 So, lesson three, torques with angles. I'm going to give you beams at angles and you're going to have to find components. It says, suppose we have a situation as follows. The beam below has a mass of 12 kilograms and a total length of 8.5 meters. If the beam is in equilibrium, what is the tension in the rope? Okay, we would consider this a C plus B level question. In other words, we would expect most of our kids to get this. It's going to look scary, relax. The first thing I'm going to ask myself is, am I going to use forces or torques? And there's a pretty simple answer. Is there a beam in this question? Torques. Torques. Then I'm going to very, very carefully label the forces that are acting on the beam. What are the forces acting on this beam? Get the obvious ones. What are the forces acting on this beam? Get the obvious ones. Gravity. Now, where am I going to put the mass of the beam? We talked last day about this concept of center of mass. So, I'm going to put the mass of the beam right there. Which way does gravity act? What other forces are acting on this beam? Get the next obvious one. Tension. Now that can't be it. And the reason I know that can't be it is this tension in my mind is pulling upwards and to the left. Is there a force to the right on my diagram anywhere yet? So that means that this couldn't be an equilibrium. It would have to be accelerating to the left. I know there has to be a force to the right. Where? Right here. That hinge, and I'll call it FX for horizontal, has to be pushing to the right. Has to be. Otherwise this couldn't be an equilibrium. And not only that, I don't think tension is taking the entire mass of this beam. I suspect that because there is a contact with the floor, there is some kind of vertical normal force. I'm going to call it FY instead of normal force, but I'm pretty sure that there must be some vertical component there. And if I tried solving this with forces by saying the sum of all my forces down equal the sum of all my forces up, this would be impossible because I have one, two, three unknown forces. I know MG I could figure out. That's another reason we have to use torques. Torque is what times what? We probably want that at our fingertips for the rest of this lesson. Torque is force times perpendicular distance. Distance from what? The pivot. That's the word we're going to use. Here's my question. How far is this force from the pivot? So how much torque will it exert? Zero. How far is this force from the pivot? Zero. So how much torque will it exert? That's why we have to use torques. We can use torques and ignore these two forces. If I had said though, give me a free body diagram. That's it. And I would absolutely take marks off if you didn't reason your way to saying there's got to be two forces. It has to be. Otherwise this couldn't be an equilibrium. But now we're going to use torques. Torque was what times what? First times perpendicular distance. You see, I got another problem. Here's my beam. Is mg perpendicular to the beam? Nope. To the ground, but that's not my frame of reference. Is tension perpendicular to the beam? Nope. So now we're going to go components. I'm going to draw a little dotted line like that. And I'm going to call it tension perpendicular. I'm going to draw a little line like that. I'll call it tension parallel. See this edge? So 40 degrees. I'm going to do the same thing with gravity. I'm going to draw a line like this. And I'm going to call it mg perpendicular. And I'm going to draw a line like this. And I'm going to call it mg parallel. And it looks like in both of these triangles, there is a 90 degree angle. There is a 90 degree angle. Oh, look up for a second and put your pencils down. I'm going to write on this diagram a whole bunch. How big? How big? How big? 90 minus 35? How big? 35. We do that again. Ready? Watch. How big? 55, because it adds to 180. And this is a 90 degree angle. So if that's 55, this is 35. I found a way to get that 35 from there to there. It won't always be that way, but you're going to have to, in your mind, drop this line down, form a 90 degree triangle. That'll give you that one and that'll give you one over here. In other words, 35 degrees. So Gordon's asking a great question. He's saying, won't it always be at the same angle as the beam is on? Only if the beam is mounted on the ground. What if it's mounted on a wall? What if it's mounted on a ceiling? So the answer is there's going to be so many different diagrams. Don't try and memorize the rule. You're going to have to kind of derive in a hopscotch and hopscotch. But the method is almost always going to be, if you have gravity, drop it straight down, because it's going to be perpendicular to the ground. It's going to give you a little right angle there. And you can usually use some basic triangles add to 180. What did I say we're going to use to solve this? Torques. I'm going to underline the word equilibrium, because that's the trigger word here. And there's a beam. So I can say this. The sum of all the torques clockwise, that's in this direction, equals the sum of all the torques counterclockwise, that's in that direction. You were away last class. You want to watch the video, trust me. I forgot to tape the video. No, never mind. I did. That was math. I had to use last year's video. Here's my pivot. So now we need to use our imagination. And I'm not looking at tension. I'm not looking at mg. I'm looking at the two perpendicular ones. These two guys. Which one would cause this to spin this way? Clockwise. There is no mg. I'm going to yell at you in love. mg perpendicular. What do you think the most common mistake on this is? Kids just want to use that. Nope. Torque is not force times distance. It's force times perpendicular distance. So it's going to be mg perpendicular. And then the second most common mistake is kids forget to multiply by the distance. What's its distance from the pivot? Well, let's see. If I look in the original instructions, how long is the beam grand total? And where am I going to put the mass of the beam? Dead center. 