 screen is unfrozen here we go. Number one multiple choice this would also be fair game as a written although I think to make it worth a written I would add a mass hanging somewhere else as well or instead of saying find the tension I might have say find a component of the wall force we'll talk about how well that's going to work meanwhile I would put the center of mass right there and thankfully that's straight down and then I would have tension now that can't be the only forces because tension is pulling to the right in this case unusually I think the wall is pulling to the left oh well that makes sense if tension's pulling it out from the wall this will be tugging it into the wall and I think there'd be a vertical component but Brandon because I'm putting my pivot right there how much torque are these guys going to give me none because torque is forced well this is not going to give me any torque because it's not perpendicular it's parallel and this here is zero from the pivot so I'm going to start out with my usual the sum of the torques clockwise equals the sum of the torques count oh hang on components mr. do it I would have to go tension perpendicular and tension parallel and because this and this are the only horizontal forces for what it's worth Zach if I ever asked you to find the horizontal component of the hinge really what I'm asking you to find is the parallel component of tension you'd walk through all the same stuff but at the end you'd find tension parallel um see the Z that's 27 degrees okay the sum of all the torques clockwise in that direction equals the sum of all the torques counterclockwise in this direction clockwise in this direction would be mg times its distance from the pivot three little hint for your test by the way there's going to be one question at least one one that I have in mind that I can remember where Zach I didn't put the pivot at the end I put the pivot somewhere around here which means even if you're at center of mass the distance won't be three you'll have to do a bit of arithmetic to figure out how far am I from the pivot okay uh and then it's going to be tension perpendicular times its distance from the pivot and in the interest of saving time I'll divide by six right now and I get tension perpendicular equals uh m 25 yet clear clear clear 25 times 9.8 times 3 divided by 6 122.5 why should I be able to do that in my head tension perpendicular equals 122.5 then I would look at this triangle here and I would say uh opposite hypotenuse sign of 27 equals tension perpendicular divided by tension tension equals tension perpendicular divided by the sign of 27 divided by the sign of 27 you had 128 is that right I'm in radians because I was going it can't be right sign of 27 is going to be around point sign of 30 is 0.5 should be an answer twice as big 122.5 divided by sign of 27 should be around 250 ish yes 270 b now let's stop how else could I change this question for a test nick it seems to me see this line here I could have told you tension could you have found tension perpendicular so you would know that and you know the six is there I could have asked you to find the mass of the beam in other words another question same diagram tell you tension say how much is the beam weight beam weight I think that's it for twists and turns number two a uniform beam of mass 25 kilogram rests on supports p and q as shown what force is exerted by support q well what are the forces acting definitely have the mass of the beam times g no other masses hanging on this so I have force p pushing up and I have force q pushing up and they want me to find force q so I'm going to put my pivot right there because that's going to make the torque from force p vanish now if I go the sum of all the torques clockwise in this direction equals the sum of all the torques counterclockwise in this direction clockwise mg times its distance from the pivot oh I almost wrote three it's not three is it whole beam is eight long so four and that equals force q times its distance from the pivot which is six so force q is going to be mg times four divided by six uh 25 oh heck I was going to try and do it in my head I'm going to use a calculator you get 163.3 repeating 163 newtons also b probably you found the next one the written question whoa yeah it's tough this puppy what I chose to do was to make a nice big free body diagram because I wanted to really mark this up you guys could probably write small enough to label this diagram but if I do that you guys won't be able to see it so what I actually did was you okay there try okay I drew a long beam I said there's my beam what are the forces acting on this beam get the obvious ones I said well there's going to be the mass of the beam times g and then there's this mass now I started to put it here but then my triangles are going to overlap so Zach just to make it easier to keep track of stuff I moved it down to here I said forget about scale I want there to be enough room that I can draw both triangles is that okay so if I redid this diagram again I wouldn't have it so close to the center so I called this mass of the load times g what other forces well this is a bit unusual we have tension going off this way I said okay it looks like we have our pivot right here by the way what do the forces on the pivot have to be I said well this is pulling left this thing has to be pulling to the right has to be and Dylan do you notice I got two downwards forces do I have any upwards forces anywhere else on my diagram right now do I have any upwards forces right yeah in fact I said you know what that plus that is going to equal that I drew it really big just to let myself know hey I'm pretty sure and in fact this thing here is experiencing a lot of force which makes sense when I looked at this diagram I thought yeah it's getting yanked down it's getting yanked that thing's ready to come out of the ceiling components I said I want mass of the beam g perpendicular parallel mass of the load g perpendicular parallel tension perpendicular parallel where that's the right angle that's the right angle and that's the right the tricky part here I think was doing the trig here's what I said look up so here is uh perpendicular and there's parallel I added a line in my mind I drew this can you see the Z how big is that angle right there 34 so you ready Ian here's what I said 38 that would it be a 38 I said that's 38 what are these two add together to give me 90 what are these two add together to give me vanishing for them mr. do it they can't see it what are these two this one and this one add two to give me 90 and what about this one and this guy next to it they also add to 90 I said look if this is 38 this is 52 this is 38 there may be another way to get there that's what I did can we come up with something better now what's your question kiddo okay there's probably other ways to get there you probably could have extended a vertical line like this and figured something out in fact that probably would have been smarter if you've done that because then that's 30 that's 50 I don't know probably some other stuff you could do anyways I convinced myself that this angle right here was 38 degrees and this angle right here was 38 degrees and huh so I went back to this drawing and I said okay here's perpendicular there's parallel oh I said if I extend the beam just a little bit how big is this angle 38 how big is that angle 38 for a candy is remember what you called that rule last year corresponding corresponding angles were ones that would slide down like snakes and ladders and overlap with the angle further on down the same ray the same line the same segment anyways this is 34 and once I had that I said ooh I see a z 30 I said 34 again this is 38 once I saw that I had a z I said okay this bad boy is 38 degrees whoo that's about as tough as a geometry section as you're gonna get probably a little tougher than you're gonna get now I was ready to do the torques the sum of all the torques clockwise in this direction equals the sum of all the torques counterclockwise in this direction here's my pivot kind of unusual what would cause the pivot to spin clockwise clockwise I said you know what the only thing tension tension perpendicular times its distance from the pivot it's the whole length of the beam from the pivot uh 3.2 that equals the mass of the load times g perpendicular times its distance from the pivot they told me this is 0.9 they told me this is 0.9 so how far is this distance here how can I figure it out 3.2 minus 0.9 right the whole beam is 3.2 the length is 0.9 so 3.2 minus 0.9 2.3 yep people nodding 2.3 and the mass of the beam times g perpendicular times its distance from the pivot oh center of mass 1.6 I'll get tension perpendicular by itself in other words I'm gonna divide this whole thing by 3.2 and I said these two triangles here these two triangles they are identical I'll figure out which trig function it is in the first one it'll be the same in the second one I said let's see mr. duick this is adjacent hypotenuse which trig function cos it's cos in both you may have noticed Evan if you've done a bunch of these that usually the mass hanging from an angled beam or gravity hanging from an angle beam is usually cos not always it really depends on what angle they give you and today we're going to do an example where it turns out to be sine but usually it's cos so this is going to be mass of the load times g times the cos sine of 38 and then don't forget to drop the distance down plus mass of the beam times g times the cosine 38 and don't forget to drop the distance down time for my trusty calculator I've forgotten the masses mr. duick 5 kilograms 2.4 okay 2.4 times 9.8 times 2.3 well times cosine of 38 degrees times 2.3 and 5 times 9.8 times cos of 38 times what was the distance for the 5 kilogram mass of the beam 1.6 double check 2.4 9.8 cos 38 2.35 9.8 cos 1.6 divided by 32 and I get tension perpendicular is point a 3.2628 for me I divided by 3.2 and I wanted to divide but I divided by 32 and I should have divided by 3.2 so is it 32.6275 32.628 tension equals here mr. duick times 10 is that it okay 32.628 but that's tension perpendicular tension let's see which trig function opposite hypotenuse sine 38 equals perpendicular over tension and again you may not have noticed that it's been almost always sine to find the tension from the perpendicular not always usually tension is going to be perpendicular divided by sine 28 equals divided by sine 28 tell me you get 69.5 times the sine of 28 let's go back to where I started from here divided by the sine of 38 tell me you get 53 53 newtons oh if you got that wrong you don't get full marks how would I give out part marks oh probably something like this I would give you one mark if I saw that one mark if I saw that two marks if I saw that four five six seven probably something like that for me yes what I would like you to do is give yourself a five not out of seven it's out of five even easier than I would probably go I would probably go one mark for that come on pen one mark for that and one mark for that one mark in answer to your question Vitaly so that makes this whole quiz strangely enough out of nine since multiple choice for some reason in physics is worth two marks a piece what I would like you to do is as we've done occasionally the person on the end of each row get out a piece of paper write down your name write down your score and pass it down but I'd like you to be able to hang on to these to study from because I won't see you between now I see you Wednesday but if I don't get these back to you Wednesday then you don't have to study from I don't know so lesson five number one for those of you who are watching at home a uniform 15 kilogram pipe of five meters has 160 Newton force applied four meters from its lower end as shown using the point where the pipe touches the ground as the pivot so this time they're telling me where I have to put the pivot calculate the sum of the torques acting on the pipe this is a bit of a twist nick because this is not an equilibrium this is not going oh some of the torques clockwise equals some of the torque counterclockwise they're not balanced and I can tell they're not balanced because I glanced at all four answers and none of them was zero so here instead of doing my clockwise equals counter clockwise I think all I'm going to do is figure out the total number of torques clockwise how much the total amount of torque counterclockwise and then I'm going to go bigger minus smaller whatever's left over tells me which way it's spinning so I picked this one I thought it was a nice twist still and instead of equilibrium this thing is spinning I want to know which way is that force enough to get it to spin upwards into the left or is it's mass enough even if they're pulling to still have it pull it down to the ground so I'm still going to start out the same way by saying hey what are the forces acting on this beam get the obvious ones Mg center of mass what else tension now tension is pulling to the left I'm almost positive that this hinge here then is pushed in fact I am positive this hinge is pushing to the right there may also be an upwards or a downwards force right there this is multiple choice so I don't care about what the free body diagram looks like anyways so even though I've talked about it I'm not going to worry about it I am definitely though going to have to go components perpendicular I know this is parallel you may have noticed Taylor I don't often label parallel unless I need it because the less I label the less confusing the diagram is and Mg perpendicular and there's the parallel and I'm going to zoom into this diagram a little bit here let's see ready Brandon how big is this angle right here yep how big is this angle 90 34 how big is this angle how big is this angle here how big is this angle tell me exactly in your head how big that angle has to be how do we know how about just 90 minus 34 because 180 minus 90 is 90 right this is 90 these two got to add to 90 as well for the triangle add to 180 so in your head 90 take away 34 no excuses what you got 56 so you said this is how big 56 what do these two angles add together to form look at them these two angles here 90 so that means the whole point of this my friend was to realize this angle here is 34 degrees as well this one's a little easier Brandon see it 42 how big yep okay so this one because Miguel they told me that we are not balanced you see how you pick that up from the question it does not say equilibrium instead it says find the sum of the torques I'm going to find torque clockwise first torque clockwise equals what would cause this to spin in this direction mg perpendicular times its distance from the pivot how far is it from the pivot 2.5 which trig function is perpendicular going to be cos the torques clockwise end up being mg cos of 34 times 2.5 and that's the mass of the beam let's crunch that please can you get your calculators out and go mass 15 15 times 9.8 times cosine of 34 times 2.5 I get 304.7 units Newton meters direction clockwise what's the sum of the torques counterclockwise well what force would cause it to spin this way tension is wrong perpendicular it's not make that mistake that's the most common mistake so I'm going to go the torques counterclockwise ends up being tension perpendicular times how far is it from the pivot let's see whole beam is 5 that's 1 4 by the way which trig function could I use to find tension perpendicular okay so tension perpendicular is going to be tension times the sine of 4 sine of 4 mr. duic sine of 42 times 4 the perpendicular sorry the perpendicular torque the counterclockwise torque is going to be 160 times the sine of 42 times 4 428.2 which is bigger counterclockwise or clockwise so who's winning can't be spinning that way can't be spinning that way it's got to be spinning counterclockwise see that's winning this is not in balance this is not an equilibrium yeah you know what if you go 428 minus 305 you get about 120 I'm pretty sure the answer is c bigger minus smaller so trying to think of different ways they can give you a multiple choice question with a twist here's an example of one okay lots of other good examples of the kinds of multiple choice questions you'll see in your big review so make sure you work on that I will say I'm having a lot of people handing those reviews in five or six or eight days after the actual test I hope those of you that are doing the reviews before the test are you finding they're helpful yeah so rest of you like easiest way to raise your test grade do the review before the test second easiest way do the homework third easiest way give me ten thousand dollars fourth easiest way pay attention during class and don't show up late example two a beam question but uh with a twist this time they want you to find the mass of the student instead of giving you all the masses and saying find the torque or find the force or find the tension here find the mass okay I'm still going to start out by labeling my forces so what are the forces acting on this beam get the obvious ones pardon me okay and I'll put that center of mass mass of the beam times g what else yeah mass of the student times g what else I can't be it otherwise it would be falling into the ground okay and this one's a bit interesting because they told me what the scale read what does the scale read forces come in pairs if the beam is pushing down with 350 newtons what must the scale be pushing back up at so this is their way of giving me a force without being blatant about it and I think there's one or two of these in the review as well there's at least one I know where they have an object sitting on a scale and they tell you either find the scale measurement or they tell you the scale measure and you know what I'm pretty sure there's another force right there I'll call it f y for force up but I don't care about it because how far is it from the pivot zero does this question suggest that this beam is moving at all no in fact looking at this picture I'm pretty sure this is an equilibrium so now we're going to use our standard torque approach the sum of all the torques clockwise in this direction equals the sum of all the torques counterclockwise in this direction clockwise so here's my pivot right here clockwise would cause it to spin this way which force or forces would do that I think yeah both the masses so it's going to be the mass of the student times g times how far is it from the pivot one point two plus the mass of the beam times g times one point five half center of mass equals 350 times its distance from the pivot three what am I trying to find here so I think Brandon leave this minus this over and then divide by both of those and that should be get me that by itself yeah I'm going to do that in one step so the mass of the student is going to be the mass of the beam times g times one point five no mr duke you're minusing that over mr duke it's going to be this 350 times three minus the mass of the beam times g times 1.5 and then divide that answer by a g and a 1.2 that should get the student by itself what did we say the mass of the beam was 25 350 times 3 minus 25 times 9.8 times 1.5 divided by 9.8 times 1.2 is that right 58 58 kilograms c before we turn the page how else could I change this question hint hint wink wink nudge so this was a twist here Miguel I said find the mass or could I have told you her mass and this and have asked you to find where she's standing in other words on this line here could I ask you to put an x right there yep I feel good about that okay and I would have no problem not putting the pivot on the end maybe putting the pivot right here so she's balancing it like a teeter totter okay just means your distances would change right now turn the page I like this question I like this question I like this question so we've seen a picture like this before but the tally read very carefully what do they want me to find in this question that's the trick to this question the horizontal force where from what do you see it in the diagram where the horizontal forces find it okay this is a question I would consider fair game as a hinge force we did a couple in our homework where I actually asked you to find the horizontal and the vertical added together tip to tail and find the resultant that's a little overkill to me but asking you to find the horizontal or the vertical component of the hinge it's not too bad and you'll see when we label this this is a nice big diagram that I think I'm going to label the diagram and not redraw it what are the forces acting on this bean get the obvious ones okay we have gravity mass of the bean times g what else mass of the load times g and I like both of these Aaron because holy smokes they're perpendicular no components what else definitely tension which way is tension pulling what two directions up and left and as soon as you said left I said that can't be the only forces then because there's a force to the left and it's not accelerating to the left there's my horizontal force pushing to the right is what I'm trying to find and I'm pretty sure that this is pushing upwards actually my last class we had more time we solved for this and we found tension is pulling so hard that as it turns out this is end up pushing downwards but we're not going to worry about that because how far is it from the pivot zero how can I find fh I can't yet but I got to do something else to my free body diagram which force do I need to find a component for okay here's perpendicular here's parallel does anybody see metallic I'm going to argue and I'm going to say really this question does not want me to find fh the horizontal component which force is this question really wanting you to find tension parallel because those two are equivalent to each other did I say I like this question okay so Evan what I'm going to do is I'm going to use torques to find tension parallel and then at the end I'll say that equals fh oh sorry I'm going to use torques to find perpendicular but once I know this I can do some basic trig and find that uh angles oh see the z 48 so I'm going to start out the same way as we have before 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 if that's my pivot point my hinge which force or forces would cause it to spin clockwise mass of the beam times g times its distance from the pivot 3 not 2 3 plus the mass of the load times g times 6 that equals the perpendicular component times 4 what's the mass of the beam 32 times 9.8 no need for components because we're already good times 3 plus mass of the load 93 times 9.8 times 6 if I divide that by 4 that will give me tension perpendicular once again strongly encourage you to follow along on your calculators in this unit most common mistakes is calculator mistakes divided by 4 and I get the perpendicular component 16 oh 2.3 anybody else yep tension perpendicular equals 16 oh 2.3 that's a force so it's newtons I'm not trying to find tension perpendicular but tally what did this question want us to find I got not uh not hinge horizontal force okay the horizontal force and I said to you I can't find it what does this question really want me to find sorry parallel okay so I know perpendicular opposite adjacent or hypotenuse opposite adjacent or hypotenuse so here's one where we're not using sine at the very end as it turns out which trig function we're using that tan of 48 equals opposite over adjacent tension perpendicular which equals fx is going to be sorry tension parallel which equals fx is going to be perpendicular divided by tan 48 conveniently I have perpendicular on my calculator divided by tan 48 and I get 1442.3 oh sig figs 1440 again trying to show you different ways I can tweak these and make them either a little bit trickier or a little easier this one a little bit trickier is it 1440 good morning okay turn the page I like number four I like number four so question four we have a beam at an angle and a rope at a different angle in fact I would argue this is very very similar to the question the last question on your take home quiz there the beam was at a yucky angle and the rope was not parallel to the beam the perpendicular to the beam so here we go what are the forces acting on this beam get the obvious ones mass of the beam times g mass of the load times g tension I doubt that's it in fact I'm almost well I know since tension is pulling left I'm pretty sure the hinge is pushing right and probably pushing up as well now what components mass of the beam g perpendicular and mass of the load g perpendicular and tension perpendicular this one's a little weird ready now let's see how we do it would be good see the z 75 how big ready how big was this one here see the other z bottom left bottom left this time I don't think we ever used that one but I said to you usually it's cosine I think it's gonna be a different trig function this time I try and use the angle that they gave me you don't have to you could have up here used 15 I just want my answers to look like their answer keys and I'll mark yours fine I'll grumble a bit and do it on the scrap piece of paper or probably by this time I have both on my answer key and I'm pretty sure this is also 75 this one's a little easier torques the sum of all the torques clockwise in this direction equals the sum of all the torques counterclockwise in that direction clockwise I got to be really really fussy perpendicular components of the masses okay mass of the beam g perpendicular times its distance from the pivot which is going to be the beam of six meters long three plus mass of the load times g times its distance from the pivot six equals what's going to spin it counterclockwise tension perpendicular times its distance from the pivot which is how far oh four they gave me that better okay well let's sit glances down no she's okay sorry but I got to teach um which trig function yeah this time it's going to be opposite hypotenuse it's going to be sine it's going to be mass of the beam g sine of 75 times 3 plus mass of the load g sine of 75 times 6 that equals tension perpendicular times 4 I'm going to divide by 4 right now what if tension perpendicular end up being when you go mass of the beam times g times sine 75 plus 3 sorry times 3 plus the mass of the load times g times times 75 times 6 divided by 4 let's see mr. duick beam is 25 85 okay I can get this 25 times 9.8 sine 75 times 3 and 85 times 9.8 times sine 75 times 6 138 4.41 3 is that my final answer what do this question want me to find okay let's go back to this triangle and let's say to ourselves all right which trig function um opposite hypotenuse sine of 67 equals opposite over hypotenuse so the hypotenuse is going to be tension perpendicular over sine of 67 that answer divided by sine of 67 you get 1504 oh sig figs 1500 newtons try to remember have i showed you these slow motion balancing movie the guy's everything's backwards reverse okay turn the page and we've done the pen and teller magic trick yes the trick truck pen and teller number five a similar question this is actually on my other version of the test but there's one sort of like this on yours we have a board and it's just laying down on those two rests and we have this bird it's slowly walking further and further to the right and eventually it's going to walk so far to the right tailor that that board is going to tip so a is asking what maximum distance x from the right hand support kind of 1.2 kilogram bird walk before the board begins to leave the left hand support suggestions leg 1 diagram absolutely one of the forces acting here get the obvious ones oh oh by the way are any of my pivots going to be right at the end so let's be really careful when we're doing distances not to get in the habit of just how far from the end of our pivot goes here they're in the middle i'm probably going to move around a little bit so what are the forces well definitely have mass of the bird times g what else mass of the bridge how long is the bridge 2.6 oh eyeball the center of mass 1.3 i'll call it m1g because b and bridge and b and bird begin with the same letter can that be all no why not yeah so um put your pencils down here's the most common mistake kids do this f right and then they do this f left and the reason that that's incorrect is if i mathematically put my pivot there don't write this down but if i mathematically put my pivot there what direction is this wanting to torque clockwise or counterclockwise what direction is this wanting to torque what direction is this wanting to torque clockwise you might not have balanced torque like you don't have balanced torque here's