6. Intro to Forces (Dynamics)

Forces in 2D

# 2D Forces in Horizontal Plane

Patrick Ford

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Hey, guys. So up until now, all of our problems have only involved forces acting in the horizontal plane or the vertical plane. But now we're gonna start to see some problems are going to combine the two. So we're gonna talk about some problems that are in two dimensions. But where the forces are in the horizontal plane, let's go ahead and check it out. We're really just gonna combine a lot of things we know about forces with a lot of things we know about vectors in too deep. All right, so we've got this 5 kg block here. We're just gonna jump straight into the example. It's pulled by two horizontal forces and we want to figure out the net force and then the acceleration in later parts, we're just gonna stick to the steps. We're gonna draw a free body diagram, and the first thing we need is the weight force. Now, if you were to look at this block that's on a tabletop, then see the weight force. You kind of have to look at the table top from the side like that. So your weight force comes down like this. This is w equals mg. Now the next thing we look for is applied forces intentions. So we have one that's along the X axis and one that's at some angle. So what does that mean? Does that mean that you have to force Is one like this and the other like this? Well, no, because remember that these are horizontal forces. So what happens is when you're looking at it from the side view, if they're horizontal, you actually can't see them. So we have to do is we have to look at this table top from the top. So imagine now from this table. You're looking at it from the top down, and your forces are going to be along this plane. So here what happens is we have one force. This is my F one, and I know this is equal to two Newtons and I have another force. It's in some angle. This F two is equal to five, and I know this is at 37 degrees. The last thing we check for is if two surfaces are in contact to see if there's any normal and friction. So we do have two surfaces in contact in the side view. So our normal force is going to point up like this and there's no friction or anything like that. So we ignore that. So here's what's going on in these problems, right? We have these two different views here. So if you have all of your applied forces, right, so you're applied forces here like F one and F two act only in the horizontal plane, meaning from the top view they're acting like this. Then that means that your normal force is going to be equal to MG because your object is going to be an equilibrium in the vertical plane, right? So basically the boxes sliding only along the tabletop. So that means you're vertical forces like wait normal are going to cancel. So what happens in these problems is that really these vertical forces, like waiting normal aren't really important, and the one that we're going to focus on are going to be the F one and F two, your applied forces. All right, so let's go ahead and get started. So in part, and we have to figure out the net force, right? We have net force like this. And so in previous videos, what we've done. Is that gonna be Sigma F? Which means you're gonna add your forces like F one and F two and in previous videos. If you have, like, five in this direction and fire in this direction that you just add them straight up like numbers and that makes 10. But the problem is, we can't do that here because we have vectors, right? We have these two arrows that point in different directions. So in this problem, what happens is that because forces are vectors, then when a force acts at some angle in two dimensions, we just treat it like any other vectors, and we're just going to decompose it into its X and Y components. And then if you have multiple forces that are acting like we do in this problem here, then your net force is calculated by using Vector Edition. And that's exactly we have to do here. So really just using a lot of the same vector edition that we've seen before with forces. So if we want to figure out the net force, our net force is actually gonna be like this. So this is gonna be our F net and so we're going to have to do is we're gonna have to add these things by their components. So I'm gonna have to figure out FX and FY and then use the Pythagorean theorem. So my magnitude of my net force is just gonna be the Pythagorean theorem of my f X squared plus my f y squared. So I just need to figure out those two numbers here, and I can figure out the answer. So this is going to be square roots. I've got some number squared, plus some number squared, and that'll equal my net force. And to do this, we just built out little table to keep track of all of our components. So that brings us to the second step. After we've drawn the free body diagram, we have to decompose all of the two dimensional forces by using sign and coastlines and things like that. All right? And we have this table here to keep track of everything that's going on. So our first force are X and Y components. Let's see this First forces just to Newton's. It's purely along the X axis. That's very straight forward. That just means that the X component of two, and the Y component is zero for the second force. This is one where we have five at 37 degrees. So what happens is our F two X is going to be five times the co sign of 37 and that makes four. So this is gonna be four here, and there are f two y is gonna be five times the sine of 37. And that's gonna make three. So this is gonna be three years, and I just add them straight up and your FX component is going to be six. And your F Y component is going to be three because all you're doing is you're just adding F one and F two the components straight down. So that means I have these numbers here I have six squared plus three squared. And if you work this out, you're gonna get +67 Newton's. So that's the answer to the first part. We got 6.7 Newtons. That is the magnitude of our net force into dimensions. So let's move on to part B now. Now what we're gonna do is we're going to calculate the acceleration in the X axis that's a X. So the way we've done this before is by using f equals M A. If we have the Net force, we can calculate the acceleration. But this net force here is in two dimensions, which means we're gonna get a two dimensional acceleration if we want to get the X, the acceleration just in the X axis that we're gonna have to use f equals M A. But we're going to be using the net force in the X axis, and that's going to be an acceleration in the X axis. So we're just really going to be looking at one of the axis, and that brings us to the third step we're gonna write f equals M A and the X and the Y axis. So, in part C, we're gonna be doing is calculating f net Sorry. The acceleration of the Y axis. So we're gonna use F net M A y. Alright, we'll get to that in just a second. So the net force in the X axis is actually very straightforward because we already know what that is in the X axis. It's just gonna be the combination of the X component of the first force and the Second Force. So really are F net in the X axis is six and this is gonna be five. And so what happens is what's so we have six equals five X and so therefore, your acceleration is going to be 1.2. That's going to be a X. Now, we're just gonna do the same exact thing in the Y axis. So here are F net y. So that number is from our table is going to be this one right here. So this is our f net X. That was this guy. And so this is gonna be our definite y. So we've got three is equal to five times a Y. And so now we've got is 3/5, which equals 0.6, and that is equal to our acceleration in the Y axis. Right. So now we're looking for the acceleration, the Y axis. We just use the net force in the Y axis. And now finally, what we want is we want the acceleration A. So in parts B and C, we calculated a X and A Y. But now we want the acceleration, which is really just gonna be the two dimensional acceleration, right? So this net force F net here is going to mean that the object is gonna accelerates in this direction and this is our two dimensional acceleration. So here we use F Net X Here we use f Net y to calculate the two dimensional acceleration we're just going to use f equals a. So this is f equals m A. In other words we have here is our f net is equal to m A. And so what we've got here is we already have that number. It's 6.7, which is equal to five times a day. And so that means that we've got 6.7 divided by five, which is equal to 1.34 that is equal to our acceleration. So we've got 1.34. There's actually another way you could have gotten this number. Remember that if you have both the components of the acceleration, then you can figure out the acceleration because it's all vector edition. Right? So what I mean by that is that your acceleration is really just the hypotenuse of your a X squared plus a Y squared so you could have also gotten, uh, you could have also done this by Ewing. 1.2 x Sorry. 1.2 squared plus 0.6 squared. And if you go ahead and work this out, you're also gonna get the same number, which is 1.34 m per second square. So there's two different ways to get to that same number. Right? So that's it for this one. Guys, let me know if you have any questions.

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