4.25. Are there any other clockwise forces? Nope. Equals. What forces would cause this to rotate counterclockwise if it could? Don't say tension. Tension perpendicular. Times its distance from the pivot. 8. See how I can put all that together? Because it says the whole thing is 8.5 long and there's 0.5 on the end. 8. What am I being asked to find? Tension. I'm going to get tension perpendicular by itself and then I'll do the trig to figure out tension. mg perpendicular. I'm going to pick this triangle up and just draw it over here. So here's mg that I know. This is the one that I want. Where this is 35 degrees. Which trig function could I use to figure out mg perpendicular? Math 12 degrees please by the way. Sorry. Cosine. I think I would say cosine of 35 equals the perpendicular over mg. By the way I know the m's cancel here but I don't want to do that because I want mg perpendicular. How would I get mg perpendicular by itself? The mg would move up to there. So I think what I can say is this. mg cos 35 times 4.25. Is that okay Megan? Equals tension perpendicular and you know what? I'm going to divide by 8 to get the perpendicular by itself. Do I know m? Check. Do I know g? Oh calculator time now. Make sure your degrees. What was m? 12? 12 times 9.8. Cos 35 times 4.25 divided by 8. You get 51.1765 blah blah blah. Okay so tension perpendicular equals 51.1765 blah blah blah. It's not what I want to find. What do I want to find? Tension. I'm going to redraw the triangle with tension. This little guy here. So that one looks like this. Perpendicular parallel I don't care about. But there's tension with this angle here was 40 degree. Yes it was. Which trig function can I use? Here. Sine. Sine of 40 equals perpendicular over tension. How would I get the tension by itself? Yeah my stuff moves diagonally. T is going to be up there. Sine is going to end up down there. Tension is going to be perpendicular divided by sine of 40. Which I still have on my calculator. Divided by sine of 40. So if you were designing this if this was a construction thing you would have to make sure that your rope could handle at least 79.6 N and of course you'd build a huge safety margin. There's the tension. Okay. That's what we're going to be looking at today. We're going to be taking Torx beams and either the beams will be slanted or the forces will be slanted or both. Now there's two approaches you can take Sally. You can find the perpendicular component of the forces which I find easier. Some teachers will teach you instead of taking the perpendicular component of the force take the perpendicular component of the distance. There are a few obscure questions where that actually works better but not really in physics 12s. I'm going to kind of gloss over that approach. I'm going to say find the component of the force. So there are two main types of beam problems. One where the beam is nice although can you see you'd still have to find the perpendicular component of tension but gravity and a mass hanging on the end with another gravity are perpendicular already. I would consider that C plus level. I like this question. I like this question. And then where the beam is also at an angle so you would have Mg down find the component. The mass of whatever is hanging on the end down find the component. Tension find the component. And again I can't give you a generic here's where you find the angle all the time reproach. Forces we always could draw remember we could always draw the diagrams identical because the ramps are always on the ground. Here not so much luck. So to solve the beam problems we're going to apply the two conditions for equilibrium. The sum of all the torques are zero except I always write the sum of the torques clockwise equals the sum of the torques counterclockwise. And the sum of all the forces is zero although I'll usually write all the forces up equal all the forces down all the forces left equal all the forces right. That's what I did here Sally when I was able to figure out there had to be the hinge had to be pushing to the right otherwise nothing's canceling out if I wanted to I could find the horizontal component of tension now nothing's canceling that out. In fact if I wanted to I could find the dead vertical component of tension and figure out how much gravity was pulling down and how much this Fy had to be okay components, components, components. Example two Find the chord tension chord tension and find the hinge force exerted by the hinge on the beam. First thing I'm going to do is I'm going to label my forces on this diagram. What are the forces acting? Get the obvious ones. Which one? Cause there's two. Which one? There's two masses in this question. What are the two masses? Mass of the beam and the hanging thing. So I'm going to go