Gravitational Forces in 2D - Video Tutorials & Practice Problems

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Gravitational Forces in 2D

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Hey, guys. So now that we've got a good feel for the universal law of gravitation in one dimension, it's time to look at how it works in two D. But we're going to see this very similar stuff that we've done before. So we're working with Net forces in one dimension. Then if we had net force, if we have multiple forces on an object, for instance, like this force right here, that was, I don't know, three Newton's and we had another force. It was acting on it like negative four Newtons, that we could figure out what the Net gravitational forces just by subtracting or adding these things together. So I get a net force of negative one Newton that points in that direction. It's a simple addition, right? Well, in this case, let's check it out. So we have a sort of triangle of masses. So all these things are a little EMS, Newton's losses, they all pulling on each other. So this bottom left one fields of force that points in that direction, and it also feels of force on it that points in that direction. So let me just make up some numbers here I'm just making this up. So let's say this was like three Newton's. And now this was like four Newtons. How would I figure out the gravitation, The net gravitational force. So these questions will ask you what the net gravitational force is. But you can't just add these things together because they think they point in different directions. And so for to solve for net forces in nonlinear arrangements, you have to use Vector Edition and because remember that the force of gravity is a force and forces are vectors, so we can always take a vector like this and break it up into components. All right, so the point is, if you have this vector here that points in some off angle, you can always decompose it into its X and Y components, just using your sine and cosine stuff from Trig before. So you have to know some angle theta and stuff like that, Right? So now we can do is now we can take these components, add them up, and then figure out what the Net gravitational forces like that. Okay, sort of like refresh ourselves all of this stuff. Let's go ahead and just do a quick example. So we're supposed to calculate with the magnitude and the direction of his net force is on the bottom mass the M one in the figure. So the first thing is, we just have to label the forces they're acting on this. So I've got a force that points in this direction between these two masses. And then we got a voice, that force that points to the right from those two masses. