 Says show work clearly blah blah blah blah blah. Okay, two forces if the two forces acting on the object below are equal in magnitude Which of the following is not? Possible in other words, which of the following is false Forces are equal in magnitude. Okay, um The object is at rest I Think that's possible. In fact, that's very likely could be traveling at a constant speed if the forces were in balance, but probably at rest so The object is accelerating to the left Can it be accelerating if the forces are in balance? Yeah, I think the answer here is Be I Think it could be at rest I think it could be moving at a constant velocity up or a constant velocity to the right as long as you had a constant velocity That means you have balanced forces number two a Mass of five kilograms is suspended from a chord as shown below what horizontal force F is Necessary to hold the mass in the position shown. Is there a beam in this question anywhere? But to be an easy question, so is there a beam in this question anywhere? No, no torques. This is equilibrium This is going to be free-body diagram and then we're going to add our forces together Perhaps in a vector triangle if there's three of them coming back to where we started from or we'll go components Let's see. What are the forces act? Well, I got good old mg down mystery force F and Tension, you know what since I only have three forces and a lovely right angle I think I'll use my trig approach here. I'll draw the easiest force first Mg the toughest force I think tension looks the uckiest and I know I have to stop exactly there Although that's really a lousy angle. I know I have to stop exactly there because the third force It tells me is exactly horizontal and that's the mystery force that I'm trying to find I Also notice if I add a vertical line right here That this angle here is 35 degrees and I see Z I'm pretty sure that this angle down here is 35 degrees and now let's see what I don't mean to find F Opposite adjacent y tangent of 35 equals opposite over adjacent I think that mystery force F is mg Can 35 I get out my calculator. I say, oh, I'm in math 12 as well better make sure I'm in degrees Speaking of which ha ha radians nice try M5 wasn't it? I think yeah five times nine point eight times the tangent of thirty five Thirty four point three Force F equals thirty four point three Newtons B. I like this question. I like this question. I like this question. I like this question Okay Now different ways I can ask this question. I Can give you this mass and you multiply by g or in this case? I gave you mg and say find the tensions I would also Jordan have no problem giving you one of the tensions say find the other one find the mass Because I'm gonna say in the triangle as long as you know one of the sides you have a pair You can find any of the other sides. I figure that's fair game. Okay So it says first of all draw a free-body diagram showing the forces acting on the knot. Okay, there's the knot. I have mg tension one Tension to Hopefully y'all got that one. I won't prop I probably won't be horribly fussy when it comes to Scale or angles. I mean be close though What is the tension in rope one no idea what is the tension in rope to no idea I Need to draw this. Let's see. Oh How many forces do I have here Ryan? Trink for components like the rock climber one that we did three, huh? Okay? Easiest force first Mg and then I'll probably do tension one which looks like it goes kind of shallow about like that and You know if I do this properly Probably a little longer if I'm thinking my scale and then tension two looks kind of like this And I want to find some angles. Let's see. I noticed that tension one has a 20 there How big is that angle right there then? Is that how big is this angle right here then and these two together add to 90 ha ha 70 got one I noticed this one 60 degrees with the vertical line 60 degrees with a vertical line. Sorry. Did I say vertical horizontal 60 degrees of the horizontal line? And I think this forms a 90. So I'm pretty sure that this here is aha 30 These two together add to a hundred. Oh 80 Okay, I've done the hard part. I've got my triangle and You'll notice who I have a pair in fact, you know what instead of writing mg They told me what mg is I'll put a little 750 right there Why not and this is the pair that I know which means it's gonna be sign law and cross multiplying and Certainly a nice change from the cosine law squaring all sorts of stuff so it's gonna be sign of 80 over what's across from it Equals now I want to find tension one. So it's gonna be this pair here Which means that it's gonna be 30 over tension one. I'm pretty sure that tension one is gonna end up being 750 sine 30 divided by sine 80 750 sine 30 divided by Sine 80 Boy, why am I yawning mr. Do it? It's first thing in the morning. You get 381 Yep 381 what Ryan Newvins? Check what is the tension in rope to it's gonna be almost identical in my approach Here's my pair sine 80 over 750 equals Tension to here's my pair sine 70 over tension to in fact I think that tension to it's gonna end up being 750 sine 70 over sine 80 Fact if I'm really lazy, I can