 So, lesson eight. The title is Various Two-Dimensional Force Problems. Just looking at some more weird ones and the first one is a classic, it's called the Lawn Mower Problem. I like this question. I like this question. I like this question. I like this question. I like this question. Hopefully the back corner heard that. The Lawn Mower Problem is where you have vertical force components other than gravity in the normal force. We call it the Lawn Mower Problem because the classic one is pushing down the angled handle of the Lawn Mower. You're pushing down as well as pushing sideways. Or if you're pulling on something with a rope, I've also seen it done this way where you're lifting something with a rope at an angle, which means you're doing some lifting and some sliding. Example one. Oh, here's the key idea. Here's the key idea. Here's what's really, really important. In a question like this, normal force is not equal to MG. Just like on a ramp, normal force was not equal to MG. In fact, normal force won't be equal to MG Coast. This is why I said to you guys, well, it's not always MG Coast or MG, most of the time. This is why I didn't have you just memorize last unit where I said I was signed. Parallel is usually signed. Perpendicular is usually coast. Example one. It says find the normal force, find the friction force, and find the acceleration. You know what I'm going to do first? Yeah, and because they drew this force coming in like that, and I like my forces leaving, I'm going to draw my dot over here. So here's my mass. You could label the diagram itself. I just don't like having things coming in. What are the forces acting on this block? Get the obvious ones. Gravity. What else? Normal force. How big? Are we pushing down or are we lifting up on this block? Look at the applied force. Are we pushing down or lifting up with the applied force? Pushing down, which means if you're pushing down harder, what's going to happen to the normal force compared to gravity? I'm going to be really fussy then, and I'm going to say I'm going to draw my normal force larger. If I was lifting up, I would, and I'd ask you to draw a free body diagram. Can't hint, wink, wink, nudge, nudge. I'll look for a normal force smaller. What else? Well, which way is friction? Which way is this thing going to slide if it can't at all? It's got to go, because you're pushing this way. It's got to go to the right if it's going to go anywhere. So which way is friction? And then we have the applied force, 200. What don't you like about this diagram? What bugs you? There's one force that's making me go, blech. Which one, Matt? The applied force. Why? Because it's at an angle. Hey, if we were clever, what would we do? Components. And that's really the solution to this. I'm going to go, I'll call this 4x and force y. You could also call them parallel and perpendicular, but I think last unit when we were horizontal, we called it x and y. Oh, what angle is that right there? 20 degrees. What's the first thing they ask us to find now that we've done our free body diagram? I don't know the normal force. Oh, but look, look, look, look, look carefully. I know two other forces the same size as the normal force. What two forces are the same size as the normal force? In other words, think what are your forces up and what are your forces pointing down? Pointing up is this, yes? What two forces are pointing down? Is that okay? Do I know M? Check. Do I know G? Check. Do I know that? Trig. Let's see. Opposite adjacent are hypotenuse. And they gave me the hypotenuse. Which trig function? Since this is the first time I've done one of these, I'm not going to take any shortcuts. I'll go to the margin here and I'll write sine 20 equals opposite over hypotenuse. We started taking shortcuts with the ramps because we'd done so many of them. We started saying, oh, it's opposite. It is sine. And it's time-sing. Here, I just wasn't sure. I knew it was sine. I just wasn't sure if I was multiplying by sine or dividing by sine. Man, if I get Fy by itself, what do I do with that 200? Okay. So if I hear you correctly, this is going to be MG plus, oh, Mr. Deweyck, why'd you put an F? Oh, no, this is good. This is good, Mr. Deweyck. Never mind. I wrote it right. Plus 200, Matt, you said? Sine 20. Is that right? There's my normal force. There's my normal force. What is 50 times 9.8 plus 200 sine 20? 558. I'd love to hear the sound of calculators being opened. So I'll write 558 because this is an answer, so I'll go to three sig figs. I'll keep this number on my calculator, though, anticipating that I'm going to use it. 558 Newtons. B. Jacob, what is B at? What's the second thing they asked me to find? Friction. Friction is what times what? I don't know the- oh, wait a minute. I do know the normal force because I just figured it out. But if we hadn't, I would have said, I don't know the normal force. Oh, but look, look, look, look, look, I know two forces the same size as the normal force. Here, since we found the normal force, we can go friction equals mu Fn, and they told me mu is 0.1, and Fn is 558. You know what? I don't need a calculator. It's 55.8 Newtons. All right, multiplying by 0.1 just moves your decimal one to the left. So