6. Intro to Forces (Dynamics)
Forces in 2D
2D Forces in Horizontal Plane
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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.
Three horizontal forces act on a box (mass = 8 kg) sitting on a smooth surface. F1 is 30 N acting at 53° counterclockwise from the +x axis, F2 is 13 N acting at 67.4° clockwise from the +x axis, and F3 is 20 N directly along the -x axis. What are the magnitude and direction of the box's acceleration?
1.5 m/s2, 14° ccw from +x-axis
1.5 m/s2, 76° ccw from +x-axis
6.9 m/s2, 1.1° cw from +x-axis
1.5 m/s2, 0.46° cw from +x-axis
Solving an Unknown 2D Force
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Hey guys, We've already seen how to solve some two dimensional forces problems, but occasionally in some problems, you're gonna have to solve for a force without knowing against magnitude or direction from the problem. So what? I want to do this video to show you how we solve for an unknown two dimensional force. Let's get to the example so I can show you what I mean. We've got these three forces that are pulling this block. We know the magnitude and direction of two of them. This first force and the second force we want to do this problem is find the magnitude of the third Force. So basically want to find the magnitude of alcohol this F three so that the acceleration of the block is 2 m per second only along the X axis. What that means is that the acceleration in the X axis is two and a Y is equal to zero if it's only in the X axis. So we want to do is find this third force here, the magnitude. And if we want to figure out the magnitude of this third force, we have to use the Pythagorean theorem. We're gonna have to figure out its components f three x and F three y and then plug them basically back into this equation here. So I'm gonna go ahead and stick to the steps. I'm trying to sell for a missing force now, but I want to draw the free body diagram that's basically already drawn for us. So the question is, Do we have to draw this third force? And the answer is, it's kind of unclear which way it's gonna point. We know that the acceleration is going to be along the X axis here, but it's not really clear from the diagram, whether it's going to be up like this or down like that, below or above the positive X axis. So we're just gonna skip Skip that for now. We'll draw it later when we actually figure it out. So remember, if we want to figure out these components F three X and F three y, then we have to decompose any other two dimensional forces because basically we want to fill out this table here of all these components and then build out an equation. So I'm gonna do that. My only other two dimensional force here is this F one at 60 degrees above the X axis. So what I can do is break it up into its components. This is going to be my F one X, and this is gonna be f one y to calculate. I'm just gonna use my normal sine and CoSine so f one X is going to be 100 times the co sign of 60. And this is going to be 50 f one y is going to be 100 times the sine of 60 which is 86.6. Remember, the signs of components are really important. So here we have positive points along the positive X axis and positive y axis. So both of these are just gonna be positive. So this is 50 and 86 6 now. F two is actually a little bit simpler because we know that F two points 70 purely along the negative y axis. What that means is that the X component of F two is actually zero, and the y component is gonna be negative. 70. Not positive. 70. All right, so now we've got to figure out what this third force is. So we want to figure out the components of F three, and to do that, we're gonna actually have to write an equation. F equals M A in the X and the Y axis. Basically, we're going to do is build out an equation vertically using these numbers right here. We know that the sum of all forces, the net force is going to be F one plus F two plus F three. So when we write this in the X axis, your some of all forces in the X axis equal m a X. So what are all your forces in the X axis? You have F one X and then f two X over here. And then what about this third force? Do we write it as positive or negative? Well, that's the whole thing. If you ever are expanding, F equals M and your X and y axis, you're gonna assume the components of unknown forces are going to be positive. Meaning you wouldn't write them with, You know, we wouldn't write them with a negative sign or anything like that. So we basically just gonna assume that F three X is gonna be positive we won't stick a negative sign in there, and this is going to be m times a X. Now we just replace everything that we know. We know that F two X actually is going to be zero, and we know F one X is going to be 50. So 50 plus F three x is equal to and then the masses 40. And then the acceleration of the X axis is to remember that we said that X is equal to two. So basically, what we know is that all of the forces in the X axis have to equal times two, which is 80. Which means that we go Saul for this F three x What you're going to get f three X. It's just a T minus 50 and that's going to equal 30. So basically, we know that this X component has to be 30 because these things have to add up to which is mass times acceleration. Now you do the same thing in the Y axis, so the Y axis are f y equals m a Y, except the y axis actually easier, because we know that the acceleration of the Y axis is zero. So basically all the four step to cancel. So this is F one y plus f two y. And then again, we're just going to assume that the components of F three are going to be positive. So this is F three y here and this is equal m times a y. Actually, that's just zero, right? So that's zero here and now. We just replaced all the values that we know. You're basically just gonna drop them and plug them in from the table. So we know this f one y is 86.6 and this is negative. 17. That's the components. Plus F three y is equal to zero. And so we got 16.6 plus F three y secret zero. So that means that F three y has to be negative 16.6. So even though we just assumed that the letter F three y was positive, what we end up getting is a negative number. So it just means that we have to fill this out on the table. This is negative 16.6. That's what you need to cancel everything out on the Y axis 20 So now here we actually have our components. We know it's 30 and negative 16.6. So if you go back to the diagram, we can see that this force is going to look something like this, right? It's going to be 30 to the right, 16.6 down. So this is F three over here. So now the last thing we have to do is we have these components and we can figure out the magnitude by just using our Pythagorean theorem. So F three y is just gonna be square roots of 30 squared plus negative 16.6 squared and you'll get 34.3 Newtons. And that's the final answer. So you finally answers 34.3. And if you wanted to figure out the direction, you could just use the inverse tangent of your Y over X components, and that would be pretty easy to figure out. That's it for this one. Guys, let me know if you have any questions
2D Forces in Horizontal and Vertical Planes
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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.