just go second function enter and change the 30 to a 70 and you get tension to is 716 newtons Okay So keep an eye out for this kind of a diagram where you see a rope another rope two different angles and a mass But having said that I would have no problem mixing and matching this Hey, I'll give you one tension find the other one and find the mass or in this case. Here's the mass Mg find both tensions or I Think it's about it equilibrium in terms of part marks Well, if you got the right answers you get full mark seven well five out of five plus one This question was only worth six marks. That's a little weird Usually the written or seven. Oh well Otherwise, I'd probably give you one mark for a good free body to triangle if I could see that somewhere, but I'll let you well Let's go find it equilibrium Give yourself a score if you're not it out of ten But part marks in case you were wondering What's something like this? Did I have part marks on here? Ha, I didn't have part marks on here Okay, give yourself some good part marks if you didn't get the right answer Give yourself a score out of ten please and Pass the quizzes in please Look up. So we've been looking at Torx. Torx is what times what? Okay, go ahead. I like the way you said it He didn't say force times distance because I have to freak if you say that force times distance is work His jewels is scaler. His energy is totally different But torque is force times perpendicular difference distance It's a cross product for those of you who want to get fancy at your multiplying vectors Which we really don't know how to do properly We said it does have direction clockwise and counterclockwise For what it's worth and we said that if your forces are not perpendicular to the beam and you use torques when there is a beam You're gonna have to find the components. We talked about rotational equilibrium We said that rotational equilibrium is when the sum of all the torques is zero. I don't deal with that Jordan I always said all of the clockwise torques and I drew a little arrow in that direction equal all of the Counterclockwise torques and I drew a little arrow in that direction look up Alright the physics of cow tipping. Yes, someone has actually analyzed this you see as it turns out if You draw a line from the top shoulder of the cow all the way to the bottom hoof Why that's sort of like a beam as a matter of fact trying to tip a cow is sort of like having the pivot point of your torque Right down there and the UV prof actually worked out the equation now if you look at this Mg hey, that's the force straight down at the center of mass Cosine theta if you were to actually find the perpendicular component you would find out that it turns out to be cosine A plus B is the length of the beam Hey, that sounds like a torque question where to get the force by itself which being is applied right there torque Is what times what Gordon? And how far away is force F from the beam a Plus B so they had that here then they divided the end of by itself It's a torque question as it turns out and then when you crunch the numbers for an average cow You find that to flip a cow you need 1360 newtons of force which is why it's probably an urban legend or a myth But some of you on a farm it may feel free to try it out too much force required. Oh Come on, that was pretty good. I felt count. No Okay Then questions from the homework Which ones would you like me to go over I'll go over a couple some of these were definitely on the nasty side Can I do which one? seven Love to Not really, but I will any before number seven okay, and I think so for most of these it's the startup like I think you've noticed once we get everything set up The equations aren't even you divide once and you've got the thing by itself and you go to your calculator and the way you go I agree. It's the setup. Okay, so here's our beam. That is a crane Beam Yeah Torques. Yeah, what are the forces acting on this beam get the obvious ones Well, there's gonna be the mass of the beam which is gonna be I think right about there I'll call that mass of the beam times G Center of mass. You okay with that and then we have mass of the load times G and Tension I can't use any of those Because none of them are perpendicular to the beam you want to be careful I some kids think oh these are perpendicular Mr. Do it I'm not to the ground to the beam the beams on an angle. We're gonna be doing some trig here So what I would do now, and I'm gonna zoom in a little bit. I think I can fit all this on one. Yeah, I Would say There's the perpendicular component There's the parallel component. There's the perpendicular component. There's the parallel component by the way Sally this rope is pulling down and to the left These forces are both down I'm positive this hinge is pushing To the right. I'm not gonna label it though put the I'll put them in a race at this second because that cancels out the left and Since these are all three down. Is there any upwards force? Where on that hint this hinge is probably doing that It has to be otherwise this crane would have to be collapsing into the ground But how far are both of these