far so good? Happy joy. All right, vendor, what's the third thing they want me to find? Okay. Oh, who's winning? Now, look very closely at our free body diagram there. Think which way this block is going to be accelerating. Who's the winning force? And don't say 200 because it's normal. Who's winning? Fx. Who's losing? Oh. So, we've got to actually, although this question looked ugly, our force equation ends up being winner minus minus, Mr. Dewick, not equals, minus loser equals ma, a fairly simple force equation. Do I know Fx? Got to do some trig. Adjacent. Which trig function? I think cosine of 20 equals Fx over 200. Our vendor, how would I get the Fx by itself? Okay. So, you know what? I'm going to go like this, 200 cos 20 minus, and friction was 55.84, if I really want to carry a few extra sig figs. Our vendor, how would I get the a by itself? Let's do that, this line, since they said find a, is that okay? And the mass was 50? Yeah. The key is a good free body diagram, Breanna. If you do that, it kind of falls apart, I think. If you don't, if you try doing all of the forces without labeling your diagram, if you think you're so smart, you can keep track of everything. If you somehow feel the one second that it takes to label your diagram is a waste of time, you're in for a world of hurt, not that I'm mentioning any names. Who came in for help yesterday and didn't have any free body diagrams filled in and was wondering what they'd done wrong? I don't know, won't mention any names. Message received? Good. Do you guys get 2.64? I don't know why you got 3.64. There you go. Easy enough to figure out. There's the lawnmower question. Did I say I like this question? There's two versions of this, pushing down which increases your normal force or lifting up with a rope which would do what? Decrease your normal force. Of course, if you're moving something, lifting up is the smart way to do it because then it slides easier. Example two, write force equations for a mass on the ramp as shown below. The force F represents a person pushing down on the mass and they're pushing exactly perpendicular to the ramp. The system is at rest. Again, I'm going to do a bigger diagram because it's also a bit too small. There's my mass. What are the forces acting on this one? Get the obvious ones. Kara, absolutely. What else? Normal force. I'm going to hold off on that for a second. Can we remember that we're going to draw one? Because I'm not sure how long it's going to be because I think that's what they're mucking around with. Is there friction? Since they put a mu there, I'm going to assume there's friction. Which way does this thing want to slide? Down the hill, so friction up the hill. By the way, we would break gravity up into mg perpendicular and mg parallel. Now let's deal with the final two forces. There's the applied force. This is an ugly or a tricky free body diagram because Connor's stuff is going to overlap. The applied force is also in the same direction as mg perpendicular. Does this make the normal force bigger or smaller or stay the same? Jacob, I think you're right. Now the normal force isn't just cancelling out that. It's also cancelling out that the normal force is going to look something like that. In terms of equations, you would say this, the normal force is cancelling out mg perpendicular plus the applied force. Really, that means it's going to be mg cosine of whatever the angle is plus I think perpendicular is cos. It is. Friction is what times what? Mu times this mess. This is why, and I think it was in this class, maybe it was my other block, when someone said friction equals mu mg, I went so ballistic because no, it can be much, much more complicated. Normal force is not always mg. In fact, if there's a ramp involved or any angles, it ain't going to be mg. You're going to be doing some trig. Who's winning? It's a trick question. Who's winning? No one. It's a tie. Another equation would be mg parallel equals the force of friction. Friction is what times what, Mitchell? What's fn? The mess. I can now work my way through whatever I needed to find. There's another example of a lawnmower type question. I would not put one this nasty on a written again. Multiple choice nasty? Sure. Fair game is a multiple choice nasty. The previous one, though, that we did, because it was on level ground, not on a ramp, that one I would feel okay putting on a written. Two, the boat problem. I have to admit, well, this is nerdily cool. It's not on your test, and I'm cutting corners. By the way, you break it into components. It's not as impressive as you think. So turn the page. You had to draw the lines through it for her, really? You just can't bring yourself to draw lines through things? The low CD, right? The third unique force problem is what's called the static equilibrium problem. This is when you have an object that is at rest. We're going to be doing an entire unit on this. You can make a little note right now that a question like this is not on this test, but I want to plant the seed today so that three units from now, when we look at equilibrium, hopefully by that time it's had a chance to grow. If an object is at rest, that means it has zero acceleration, and therefore balanced