Big sneeze online Again Newton meters, which one's bigger because these aren't balanced this time which one's bigger That Wrong wrong Yeah, Gord said it see I hope you didn't actually pull out your calculator Hopefully you glanced at the answers and said oh, yeah about a hundred and twenty as opposed to actually crunching the numbers So see so there is an example of a can I call that a curveball not exactly? But Kellen although I think there is one like that somewhere in your review now that I think about it I think there was one where it simply said find them find the net torque and it wasn't zero But hopefully you're getting comfortable now with components and beams at angles turn the page a Student stands on a uniform 25 kilogram beam. Oh Another late person have to pause my lesson So we've seen diagrams like this before clearly a beam Two things on the end and a mass in the middle the only twist here is instead of saying find the force on the Left or the force on the right or both of them say find the mass of the student, okay? Thankfully, it's a horizontal beam that makes things way easier because when it comes to gravity that's gonna mean no components It's gonna be perpendicular So we're gonna start out. I'm going to assume this beam is an equilibrium. They got a woman standing on there Let's assume if it was spinning she wouldn't be standing on there What are the forces acting on this beam get the obvious ones? gravity where Okay, and that's the mass of the beam I'll call that mass of the beam times G. What else? Mass of the student ms times G What else? Okay, this it's telling me the scale is feeling the beam pushing down at 350 newtons now we're not drawing the force on the scale But if the beam is pushing down at 350 newtons forces come in what? So there must be an upwards force right there of 350 newtons and I could call it normal force but since they gave me a number I'll just put the number there one less variable to deal with if that's okay, Evan and There's also I'm sure a force up right here And that's a problem because there's two things I don't know the mass of the student and this force or it would be if I tried solving this using forces By saying everything up equals everything down But with torques of course Evan, I'll put my pivot right there Gets rid of one unknown force Is that okay? Good try this on your own I'll freeze my screen see if you can figure out as a little twist what the mass of the student has to be no components But it's gonna be the some of all torques clockwise equals some of all torques counterclockwise When you're done, let's look up. You can look up. I'll unfreeze my screen every so often We're just trying the second question on our own and then I'll unfreeze my screen see if you can get it What was the mass of the beam? 25, thank you See if you get what I got. Oh Have I gone on a rant about the calculator errors that drive me crazy in this unit? By the way, I find Justin in all honesty The top of the fraction is always complicated enough that you may have noticed I used to always just put everything on the top and brackets everything in the bottom and brackets more and more when I'm typing On my calculator in this unit. I do the whole top I press equals and Then I just divide by the bottom. I It I found it was too many brackets to consistently Easily keep track of and because you're almost always only dividing by one thing or at the most in this case two things I figured and I'm not gonna worry about brackets so much I'll just type in the whole top enter and then divide by whatever's on the bottom enter So I did that here. I did top line got six hundred eighty two and then I divided by nine point eight times one point two 58 yeah, okay. I probably If I was in a rush would have crossed out D anyways because 89 kilograms. That's a pretty heavy student person That's a football player weight. So Message for this question Kellen is on your test. I won't always say find the tension or find the force I may say hey find the mass that's gonna find the mass of the beam would be an interesting one. Okay, or Find the weight of the beam, which would be mg of the beam that would be an interesting one But I don't always just have to define tension Turn the page I Like number three. I like number three. I like number three. I like number three number three is a nice question I like number three. This is about as tough as I can ask you. I think says this a Six meter uniform beam of mass 32 kilograms is suspended horizontally by a hinged end in the cable a 93 kilogram object is connected to one end of the beam and that wants me to find the force on the hinge Now the real answer here is I can't find the force on the hinge directly I think what I'll try and find is it's horizontal component It's vertical component. I'll add those together tip to tail and I'll find a resultant, but to do that That do a lot of torques and you can see seven marks. This actually was from a provincial. I think this was from the 2001 Provincial I think so this is also considered fair game now it was one of the tough questions that year So there were other easier written questions as well I think the written question on energy was a straight roller coaster heat question, which was almost plug-and-chug So, okay Hmm suggestions, I have no idea what you're saying. I Think I'm gonna put the pivot right