from the pivot? So how much torque will they exert so I'm gonna nuke them But if I did give you this and say pick the best free-body diagram or I asked you to draw the free-body diagram And there's gonna be a perpendicular component of Tension where that's the right angle and that's the right angle and that's the right angle I'm gonna do the bottom the g's first How big how big how big This they all have to be 45 this time because it's these two have to add to 90 and these two have to add to 90 In fact, usually you'll find this angle and this angle are the same So I'm gonna put that in there 45 45 that one wasn't too bad the ugly one's gonna be this other Yeah, how big I mean That's not a proper Zed. I can't use my Zed rule there because the Zeds had to be parallel rats, oh How big how big? What do these two add to ah Gives us one more angle. I'm gonna try something. So it's 180 minus 45 135 math 12 coming in handy here a little bit. So I'm actually gonna label this 135 ha ha ha One two I now know this angle. How big how come I know this angle? triangle That's 165 so 180 minus 165 15 degrees and that seems about right in terms of its size So I think I've done it. Okay, and now I say to you my child Now I see a proper Zed. Oh That right there is 15 degrees Just to unclutter my diagram because I can I'm gonna get rid of these guys now because that was they're gonna get in my way But you know on my homework to be stay there. So far so good. Okay Whoo, I Gotta go back to the small screen now By the way, this is this I would consider this fair game as the nasty multiple-choice question little over the top for a written but only a little equilibrium The sum of all the torques clockwise Equals the sum of all the torques counterclockwise Here's my pivot which forces would cause it to spin in the clockwise direction perpendicular which I didn't bother labeling on there. I guess I should M beam G perpendicular M L G perpendicular. I kind of got it fit in there. Sorry So I'm gonna have The mass of the beam G perpendicular times its distance from the pivot. Oh The beam is six meters long center of mass Three plus The mass of the load G perpendicular times its distance from the pivot and that equals Tension perpendicular times its distance from the pivot which conveniently is also six trig M G perpendicular and M G. I think it's cosine Right, I think this is gonna be the mass of the beam G cosine of 45 Times three now you're in math 12 Sally the nice one of this because it's 45 degrees Sign and cosine give you the same answer. So even if you got it wrong and you sign you'd still get the right answer by a fluke triangle Plus mass of the load times G times the sign, Mr. Do it cosine. We do it correctly Cosa 45 times six that equals tension perpendicular and Sally I'm gonna divide by six right now and get the tension perpendicular by itself Could you take it from here once you're done? Then you have to figure out which how tension and tension perpendicular are related I think they're related through sign. Okay, I'll let you take it the rest of the way I'm I think most people it's this first line. That's the challenge right getting all that set up in the trig Okay Any others? Okay Last lesson Short unit partly because I snuck a lot of the forces equilibrium stuff in earlier when we did ramps and things For So all we're going to do today is we're going to look at a famous pork question at an angle and Then if I have time today, I would like to go through your energy test You Missed the physics of count tipping no Well, I guess you could The latter question leading a ladder against the wall a ladder seems to me Gordon is a beam When you lean it that's James to me. It's kind of a torque at an angle Yeah Well Gordon said he's gonna fall what we're really going to ask ourselves is What are the physics of a ladder sliding out? When is it safe and when is it not? So it says draw a force diagram for the ladder below and Identify the best location for the pivot point. We do have to add one thing We have to assume the vertical wall is frictionless Otherwise, we'll have one too many unknown forces and we can't solve this. So here we go Here's our diagram What are the forces acting on this ladder get the obvious ones? Okay, the mass of the ladder Where am I gonna put that? Center of mass, so I'm gonna go like that and I'll call it MLG for mass of the ladder what else? I think it's at least one more kind of obvious one for me The weight of the dude. I'll call it M2g Can't bring myself to label him as the dude and it could be a female to a stick figure What else did by the way, can this be it no Because if that was it this ladder would have to be accelerating which direction down is it Okay, I think there is a ground force here And you know what we've given that force a name in the past. Let's call it the normal force I could also call it FY Okay Gordon is saying I think I need friction between the ladder and the wall because it seems to me You can't set up a ladder on ice I agree with you, but we're gonna have to prove why we need it because right now This is technically in balance like right now all of my forces cancel out by the way This is probably equal to both of those Right It's touching a wall here It's leaning against the surface here Here also just like with the normal force here also forces come in pairs If the ladder is pushing against the wall, what's the wall doing? So I'm gonna go like this and I'm gonna call this Normal force number two. That's my force to the left and by doing that then Gordon I can tell you I agree there is friction down here But I can tell you which way friction has to point this friction have to point to the left or to the right To the right to cancel out normal force number one In fact, that's the key idea to solving this question. The key idea to solving this is to recognize Oh, you know what look up. Let's all put a one right here next to the first normal force the key idea to recognize is that Normal force number one is going to be equal to the mass of the ladder times g Plus the mass of whatever else is on the ladder times g Those are an equilibrium the sum of the forces up equal to some of the forces down and Friction equals Normal force number two. I'm not saying friction equals mu times a normal force I'm just saying look those two forces are the only horizontal forces so they got to cancel each other out Otherwise this ladder would have to be accelerating. Oh Normal force two did I put a one there? That was silly normal force two Okay That's our diagram If We want to solve this where would the best place to put a pivot be I think if I put it here I cancel out two forces when it comes to torques Then what I would do is I would find the perpendicular component the perpendicular component the perpendicular component and And I think if I know both masses I can find normal force too And then if I need to usually they want you to find the friction force Or they might take it one step further and say find the coefficient of friction And then you would say friction equals mu times this normal force Kind of like example two Says find the value of the coefficient of friction for the ladder on the ground Little tie to your Dylan because we don't have somebody standing on the ladder to give us an extra mass No, the extra mass is just really one more term more than anything Let's label our diagram. What are the forces acting on this get the obvious ones? So we're gonna have mg mass of the ladder times g I'm gonna have Now I called it a normal force. You can also call it if you want to f wall or Anyways, it's technically a normal force. So I'm gonna call it normal force number two But you'll see when you're looking at my answer key. I'm not consistent Sometimes I called it force of the wall. Sometimes I might have called it force x horizontal. Whatever Normal force number one, and I'm pretty sure Brendan these two cancel each other out and Friction, and I'm pretty sure that these two cancel each other out. What do they want me to find here? okay friction equals mu times the normal force number one So if they want me to find that that's the only equation we have with mu in it So if they want me to find the coefficient of friction mu, I'm pretty sure I'm gonna end up going mu equals friction divided by Normal force number one. I guess me again. I'm gonna try and find each of these and then divide them Well, let's start. I think I'm gonna put my pivot right there And I'm gonna use torques But I need to go perpendicular so here's mg perpendicular and Let's zoom in just a little bit here Tyler imagine this line goes straight down and touches How big is this angle? How big is this angle if this goes straight down and touches? This is 90 60 and then the one next to it 60 because these two add to 90. Okay. We've got our 60 Yeah, oh And the upper one I can kind of also find the perpendicular component. It's gonna go like this Here's gonna be f normal to perpendicular and parallel See the Z. Okay So Dylan Lesser, how big is that angle? and See the Zed again. Ah How big is that angle? 60. Okay, that one is gonna be two Zs Okay Where that's the right angle? All right torques How do I know torques? There's a beam The sum of all the torques clock wise that's in that direction Equals the sum of all the torques counter clock wise that's in that direction If that's my pivot point if it could magically somehow rotate right around there I know it can't there's a wall in the way, but use your imagination. What would cause it to rotate in this direction? Sally you're right okay Mg perpendicular times its distance from the pivot. How far from pivot is it? Sorry 2.5. Did you say are there any other clockwise torques? I don't see any Equals counterclockwise. I think normal force number two perpendicular Times its distance from the pivot five By the way, how far is it from the pivot? Zero no torque Gordon said wouldn't normal force one make it spin to and ah That's why I put the pivot there because I cancel up both those which I don't know anyways very nice By the way, sometimes every once in a while you'll see them give you a beam with no length None call the length L So you would say this is L What would you call this? L over 2 you'll find you have L's and everything and since L's