forces, it means your net force is zero. Example five, it says, find the tension in both the cables. What are the forces acting on this object? Get the obvious ones. Is there a normal force? Is this object touching a surface at all? Nope. It's hanging from two cables, some kind of a sign. This is how they strung it up. What are the forces? Since they call that cable one, how about I call that tension one? Since they call this cable two, how about I call it tension two? The first step is being able to label it. Here's the second step, knowing what the net force is. What is the net force? Zero. Now, put your pencils down for a second. Here's where I'm going to get a little tiny bit clever. I'm going to add all three of these forces together, tip to tail, tip to tail, tip to tail. I always start out by drawing the easiest force first. What's the easiest force? I'm going to draw that first. Don't write this down yet, just watch me. I always start by drawing the toughest force next. What's the, out of the remaining two forces, which one looks uglier? Tip to tail, and I know I have to stop right there. You know how I know I have to stop right there, because when I add the third force, I know it's exactly horizontal, which means, Trevor, I can't go this far because that wouldn't give me a horizontal, or I couldn't stop right there because that wouldn't give me a horizontal. That's why I draw easiest first, toughest next, and I can almost always figure out when to stop. I'm going to add the third force, tip to tail. If this was a displacement, how far have I traveled if I end up back where I started from? If this is forces, and I add all three forces together and end up back where I started from, what's my net force? That's the picture. Draw that. Why is this so nice? By the way, you can solve this with components, but because I know it's at rest, unlike Cara, I know that as it turns out, I can draw a closed triangle, because that's what at rest means. No net force, no net displacement. It's going to be a triangle that comes back to where it starts from, which means I think it's way easier to solve that than break everything into components. All I need to do is figure out where the heck this 35 goes. Well I noticed this 35 is tension one next to a vertical line. Is that tension one next to a vertical line right there? Pretty sure that's 35 degrees, opposite adjacent to hypotenuse, adjacent to hypotenuse. And the last one, what do you want to find first, tension one or tension two? Not a trick question, just decide. Tension one? Now, the only thing I know is MG, so which trig function am I going to use for tension one? And since this is a new triangle, I'm going to write, sorry for tension one, you say sine? Cosine, yes, yes, yes, yes, yes. I'm going to write everything out carefully. Cosine of 35 equals MG over tension one. Say, how would I get tension one by itself? I think this is where I'm going to cross multiply or move stuff diagonally. I think tension one is going to be MG, not times cos this time, but divided by cosine. What do you get? 10 times 9.8 divided by cos of 35. Two or three sig figs, 140, is that right? 120, 119 point what? Okay, so 120, sure. Zay, redeem yourself. Which trig function will I use to find tension two? 10, I could use sine, but why use a possible wrong answer to find another wrong answer? Let's use what they gave me, because it's the same amount of writing. 10, 10 of 35 equals opposite over adjacent. How will I get tension two by itself here? I think multiply by MG, right? 10 times 9.8, 10, 35. 68.6? By the way, you could also have used Pythagoras, but then you're using this answer, which might be wrong, but a squared plus b squared does equal c squared. I figure by now we're comfy enough with trig, because we've done so much this unit that trig is almost like falling off a log, but not falling off a logarithm. Jacob, is he alive? He is. I just had my doubts. Turn the page. Example six says draw the vector addition of forces in example five, and notice that if the object is at rest, the forces add to give zero. We actually did that in solving this. A lot of teachers will teach to use components. In other words, they would say break tension one up into this way, and this way, and find some, you know what, I think do the trig in one fell swoop. So example six, already done. I like example four in that it's really a neat, nerdy kind of a question. I'm just going to talk about it, but we're not going to actually analyze it, because it's a bit beyond the scope of what we're doing. The accelerometer problem. If you were to accelerate this train car in this direction, this weight that was hanging would get pushed back. Well, no. Inertia would make it want to stay where it was until it reached an equilibrium point where the acceleration of the train was exactly canceling out the component of gravity that was pulling it down. You can calculate it, and it's not that tough. I just don't have time today, so we're not going to worry about example eight and example nine. We're going to do this, number one, number two, number three, and now you're not going to get all the rest of class because I want to show you a couple of videos, which is why I paused.