there. I Agree and then what? Let's label our forces. Okay The beam has mass by the way every once in a while to make the question easier They'll say the mass of the beam is negligible They'll go to our magic physics world and give you a massless beam to give you one less torque to deal with I've seen that happen. I don't think I did all the tests, but in case And that's gonna go center of mass right about Mass of the beam times G What are the forces? tension Like the feeling you get when you haven't done all the homework next What else You're missing an obvious. Oh, sorry. Yeah, okay. Yeah, I got this one I'll call it mass of the load times G Or you call it MO for object whatever just some way that you can keep them straight in your equation, right? That can't be it why can't that be it? Which way is tension pulling what two directions? Up I'm okay with because up could possibly cancel out those guys, but what Megan also to the Okay, it's pulling to the left There has to be a force to the right right there and I've called that usually fx for horizontal And I'm almost positive Jordan that there's also a vertical component right there I'm pretty sure when you hang something that yeah the tension is carrying most of the mass But right here it's taking care of some of gravity, too How far are both of these from the pivot zero so they're not going to give me any torque I got to do one more thing before I can start I like this I like this Yeah What am I going to have to do with tension components perpendicular and parallel And this time tension parallel is going to become important You know why Gordon It's equal to fx. In fact, I can't find fx Can't But I can find tension pair parallel and And Say they're equal. In fact, I'm not going to find tension at all. I'm going to find perpendicular Which is going to help me find fy. I'm going to find parallel which is going to help me find fx tension itself in this question Not useful Because it's at an angle and everything else is componentized if that's a word so Let's see The sum of all the torques clockwise Equals the sum of all the torques counterclockwise And I'll continue down here Clockwise clockwise clockwise. Oh these two The mass of the beam times g times its distance from the pivot Now before we go further look up everyone The only reason it's three is because the pivot is at the end What if the pivot was right there? I'm just making that up. What would the distance be? The reason I'm saying that is there is somewhere on your test going to be one Where I don't put the pivot at the end and one of the most common mistakes kids just look at the length of the beam Oh, uh, distance is four if it's eight meters long or it's a 10 meter long. The distance is five Well, only if the pivot was right at the end and it's not it's just to the right of the center Is a matter or just to the left of the center. So just make sure you think Uh, but this one is definitely going to be three And then we have mass of the load times g times its distance from the pivot Which is definitely six because it's end to end And that equals tension perpendicular Times how far is it from the pivot? You know what I'm going to do since I want to find tension perpendicular I'm going to divide by four on this line. I almost never do that, but I'm scared. I'm going to run out of a room here Okay In fact, I can almost go to my calculator now because I know mass of the beam and I know mass of the load So why not? Of course, as soon as I open my calculator, I cover up the data 32 and 93. Okay 32 times 9.8 times three and 93 times 9.8 times six And just and like I said what I've often started doing because they're long yucky fractions I just hit equals divide by four And my right is tension perpendicular 16 oh two point three Someone else get that too. Okay, and I'm not going to round off because this is not final answer 16 oh two point three and this is tension. So newtons. It's a force You see I can now find f y this vertical up Plus this vertical up has to equal what the two downs Let's write that down I'm going to write little f y colon here just to say here's where I started to solve for f y So when I look at my work later on Jordan, I know what the heck I did and we said this Tension perpendicular up plus mystery force f y equals the mass of the beam times g plus the mass of the load times g That means that f y is going to be mass of the beam times g Plus mass of the load times g minus 16 oh two point three mass of the beam 32 times 9.8 plus mass of the load 93 times 9.8 Minus my last answer which is conveniently stored on my calculator Whoa Did I do this wrong? 