appear and everything. You know what they do they cancel I Haven't seen that for a long time, and I don't think I did that on your test But I'm positive there is one in your big review Okay Mg perpendicular which trig function? Coast as it turns out again kind of nice So this is going to be m g Cosine of 60 Don't forget the distance 2.5 equals Normal force number two perpendicular. Yeah, I'm gonna divide by five right now and get the force by itself 10 kilograms 10 times 9.8 is 98 2.5 divided by 5 is a half half of 98 is 49 And I'm pretty sure the cosine of 60 is also a half. Tell me do you get? 24.5 exactly try it Whoo, did some trig in my head to a decimal place is 24.5 bang on yeah It's the one two root three triangle That careful. That's not what we want. That's the perpendicular component So now let's walk up to this triangle here. How is the perpendicular component related to the normal force? Oh remember the normal force number two is the hypotenuse How are they related? Sign by the Brendan I know what you're thinking mr. Dewick I notice you do the trig for this one, but you don't do the trig for this one in the same line I guess I could have done the trig here and as I got better I did in my answer key, but I made so many dumb mistakes on that last triangle I got paranoid and almost always did the trig for this guy on a separate line I don't know why it just kind of I got stung enough times that I started doing it that way so sign of 60 Equals normal force number two perpendicular Divided by normal force number two I'm running out of room here normal force number two ends up being 24.5 divided by sine 60 is that right? No, yes. Yes people nodding 28.3 all this and we're nowhere near done I'm running out of room folks. I'm gonna have to move up here. Sorry Now here's what we said We said that Mu was equal to friction divided by Normal force number one. I don't know friction. Oh But look look look look look I do know another force the same size as friction what Normal force number two twenty eight point three Divided by Normal force number one. I don't know normal force number one. Oh, but look look look look look I know another force the same size as normal force number one Which one? MG right those are my only two verticals I think new the coefficient of friction and sorry that I kind of looped around like this ran out of room Not very organized mr. Duke, but I think the coefficient of friction is gonna be this number divided by Please tell me I gave you the mass I did ten times nine point eight divided by 98 and I get a coefficient of friction of point two eight nine Actually Jordan, that's the minimum coefficient of friction stronger would be better right now that ladder It was a coefficient of friction of that. Yeah, I mean you drop a feather on it It's gonna slide it's gonna the bottom is gonna slip out everything to come crashing down by the way If it did come crashing down Which way would this part of the ladder move? Down which way would friction between this part of the ladder and the wall be acting up? That's why we had said we have to assume the wall is frictionless Otherwise we'd have one more force pointing up and we have two unknowns up here. We can't solve it Not at this stage, but I did point that out. Okay, make sure you what? Yeah, in fact You can actually find instead as a neat question I don't think I'd be I don't consider it fair game just as a nerdily neat question You could find the minimum safety angle and that I'm sure companies have done lots of research on Example three Each ladder below is set to climb from the ground to the top of a telephone pole Which ladder will be the most stable one two or three and convince me look at ladder one. What's wrong with ladder one? Okay, if you're right here It is possible that your center of mass could extend past the base We talked about how balance is keeping your center of mass above your body above the base so No good Here could slip would you slip standing right here? It's actually as the further you walk the further you walk The bigger this normal force gets The bigger the friction force has to get and if it can't be big enough it'll slide Okay explain your answer using principles of physics I'd probably do a good free-body diagram and walk through this, but I'm not going to worry about that. Oh There you go as the painter climbs this ladder at what position is he most likely to slip a B or C and Stay on this page Can you see that if I added an extra mass? Still be dividing by five, but I would just have plus mass of the painter G Cosign of 60 and then times their distance from the pivot which means the further they get from the pivot The bigger a number you're multiplying by the bigger an extra force you've got up here The more it is that you're likely to slip turn the page homework number one or Seven I like number seven. I like number seven. I like number seven and I consider finding the force from the hinge on Number seven because it's a horizontal beam that I consider fair game if the beam was crooked That's the one in our notes that we said. Oh, that's overkill and or the diving board question is nice