like this Well this says beam this says load so how about MBG and how about MLG and I'll do a capital L so it doesn't look like a one cause a lowercase L looks like a one. Are those perpendicular to the beam? Are they perpendicular to the beam? Ha ha Less work. What other forces are acting on this? Get the obvious ones. Okay so I have tension which way is tension pulling in which two directions? Up and to the left this can't be an equilibrium if that's it then. It is possible that the up could be cancelling out both gravities but it's also pulling to the left this would have to be plowing through the wall Newton's laws I have an unbalanced force So I know there has to be I'll call it an FX right there pushing that way and Sally although it's possible that this chord if I find the vertical component is actually cancelling out both of those I doubt it I also suspect there's an FY right there which means just like my previous diagram Jordan there's three unknown forces tension FY and FX is there a beam in this question? Torques and you know where I'm going to put my pivot? I'll put out two of the forces because if I put my pivot where it's drawn and they won't always draw one for you you'll have to think a little bit two of my torques vanish because torque is force times perpendicular distance how far is this force from the pivot? Zero how far is this force from the pivot? Zero So I think the only addition I'm going to do is the component of this guy now let's change colors here's tension perpendicular here's tension parallel horizontally what can you tell me about the two horizontal forces? how are FX and tension parallel related? more specific than what must their magnitudes be say you know how I'm going to find FX? I'm not you know how I'm going to find FX? I'm going to find tension and once I find tension I should be able to find the parallel component from the triangle so I'm going to start out finding tension are we in equilibrium? seems to be doesn't say it's rotating so the sum of all the torques clockwise that's in that direction equals the sum of all the torques counterclockwise in that direction which force or forces would be causing it to rotate this way? so I can't hear when I go like this and you say the same volume it doesn't help Sally what did you say? okay this one M L G now that's the force that we're doing torques so times its distance from the pivot how far is it from the pivot? yes you'll need to look at the diagram and read the instructions they won't always blatantly give you the distance you might have to go hunting for it how long is the pivot? how far is it from the pivot? okay I thought you said 0.2 0.82 you said? I agree are there any other clockwise torques? oh the mass of the beam times G times what? its distance from the pivot how far? center of mass are there any other clockwise forces? nope did I say forces? torques I should have said alright what are my counterclockwise torques? tension perpendicular times its distance from the pivot 0.82 do I need to do any components here? because this is a lot of writing look up can I do that on the same line even though it's my first line of writing to get the tension perpendicular by itself it's a lot of writing and you're going to find by the way in terms of writing like amount similar to the momentum at angle questions but I think you'll find the math is easier it's so ketotric so the arithmetic component nowhere near is bad let's plug in numbers mass of the load 0.7 times 9.8 times 0.82 plus mass of the beam 0.22 times 9.8 times 0.41 all divided by 0.82 that equals tension perpendicular what is tension perpendicular please? at foot top and brackets 7.928 7.938 anybody else? yep 7.938 I'll carry it on my calculator or if I can't I'll write down a few extra sig figs 7.938 what? sorry exactly 7.938 oh wow now let's find tension Dylan see the zed is the 30 going to go top left so let's do that 30 often for a complicated diagram especially one where the beam is on an angle I would redraw this triangle because I've made so many dumb mistakes along the way since this is a nice big diagram and it's horizontal I think I'm just going to go 7.938 and I'm going to say hey which trig I think I get it from this triangle here without redrawing it which trig function will I use to find tension sine now again it's not always consistent I think you'll notice last time what did we use to find MG COS here we used sine for tension it's not always going to be consistent which is why unlike in forces when we kind of did a lot of the trig in the answer I'll redraw the triangle or do the trig so if I hear you correctly it's going to be sine 30 equals perpendicular over tension tension is going to be perpendicular divided by the sine of 30 you get 14.8 what's 38 times 2 Mr. Dylan do you get 14.876 sorry 15.8 15.8 yes 15.876 who's in math 12 who's done special triangle lesson sine of 30 is 0.5 I remember that from my lesson block H hasn't had it yet so what was it 15. what 876 now since tension is one of the answers now I'll sig fig this one and I'll go 15.9 newtons and I'll put a box around it but now I'll also answer the next part so they want attention what else did they want hinge force look up look at your vertical forces on my diagram can you find the two vertical forces pointing upwards yep can you find the two downwards forces are they all nice and in the same direction like all vertical now go to forces I can now