32 times 9.8 plus 93 times 9.8 minus answer did I do something wrong here? Now I'm more you got 60, you know too as well folks What does that mean? For me Well, I think it means that this is pulling down, but that doesn't make physics sense to me Let me think here for a second. I mean it's possible You could have tension so tight that this would be pulling down But that would be if you were like pulling that way really really really really hard If you're pulling this way, I think it would want to rotate still um Have I done something silly here? mass of the beam I stole this from a provincial so the numbers should be good Sorry Well, it's it's Gordon what was that? You're saying maybe this is so heavy on this end That the rope pulling up is it's so heavy over here that it wants to spin this way And so this is having to push down and that's brilliant if I can convince myself that that's right Have I made a mistake somewhere you got did you get 602 as well? What do I have sorry Because we want to find the force on the hinge, which is going to be that plus that Ha Yann who reads English better than me bless your heart She says all they want is the horizontal force not the vertical force not the combined force I should have read the question carefully. I thought fh. I read it really quick stood for f hinge, but they want the horizontal force don't they? So do I need to find f y No, gordon actually a long time ago already told us mr. Dewick. I know what the horizontal force is What's the horizontal force? How big is fx? You told me what it was the same size as quite a while ago Uh not perpendicular parallel what I need to do is say this uh And I bet you they didn't care about the rest of the numbers Which is why we were getting a bizarre torque going down fair enough fx equals Tension perpendicular parallel you got me doing it now. I don't know tension parallel. I know perpendicular Okay, so that's 48 degrees up there see the z. How are these two related salvage this mr. Dewick 10 yes 10 of 48 equals perpendicular over parallel so that means that tension parallel is going to be perpendicular divided by 10 48 mr. Dewick read the question properly The horizontal component fx or fh for horizontal is going to be this divided by that It's going to be 16.0 16 or 2.3 divided by tan You know what I'll leave this lesson online Just as a humility exercise so my kids know that I still make mistakes Of course that means people from other schools and across the province or watch. Ah I make mistakes get over it uh 1442 1443 oh sig figs 1.44 times 10 to the Thank you. Yeah third newtons Okay, and you know what then I'll revise this from an a plus question to an a minus question I would consider this this I would consider fair game Two more we're done Two more three more we're done This here is about as difficult as the torques can get because we have the beam not horizontal We have the tension at a different angle from the beam Okay, in fact, I gave you one this is similar to what we did on the quiz on the last question Except there I had the beam coming down from the ceiling, but it was the same idea the rope and the beam were all at different angles so Do you guys want to try this one on your own or do you want me to do this? Tyler shook his head. You want to try this on your own? You want me to do this? Do we do it together? Okay, let's walk through it This is fair game one of these Six of these no one of these Six meter uniform beam of mass 25 kilograms is suspended by a cable as shown 85 kilogram object hangs on one end. What's the tension in the cable? Okay Force is acting on the beam The mass of the beam times g the mass of the load times g tension Oh, and there's probably a vertical and a horizontal, but who cares because there's zero from the pivot Remember I said sometimes they give you a different angle Most of the time they've given us the bottom angle this time. They gave us the top one. Okay Let's first of all draw in our components There's the mass of the beam g perpendicular Parallel there's the mass of the load g perpendicular parallel There's tension perpendicular Parallel and you'll notice because I'm not using the parallel I didn't even bother labeling them because it's less to think about Okay, the top triangle here isn't too bad. I'm going to extend it though. So I don't run into stuff the top triangle here Isn't too bad because as I look at this this angle here is 67 degrees. I see a zen. Okay ready Let me make this nice and big Let me make this nice and big. I thought I said see the zed see it So how big? 75 what do these two add together to? Now you could either go There's a zed here, which means this is 75 and then say there's another zed here, which means That's 75 which is going to give us a different angle for a change Okay Or you could say that's 75 if you want it to have the angle at the top so you could use cosine again then this would be Be what? 15 which one you want to do. I don't care That's 75. So that means That's 75 down there and that's 75 down there And that didn't turn off Try that again that's 75 down there and that's 75 down there two zeds Not a nice twist. Is that okay? You guys follow how I how we got that? All right, if you're really not sure Extend lines at horizontals and verticals and level with the beams and eventually you'll be able to hopscotch your way there I have the advantage because I can turn that eraser thing on and off so I can mark up the diagram and have everything vanish um If you're really not sure redraw part of the diagram often the margins somewhere and mark that one up like crazy Until you find the angle and then transfer it back to your nicer diagram Uh, what's the oh tension uh torques mr. Dewick torques we say The sum of all the torques clockwise that's in that direction Equals the sum of all the torques counterclockwise. That's in that direction Which force or forces could cause it to spin clockwise? I think the masses so we have mass of the beam g perpendicular times its distance from the pivot plus mass of the load g perpendicular times its distance from the pivot Three and six