my argument as you walk further and further and further out what happens to this plank in fact the equilibrium point is what it just lifting off and i'm going to argue if it's just lifting off if that's the maximum distance that is zero is zero it just wants to lift off airy so there's nothing holding it down and there's no force pushing it up if it just lift does lift off for a split second there's no more normal force we're talking about that threshold right there is that okay let's go yell at people in the hallway let's go see who that is silly people did you stand oh my very presence but do you see the argument here what what i'm saying is if this is just about to tip that's just about to lift off the ground right there because it's clearly going to pivot right here it was just about to lift off the ground right there what's my net force for that split second nothing down nothing up it's just about to become weightless okay nice little twist here i don't know if you ever walked out like when you were a kid at the end of the long plank how forth did you walk out before it started to tip or something that's what we're talking about here or even if you've been standing on a piece of wood that was balanced and you can move over and get it to hit back down on the ground again and then balance it back in the middle again like a cheetah totter and move over to get a hit back down again what we're really saying then is where does that equal to be sorry folks a bit of a delay there torques now here's the nice thing about this question components yes or no how about here how about here in fact i'm going to put my pivot right here and all i'm going to have then is since that's zero the sum of the torques clockwise equal the sum of the torques counterclockwise if that's my pivot and this is zero clockwise clockwise clockwise clockwise which force okay mass of the bird times g times how far from the pivot yeah what are we going to put there look at the diagram oh x that equals counterclockwise the mass i'll call it mass one the mass of the beam but i already used b for bird times g times here's we got to be careful you have to do a little bit of thinking how far is it from the pivot a lot of people tailor will just go oh 2.6 center mass 1.3 that's only if the pivot's at the end and it's not how long is the whole beam 2.6 so here's the question how far from here to here from here to here tailor 1.3 that is half of the distance center of mass real question i want to figure out is what's that distance so let's see what's the total distance from here to here 1.8 and you told me this distance was what tailor so what's this 0.5 0.5 okay from here to here is 1.8 from here to here is 1.3 center of mass from here to the pivot which is where we put it 0.5 for me for me how far from here to here so where am i going to put the mass of the beam always center of mass so it's going to be 1.3 from either end problem is our pivot's not on the end we're gonna have to do some tweaking of the numbers here okay so i knew this was 1.3 and 1.3 because center of mass but then i said oh from here to here is 1.8 from here to here is 1.3 what's left i'm going to give you at least one diagram on your test we're going to have to kind of move stuff back and forth and add a little line to figure it out okay having said that fairly nice equation x is going to be mass of the beam times g times 0.5 divided by mass of the bird times g mass of the beam 0.75 times 9.8 times 0.5 divided by mass of the bird 1.2 that's a huge bird go back again then some oh i also noticed something else what cancels now what that means is if you were to somehow magically transport this to the moon it would still be the same distance the bird could walk out or if you transported it to mars or to jupiter wherever in the universe that distance is not going to change it doesn't depend on the gravitational field it just depends on the ratios of the masses to each other so this is going to be 0.75 times try that again 0.75 times 0.5 divided by 1.2 come on and i get 0.3125 x equals 0.31 units meters you can walk out 31 centimeters and that's just past 31 centimeters is when the whole thing will start to tip b how many marks is part b worth two do you think they want me to go big torque question in fact i think we can just say this the force is up equal the force is down because it's an equilibrium because it says it just wants to start to move but it hasn't started moving yet what are the forces up only one force right what are the forces down look what about the force on the left ah we said at that split second it's zero so it's going to be 0.75 times g plus 1.2 times 9.8 and i get 19.1 newtons turn the page we're not done by the way i should have said by the way in part a uh are we okay oh we are okay never mind which one was the one that was bugging me oh number five here i don't know when did i say the test was several times there you are no no what test for math or physics yeah i already said it today did i not pretty sure yeah you came a little bit late but i think i announced it after he came here no okay hey look up i'm gonna argue that this diagram here this idea where they're asking you to find the distance very similar to this one here you guys are zoning out on me there's about 10 minutes left i'm gonna give you 10 minutes to work on the big review and that's your homework i'm not gonna do example six if you want to find example six the lesson the video lesson online has it done you can try it there okay so i'm just gonna say homework work on unit review matt taylor matt taylor i'm also i'm doing a formal one today where i've cleared my calendar just for the physics kids uh i'm also gonna be around wednesday and friday yes you may