say the sum of all of my upwards forces has to equal the sum of all of my downwards forces oh and I'm also going to say the force to the left equals the force to the right why didn't you put a sum of there because I noticed in my diagram there's only one force and one force so I didn't feel like wasting my time writing the sum of one it's not a sum so down if I hear you correctly is m beam g plus m l g and up is tension perpendicular plus f y so can I find f y yep how much more room have I got good f y is going to be the mass of the beam 0.22 times 9.8 plus mass of the load 0.7 times 9.8 minus tension perpendicular 7.938 what's the vertical component of the force that that hinge is putting out what do you get so it is putting out a force in other words the cable is not supporting all of the mass taking care of all of gravity what was it one point 1.078 and I'll carry extra sig figs is not my final answer and let's see if we can figure out um oh forces left equals forces right I think what we're saying is forces to the get your lefts and right correct mr do it I think what we're saying here is tension perpendicular equals f x no not tension perpendicular mr doc tension parallel this is why it helps to have the diagram in front of you yeah tension parallel equals f x hey how can I figure out how big tension parallel is now I could use cosine if I do that I'd use tension and I'd be a little worried because I used this to find tension in case I got this wrong I might actually choose to use tangent to hedge my bets although I've done so much math each of these is a calculated value you said cosine go cosine now over here I would write cosine of 30 equals parallel over tension how would I get tension parallel by itself okay and I'm going to stick that over here f x the horizontal component on the hinge is going to be tension cost 30 and tension was and I'm going to use the 15.876 if I can 15.87 point mr. do it point 16 cost 30 13.749 so if I hear you correctly this is what I think we're saying on that hinge we're 1.078 up and plus tip to tail we are 13.749 to the right you know what I want Dylan the resultant that's the overall magnitude and direction force on the hinge this is force hinge do you see what I meant when I said these are long like the momentum ones but the arithmetic isn't too bad because this is going to be Pythagoras and Sokotoa not sign long not cost law but holy smokes we have done a chunk of writing here yep this question like this question let's do Pythagoras first of all so force on the hinge equals this squared plus 1.078 squared square root let's try that again you get 13.8 190 0.1976 square root of that yeah 13.8 newtons which trig function tan theta is going to be the inverse tan of opposite over adjacent 85.5 86 degrees in other words this thing is almost level but not quite this force is almost straight out but not quite I'll go 85.5 degrees how will I write the direction component I can't really say east of north because this is we're looking straight on to a diagram not from above am I wrong oh yeah I'm going to even say that I would accept that 85.5 degrees with the wall is that clear 85.5 degrees with the wall like that tricky yeah very handy though because you can put the pivot wherever you want to you put it where you don't know several forces and they offer no torque so very useful for solving and doing doing questions when the beam is not horizontal the problem becomes more complex since we often need to consider one set of components for the torque equation and a different set of components for the force equations says find the chord tension and find the hinge force put your pencils down don't write this down look up you see to find the hinge force once I had written the perpendicular component the perpendicular component the perpendicular component once I got all those I would have to erase all of this and instead find this component and this component which are not perpendicular to the beam this component is straight down this component is straight down so I'm okay with that but I'd have to redo this question I'm not going to give you a question where the beam is at an angle and the chord is at an angle and I want you to find the hinge force I will however say find tension and that's what we're going to do tension to me here fair game find the hinge force overkill so let's label our beam torques let's label our forces get the obvious ones ah good old mg so I'm going to have mass of the beam g down and I encourage that you write it on the left of your arrow because you're going to be doing a triangle on the right of your arrow and it's going to get in the way of and I have mass of the load times g straight down which means even before I go anywhere else I know ah component component I'll do that later what else obvious force tension which way is tension pulling which ways is tension pulling up and left since it's pulling to the left I definitely know if by the hinge there it has to be pushing to the right otherwise it couldn't be an equilibrium Jordan I got an unbalanced force looking at this sharp angle do you think the upwards part is canceling out both of those I doubt it I also think there's an upwards component right there too those are the forces on this diagram who cares the free body diagram I would care but did this question specifically say draw the free body diagram well then I could have left those off because if I'm solving