Right center of mass and right at the end and that equals Clock counterclockwise tension perpendicular times its distance from the pivot uh six mr. Dewick. No How far is tension from the pivot four? Trig I'll use this triangle because it's bigger this one's small and yucky But this one's nice and big and they're identical triangles so whatever trig function I do here is going to be the same one here. Which trig function am I going to use here? Yeah, this time it's sine So for both of these it's going to be mass of the beam g sine 75 Times three plus mass of the load g sine 75 Times six Equals and I'll do the trig for perpendicular at the very very end because that's too much trig But I am just going to divide by four on this line and get the perpendicular by itself because it's my last step I got room good Tension perpendicular is going to end up being what was the mass of the beam 25 and 85 25 times 9.8 sine 75 times three plus 85 times 9.8 sine 75 Times it was six right all divided by four Justin once again, I'll type in the entire top line probably and just press enter and then divide by four 25 times 9.8 times sine 75 closes bracket times three plus 85 times 9.8 times sine 75 closed bracket times six I'll divide it by four You get 1384.413 blah blah blah okay So tension perpendicular is 1384 Newtons And I have room over here and I'm running out of room at the bottom so I'll move over I want to find tension In this triangle how are perpendicular and tension related? Which trig function? You also sine okay Uh sine of what I totally forget 67 sine of 67 equals perpendicular over tension Tension equals perpendicular divided by sine of 67 Final answer and I conveniently have perpendicular on my calculator. So I'll just divide that by sine of 67 1500 1500 Newtons Hey Okay Two more we're done A 0.75 kilogram board of length 2.6 meters Initially rests on two supports as shown So it's just sitting there. It's not nailed down What maximum distance x from the right hand support can a 1.2 kilogram bird walk Before the board begins to leave the left hand support in other words How far to the right can that bird walk before the whole thing flips over? Oh Little twist because here we're being asked to find the distance Definitely a beam. Yes Definitely torques. Yes How are we going to start out? How have we started out almost every torque question? Lay little horses, okay And uh ty I'm going to be very careful with my distances because are my pivots on the end? Ah, okay. Definitely going to notice that. What are the forces acting here? Get the obvious ones Okay, and that looks about center of meh. Well, you know clothes. I don't care Probably really over here. But anyways mass of the beam times g what else ty? bird Ooh, I got b for beam and b for bird. I'll call it m1 g Pretty heavy bird 1.2 kilos That's a That's a big bird Uh, can this be an equilibrium right now? No because all my forces are down. There's got to be some upwards forces Where? Well, I think right there we would have I'll call that fr for fright Now how big is this force if we're just starting to tip? zero There if he's further this way, I'm sure it's pushing up taking some of the mass of the plank But if it's just starting to tip if we're just starting to begin to tip What we're really saying here, Justin is the force on this has just become zero and now it's ready to pivot around there So I'm going to make a little note here The force is zero because we're just starting to tip that makes sense That that's the extra little bit of reasoning we have to do here. Gilemon torx Um Where am I going to put my pivot? I think I'm going to put my pivot right here because I don't know this force The sum of all the torx clockwise Equals the sum of all the torx counterclockwise Clockwise so if that's my pivot, what could cause it to spin clockwise? Don't use red mr. Do it. It's so black M1g times how far is it from the pivot? Oh That's what we're trying to find. There's my x equals What could cause it to spin counterclockwise? mass of the beam times g times Okay, I'm going to have to do some math here. How far is it from the pivot? How long is the beam grand total? 2.6 and that is that far right center of mass is This distance here 1.3 How's that help? Well, let's see What's this total distance here? What's the total distance here? Tyler 1.8 and how far do I want to be from the end? 1.3 so how far am I from here? It's going to be 1.8 minus 1.3 like I said to you I will not give you all of the information directly in a diagram You're going to have to do some of that remember that grade eight and nine measurement stuff where you are doing perimeter And you're gonna have to do some of that same skill for some of your diagrams And if you're not sure Do what I did add that extra line because it makes it way easier to spot stuff. So I said, okay whole thing center of mass right there Oh 1.8 1.3 So 0.5 Oh, this is actually an easy question to solve Five marks. Well, the key is realizing that because you're about to tip this is going to be zero I saw this question a similar one like this a few years ago A lot of kids wanted to put a force right here because they were Up they wanted one more torque So we're going to have this x equals The mass of the beam times g times 0.5 divided by m1 g as it turns out Unusually what cancels Not the masses because they're different masses