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2D Forces in Horizontal and Vertical Planes

Patrick Ford
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Hey, guys. So in the last call videos we were looking at to use two dimensional forces problems. We had two D forces on a horizontal plane only. So if you were looking at a table top and you were looking down like this, those forces would be pointing just along the horizontal. We're gonna keep on going with that. But now we're gonna take a look at some problems where you have two dimensional forces in the horizontal and vertical planes. It's very similar. We're just gonna 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 gonna assume there's no friction. But we want to figure out the normal force on the block. So we just stick to the steps. We're gonna draw the free body diagram, so the first thing we do is check the weight for us. So if we have this block that is sitting on the table like this, you're gonna look at it from the side and the weight force is gonna be acting vertically, so it's gonna point down like this. So you gotta wait for us, which is equal to MG. We can actually calculate that real quick. It's 5.1 times 9.8 this out. You're gonna get exactly 50. However, because this points downwards that actually have to pick up a negative sign. It's negative. 50 Newtons. Alright, 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're looking at this object from the side like this and the horizontal is going to be this way, so 37 degrees above the horizontal means an F force that looks like this. So are applied Force actually points in this direction here. We know this is f A and this is gonna be our 10 Newtons. And we know this is 37 degrees, all right, and then there's no tensions, but we do have two surfaces in contact. That's blocks on the floor. So that means there's gonna be some normal force like this. So there's a normal force, all right. 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 are two dimensional forces. I just want to point out one thing In previous problems when we had forces in the horizontal plane, we have to look at these two different views the side view and then the top view. Well, in this case, where you haven't 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? A little bit simpler. Alright, So lets decompose are two dimensional forces. We've got this applied force at 37 degrees. Somebody's just gonna draw its components. So this is going to be my f A X and this is my f a y Alright. And we can calculate this because we have the magnitude and the angle, right? So f a X, remember, X goes with co sign. So this is gonna be 10 times the coastline of 37 and that's eight. We know this is eight and then our f a y. We know why it goes with a sign. This is going to be 10 signed 37 that's gonna be six. So we know that this is equal to six. And both of these are positive because they point up and to the right, right, these are positive. Okay, so now we just have to write out our ethical 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 X and y axis. So in the X axis, we've got some of all forces in the X equals m a X. We've got some of all forces in the Y axis equals m a Y. Now, if we're trying to look for the normal force, right? Remember that this normal force points purely along the X or the Y axis, right? This normal force acts only along the y axis. So we're gonna we're just gonna start with that equals m a 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 and the Y axis. That's f a Y. That points up. Remember, we're not going to use F A because that's a two 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 a So we're trying to figure out our normal force. We're gonna have to figure out everything else in the equation. We know what the Applied Force component is. We know this is six. We know RMG 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 six that's basically pulling upwards. But the weight force is going to be stronger. So that means that thing is definitely gonna sit on the ground like this, right? It's just going to stay there. So that means that in the Y axis, only this object is at equilibrium. So this acceleration is equal to zero. 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 gonna cancel. So we have. Our normal force is equal to when you move everything over to the other side. M g minus F A y. So this is going to be our 50 minus r six and this is going to be Newton's. So we actually just seen that this looks very similar to what we've done in our equilibrium problems. So what? In these kind of problems, if you're applied, forces are gonna act partially or completely vertically right? We've got this force that is acting partially, vertically. It's that's a two dimensional angle. Then your normal force, as we've seen, is not going to be equal to your mg. I'm just gonna 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 FAA 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 40 for 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 and that means that there is no surface push anymore. The block is gonna go flying like this. And so your normal force is equal to zero. All right, So in most problems, though, what's going to happen is that 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 in the y axis. So that just means that your forces and the Y axis or to cancel and your acceleration is going to be zero in the Y axis. All right, so just a little comment there. Let's go back and, uh, figure out the second part of this problem here. So 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 zero because it's the neck Librium. So what that means here is we still have some left over forced, right? We still have some components of this applied force that's going to pull this block to the rights. So that means we're really trying to find here is the acceleration in the X axis. So here, in part a, we used f equals M A in the X and Y axis. Now we're just gonna use f equals M E and the X axis. Right? So we've got our f equals m A. There's really only one force to account for, right? There's one force, and that's the F A X, And this is equal to m a X. So we've got this is eight. We know this is gonna be five 0.1 times a X, and so what this means is that your acceleration in the X axis is 1.57 m per second squared, and that's the answer. All right, guys. So that's it for this one. Let me know if you have any questions.