this with torques how much torque will they exert nothing why? because torque is forced by the perpendicular distance this guy isn't well they're not perpendicular to begin with I'd have to find components but besides their distance is zero so who cares alright perpendicular component of this guy tension perpendicular there's 90 where's the 55 gonna go here top or here bottom top see the Z hopefully you're developing an appreciation for the geometry that you learned last year how handy it is in physics 12 here I'm gonna go like that like that where here is ML perpendicular G and here is M beam perpendicular G ready Ryan? watch look up see the 90 right there you gotta extend the line in your mind use your sensitivity imagination how big how big 90 minus 28 how big 28 28 but sometimes most of the time that's the most common one so that's 28 degrees and once I have it in this triangle I'm pretty sure that's 28 as well because both masses were hanging in the same direction to begin with so torques the sum of all the torques clockwise in that direction equals the sum of all the torques counter clockwise in that direction using my imagination if this thing could rotate what would cause it to rotate around the pivot in this direction sorry no I gotta be fussy not MLG perpendicular perpendicular right this guy not that guy this guy so MLG perpendicular times its distance from the pivot plus I think there's another counter clockwise torque MBG perpendicular times its distance from the pivot seven or mass I think that's all the clockwise torques counter clockwise tension perpendicular times what its distance from the pivot which is 5.9 I'm gonna do this right away let's redraw one of these triangles here and it'll be the same trig function for the other one because can you see these two triangles are pretty near identical so let's over here in the margin there's MG there's MBG perpendicular and this angle here was 28 degrees which trig function COS so I think it's gonna look like this the mass of the load times G times the cosine of 28 so we're talking about common mistakes third most common mistake Ryan we're so proud we found the trig function we forget to drop the seven down the number of times I see kids forget to drop the distance down is stunning cause there's a big of the trig plus mass of the beam times G times also COS 28 times don't forget to drop the distance down all divided by 5.9 that's the perpendicular component of the tension go to it what do you get for a tension perpendicular what do you get because I think you have all the masses in your diagram that you need right they're there somewhere 14.6 anyone else uh oh what do you get 164 anybody else hmmm oh we suspected that a while ago Justin thank you captain obvious ready mass of brackets cause there's a bunch of stuff on the top mass of the load 8 times 9.8 times COS 28 close bracket times 7 plus mass of the beam 12 times 9.8 times COS 28 close bracket times 3.5 close bracket divided by divided by 5.9 I get 143.7259 of course that's not the right answer either why do they want me to find tension perpendicular nope what do they want me to find tension which trig function what was the angle 55 sine of 55 equals the perpendicular divided by the tension you know what I think the tension is going to be the perpendicular divided by the sine of 55 we just happen to have the perpendicular still stored on my calculator divided by the sine of 55 175.5 oh I guess sig figs now 175 consider the beam below it says what will be the direction of the vertical component of the hinge force fy fy is up as in diagram 1 fy is down as in diagram 2 fy is up or down Ryan says what hey convince me even if we went max case best case scenario and instead of having the tension at an angle we had it straight up could it be an equilibrium or would there have to be a force lifting up so imagine if you had the tension going like this and you removed the wall what would happen to this thing it would drop it would accelerate it would not be there has to be a force pushing up right there gotta be the only way that you could do this would be to find the center of mass for the beam combined with that you could hang the string at a magic point right there and get perfect balance you can hang a string from something from a beam and you can find the magic spot and it'll balance that's what those baby mobiles do all the time that they put in cribs where they have the things hanging on the beam and the kids can put them whatever those are all carefully balanced out can be done but yeah that's that what would I do for my answer probably a free body diagram and then show that the forces up couldn't equal the forces down unless you had it going FY to the up number one number two number three four one two and three one two three and four oh I was going to say no angles actually I lied now two does have an angle I think the rest everything is nice and perpendicular five six except for number six I'm going to cross out the force from the hinge seven holy relax it looks far uglier than it is by the way what mechanical device is that a physics stick figure diagram of crane that's that's how the cranes have their setup okay so it's actually very useful and applicable eight this one and eight this one so there is some torques we looked last day I believe we saw the gentleman from Stanley Park yes okay what have I got for you this time have to stop my video first