what cancels Yeah, what we're saying is This would work on the moon as well. It's the same distance on the moon is on the earth Or the same distance on maras as on the earth or jupiter as on the earth We get Hopefully I get a positive answer Uh 0.75 times Let's see here 0.75 times 0.5 divided by 1.2 0.3125 so 0.312 or 0.31 0.31 meters does that make sense that's reasonable Certainly the answer can't be bigger than 2.6 because that's the length of the beam So yeah, it seems good b What force does the right hand support exert on the board at that instant? Oh What force is right here? How many marks is this worth? I think f up equals f down I think It's gotta be because for two marks jordan I don't think they want me to pull out torques or move a pivot or for two marks I think those are usually kind of plug-and-chug uh mass of the beam 0.7 0.75 times 9.8 plus 1.2 times 9.8 19.1 newtons Would the force be changing? Yeah as the bird moves It's changing the torques the force on each of these beams will be changing in fact as the bird moves left The force here would be going up because as it moves right it hits zero And then you can even almost think about when it starts to tip the force here is sort of negative It's pushing it up and away because it's going to spin So yeah number six Very similar to number five In that it looks like they're asking me to find A distance By the way see the font has changed that tells you this was from the 80s Before they had computers printing these up when they actually used to use physical type printing press machine Says consider the bar a b to have negligible mass. Oh There there's our magic bar with no mass How far from point a must a downward force Of 40.5 newtons be applied In order to achieve rotational equilibrium Good question Which side has more mass on it? Which side has more torque on it? Well, the nice thing is because we're horizontal. I can really quickly figure this out This side i'm assuming that's my pivot has that times that torque five times 2.1 10.5 This side has 3.5 times 4.3 Which side has more torque? Which side needs help then Okay, you know what I just figured out by doing that the force is somewhere here Except can I draw it like this instead because this is how we're used to seeing it mystery force f That's the distance A needs some help It could be further over. I don't know but I know it's to the left of the pivot That was my first thing I wanted to figure out By the way, if I guessed wrong if I'd put it to the right of the pivot I would know because when I tried to crunch the numbers I'd get a negative distance and that would say goofed Start over other side Now I think though this is almost similar Maybe even a little bit easier to the previous question. We're going to find a distance Oh, what are the forces acting on this beam get the obvious ones? I'll call it ma g mb g And wonderfully enough Are these all perpendicular? No components It's almost straight plug and chug. Uh, let's start out the sum of all the torques clockwise in this direction Equals the sum of all the torques counterclockwise in that direction clockwise clockwise 3.5 times 9.8 Times 4.3 Right mg times its distance from the pivot equals Counterclockwise counterclockwise. By the way, I skipped the step. Normally I wrote the variables, but I can tell you guys It's tight. You're tired. It's been a long class. So I stuck the numbers in right away as well. Is that okay? Counterclockwise, I would never do this on a test I'd always show the extra work because you don't want to do the sloppy mistake counterclockwise ma g uh five Times 9.8 times its distance from the pivot plus 40.5 was that kilograms or newtons So do I need to multiply it by 9.8? Nope 40.5 times x And I think I said to you at the very beginning of this unit one of the things I like about this unit the setup is tough But the equations you get at the end are usually pretty easy to solve I mean the most we've been doing is dividing by a number or two or subtracting something over but nothing really really high tech It's going to be this minus that answer divided by 40.5 3.5 times 9.8 times 4.3 minus 5 times 9.8 times 2.1 divided by 40.5 1.1 By the way, what could not my answer be bigger than 2.1. So this is built in error checks as well 1.1 seems pretty good almost worked out bang even that makes me think Oh, I bet you that's where they put it to start out to do the math So I think I'm right The last thing I think I'm Again, my brain works weird when I calculated very quickly the torque here and the torque here I think I went mass times distance. I don't think I included the 9.8 I don't think I included the 9.8 not that it made a difference in terms of figuring out which one was bigger But I should have Yeah, oh how far from point a Ah Someone else can read better than me. So if x is 1.1, how far from point a Therefore 1.0 meters from point a Currently I didn't drink coffee this morning But I hope that kind of helped to crystallize details kind of locked in what we're going. Okay, and there's your test You can think about what kind of using principles of physics right to explain questions there might be There are some good suggestions or some good hints in your big review package some of the ones that are there Easy to tweak and give you as well I'm going to hit stop the rest of the class is yours to work on the review