You push a 5.1kg cart along the floor with an unknown force F at 30° below the horizontal. Using a scale, you know the Normal force is 70N. What is the horizontal acceleration of the cart?
Wind Force on a Dropped Box
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Hey, guys, let's take this one out. So you got this 2 kg box, right? So there's 2 kg box like this. We're on a rooftop or something like this. We're basically gonna drop it straight down, so there's basically just gonna be some weight force once you drop it. However, there's also going to be a horizontal wind force that pushes this thing horizontally like this. We know this wind force is going to be three Newtons and we want we want to do. We want to figure out the direction of the boxes acceleration. So I've got one force that acts horizontally. Another one acts vertically, and I want to figure out what is the acceleration of this box. Well, basically, if you think about it, you get these two forces one to the right, and one down is going to produce an acceleration that acts this way like this. And I want to figure out what is the direction of this data here or what is the direction of this A. All right, So what I wanna do is first I want to draw the free body diagram. So I got this like this. I've got my weight force. This is my mg. If I go ahead and figure out this is two times 9.8 and this is actually gonna be 19.6. We also have our wind force R F W, which is to the right, and that's three Newtons and that's basically it right. There's no tensions. And because this thing is traveling in the air, there's no normal force or friction. So the next thing I have to do is that one. I would have to decompose all my two dimensional forces, but I actually don't have any, so I can just skip that step. What I want to do is one of right f equals m A in the x and y axis. So if you want to figure out the acceleration here, the direction of this acceleration. So this is data A and the way I would do this is by getting the tangent inverse of the Y components over the X component. So in order to do this in order to get a Y and a X, I'm gonna have to write f equals M A in both X and y axis, right? So in my X axis I've got f equals M a X and the y axis. I'll have f equals m a y. So now, in the X axis, the only force I have is f w. So this is f w equals m a X. So, really, this is just three equals two times a X, so X is equal to 1.5 m per second squared. And that's one of the That's one of the numbers, right? That's one basically one half of the puzzle. Now I just need to figure out what a Y is equal to. So remember that this thing is going to be pushed in the horizontal and the vertical, so you can't assume that the acceleration zero on either axis. So what are my only forces that are acting in the Y axis? I basically just have my negative mg, and that's just because I'm going to use the convention that up into the right is positive, like I usually do, So I've got negative M G equals M A Y. You'll see that the M's cancel and basically you're a Y is equal to negative G, which is just negative 9.8 m per second square. This makes sense because of the only downward force that's acting on this object is the weight force. Then you're just gonna accelerate at 9.8. That's what everything does, right? So basically, I got those two numbers, So whoops. Then I've got my negative 9.8, and now I can just go ahead and sulfur data a sofa A He's just gonna be tangent. Inverse. Now I've got my A y component, which is negative 9.8 divided by my ex components, which is 1.5. If you go ahead and work this out, you're gonna get negative 81 degrees. This negative just means that it's below the horizontal. And that's what we should expect because it points downwards like this. So that's our answer. 81 degrees below the horizontal. Let me know if you guys have any questions
Additional resources for Forces in 2D