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 going to start to see some problems that are going to combine the two. So we're going to talk about some problems that are in two dimensions but where the forces are in a horizontal plane. Let's go ahead and check it out. We're really just going to combine a lot of things we know about forces with a lot of things we know about vectors in 2D. Alright? So we've got this 5-kilogram block here. We're just going to jump straight into the example. It's pulled by 2 horizontal forces. And we want to figure out the net force and then the acceleration in later parts. We're just going to stick to the steps. We're going to draw the 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 and see the weight force, you kind of have to look at the tabletop from the side like that. So your weight force comes down like this. This is \( w = mg \). Now the next thing we look for is applied forces and tensions. 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 2 forces, 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 what we have to do is we have to look at this tabletop 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 force. This is my \( f_1 \), and I know this is equal to 2 Newtons. And I have another force that's at some angle. This \( f_2 \) is equal to 5, and I know this is at 37 degrees. The last thing we check for is if 2 surfaces are in contact to see if there's any normal and friction. So we do have 2 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 2 different views here. So if all of your applied forces, right, so your applied forces here like \( f_1 \) and \( f_2 \) 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 at equilibrium in the vertical plane. Right? So basically the box is sliding only along the tabletop, So that means your vertical forces like weight and normal are going to cancel. So what happens in these problems is that really these vertical forces like weight and normal aren't really important, and the ones that we're going to focus on are going to be the \( f_1 \) and \( f_2 \), your applied forces. Alright? So let's go ahead and get started.

So in part a, 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's just going to be sigma f, which means you're going to add your forces like \( f_1 \) and \( f_2 \). And in previous videos, if you have, like, 5 in this direction and 5 in this direction, then you just add them straight up like numbers and that makes 10. But the problem is we 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 addition. And that's exactly what we have to do here. So really just using a lot of the same vector addition that we've seen before with forces. So if we want to figure out the net force, our net force is actually going to be like this. So this is going to be our \( f_{\text{net}} \). And so what we're going to have to do is we're going to have to add these things by their components. So I'm going to have to figure out \( f_x \) and \( f_y \) and then use the Pythagorean theorem. So my magnitude of my net force is just going to be the Pythagorean theorem of my \( f_x^2 \) plus my \( f_y^2 \). 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 a 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 2-dimensional forces by using sine and cosines and things like that. Alright? And we have this table here to keep track of everything that's going on. So our first force, our x and y components, let's see. This first force is just 2 Newtons. It's purely along the x-axis. That's very straightforward. That just means that the x component of 2 and the y component is 0. For the second force, this is one where we have 5 at 37 degrees. So what happens is our \( f_{2x} \) is going to be 5 times the cosine of 37 and that makes 4. So this is going to be 4 here. And then our \( f_{2y} \) is going to be 5 times the sine of 37 and that's going to make 3. So this is going to be 3 years. Now you just add them straight up and your \( f_x \) component is going to be 6 and your \( f_y \) component is going to be 3 because all you're doing is you're just adding \( f_1 \) and \( f_2 \), the components straight down. So that means I have these numbers here. I have 6 squared plus 3 squared, and if you work this out, you're going to get 6.7 Newtons. So that's the answer to the first part. We got 6.7 Newtons, that is the magnitude of our net force in 2 dimensions. So let's move on to part b now.

Now what we're going to 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 \( f = ma \). If we have the net force, we can calculate the acceleration. But this net force here is in 2 dimensions, which means we're going to get a 2-dimensional acceleration. If we want to get the \( a_x \), the acceleration just in the x-axis, then we're going to have to use \( f = ma \), 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 axes, and that brings us to the 3rd step. We're going to write \( f = ma \) in the x and the y axis. So in part c, what we're going to be doing is calculating \( f_{\text{net}} \) sorry, the acceleration in the y-axis. So we're going to use \( f_{\text{net}} = ma_y \). 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 going to be the combination of the x component of the first force and the second force. So really our \( f_{\text{net x}} \) in the x-axis is 6, and this is going to be 5. And so what happens is whoops. So we have 6 equals 5 \( a_x \). And so therefore, your acceleration is going to be 1.2. That's going to be \( a_x \). Now we're just going to do the same exact thing in the y-axis. So here our \( f_{\text{net y}} \). So that number is from our table. It's going to be this one right here. So this is our \( f_{\text{net x}} \). That was this guy. And so this is going to be our \( f_{\text{net y}} \). So we got 3 is equal to 5 times \( a_y \). And so now what we've got is 3 over 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 in 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 going to be the 2-dimensional acceleration. Right? So this net force, \( f_{\text{net}} \) here, is going to mean that the object is going to accelerate in this direction, and this is our two-dimensional acceleration. So here, we used \( f_{\text{net x}} \). Here, we used \( f_{\text{net y}} \). To calculate the 2-dimensional acceleration, we're just going to use \( f = ma \). So this is \( f = ma \). In other words, what we have here is our \( f_{\text{net}} \) is equal to \( ma \). And so what we've got here is we already have that number. It's 6.7, which is equal to 5 times \( a \). And so that means that we've got 6.7 divided by 5 which is equal to 1.34 and 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 it's all vector addition. Right? So what I mean by that is that your acceleration is really just the hypotenuse of your \( a_x^2 \) plus \( a_y^2 \). So you could have also done this by doing 1.2 squared plus 0.6 squared. And if you go ahead and work this out, you're also going to get the same number which is 1.34 meters per second squared. So there are two different ways to get to that same number. Alright. So that's it for this one, guys. Let me know if you have any questions.