Hey, guys. So in the last couple of videos, we were looking at these 2-dimensional forces problems where we had 2D forces on a horizontal plane only. So if you were looking at a tabletop and you were looking down like this, those forces would be pointing just along the horizontal. We're going to keep on going with that, but now we're going to take a look at some problems where you have 2-dimensional forces in the horizontal and vertical planes. It's very similar. We're just going to jump straight into the problem here. So we have a block that's on the floor. It's being pulled by some force. It's 37 degrees above the horizontal and we're going to assume there's no friction, but we want to figure out the normal force in the block. So we just stick to the steps, right? We're going to draw the free body diagram. So the first thing we do is check the weight force. So if we have this block that is sitting on the table like this, you're going to look at it from the side and the weight force is going to be acting vertically. So it's going to point down like this. So we've got our weight force which is equal to mg. We can actually calculate that real quick. It's \(5.1 \times 9.8\). If you work this out, you're going to get exactly 50. However, because this point's downwards, it actually has to pick up a negative sign. It's negative 50 newtons.

We know that there's an applied force. This is 10 newtons at 37 degrees above the horizontal. So what does that mean? Well, if you were looking at this object from the side like this, the horizontal is going to be this way. So 37 degrees above the horizontal means an applied force that looks like this. So our applied force actually points in this direction here. We know this is \(F_A\) and this is going to be our 10 newtons and we know this is 37 degrees.

Alright, and then there are no tensions, but we do have 2 surfaces in contact. This block is on the floor, so that means there is going to be some normal force like this. So there's a normal force. Alright. And then there's no friction. Right? So we're done with the free body diagram. Now before we get into the second step, we know we have to decompose our 2-dimensional forces. I just want to point out one thing. In previous problems, we had forces in the horizontal plane. We had to look at these 2 different views, the side view and then the top view. Well, in this case, where you have forces in the horizontal and vertical, you actually don't need this top view. We've accounted for all of our forces in just one diagram. So that's really good about these problems. They're a little bit simpler.

Alright. So let's decompose our 2-dimensional forces. We've got this applied force at 37 degrees. So I'm just going to draw its components. So this is going to be my \(F_{AX}\), and this is my \(F_{AY}\). Alright. And we can calculate these because we have the magnitude and the angle. Right? So \(F_{AX}\) - remember X goes with cosine, so this is going to be \(10 \times \cos(37)\) and it's eights. We know this is 8. And then our \(F_{AY}\), we know Y goes with a sin. So this is going to be \(10 \sin(37)\), and that's going to be 6. So we know that this is equal to 6. And both of these are positive because they point up and to the right. Right? So these are positive.

Okay. So now we just have to write out our \(F = MA\). Remember, we're trying to figure out the normal force that's acting on the block. So we've got to write it out in the x and the y-axis. So in the x-axis, we've got the sum of all forces in the x = \(M A_X\). We've got the sum of all forces in the y-axis = \(MA_Y\). Now if we're trying to look for the normal force, right, remember that this normal force points purely along the y-axis. Right? This normal force acts only along the y-axis.

So we're just going to start with F = MA in the y-axis because that's where that force belongs. Right? So we've got our normal force which points up, we've got the components of our applied force in the y-axis, that's \(F_A_Y\) that points up. Remember, we're not going to use \(F_A\) because that's a 2-dimensional force. We have to use this component that's just in the y-axis. And then we have our mg which points downwards and this is \(MA\). So we're trying to figure out our normal force, then we're going to have to figure out everything else in the equation. We know what the applied force component is. We know this is 6.

We know our mg is 50 down, so we just need to figure out what's the acceleration. Well, just like we did for the equilibrium problems, we're going to take a look at our upward forces and downward forces and see which one wins. So here we've got our components of our applied force, the 6 that's basically pulling upwards, so that means that in the y-axis only, this object is at equilibrium. So this acceleration is equal to 0. It doesn't go flying upwards and it definitely wouldn't go crashing through the floor. That wouldn't make any sense.

So what that means is that this object is at equilibrium and all the forces are going to cancel. So we have our normal force is equal to, once you move everything over to the other side, \(mg - F_{AY}\). So this is going to be our \(50 - 6\), and this is going to be \(44\) Newtons. So we actually just seen that, this looks very similar to what we've done in our equilibrium problems. So what in these kinds of problems, if your applied forces are going to act partially or completely vertically, right? We've got this force that is acting partially vertically. It's ataxial dimensional angle. Then your normal force as we've seen is not going to be equal to your \(mg\). I'm just going to quickly go through these different scenarios that you might see. We've actually seen a lot of this before. There's really just a couple of things you might see. You might see a situation where \(F_A\) is partially pushing the block into the ground. And so what we've seen here is that when you break up all of these forces into their components, then your normal force has to cancel out both of these forces. So if you're pushing down, your normal force is going to be greater than \(mg\). We've seen that before.

And if you're pulling up but not enough to lift just like we did in this example over here, right, a situation like this, then what happens is that your normal force is going to be less than \(mg\). That's exactly what we saw here. Your normal is \(44\) where your \(mg\) was \(50\). And then finally, if you're pulling up and it's enough to lift, meaning your upward forces are greater than your downward forces, then that means that there is no surface push anymore. The block is going to go flying like this, and so your normal force is equal to 0. Alright? So in most problems though, what's going to happen is in most problems, your upward forces are going to be less than your downward forces. Basically, these two scenarios like this. And so that means that the object is going to be in equilibrium just on the y-axis. So that just means that your forces on the y-axis are going to cancel, and your acceleration's going to be 0 on the y-axis. Alright? So just a little comment there. Let's go back and figure out the second part of this problem here. So in the second part, what we're trying to do is figure out the acceleration. So what does that mean? If we're trying to figure out \(a\), well, we just finished saying that the acceleration in the y-axis is going to be 0 because it's in equilibrium. So what that means here is we still have some leftover force. Right? We still have some components of this applied force that's going to pull this block to the right. So that means what we're really trying to find here is the acceleration in the x-axis. So here in part a, we used \(F = MA\) in the y-axis. Now we're just going to use \(F = MA\) in the x-axis, right? So we've got our \(F = MA\). And there's really only one force to account for, right? There's one force and that's the \(F_{AX}\) and this is equal to \(MAX\). So we've got this as 8, we know this is going to be \(5.1 \times a_x\). And so what this means is that your acceleration in the x-axis is \(1.57\) meters per second squared. And that's the answer. Alright, guys. So that's it for this one. Let me know if you have any questions.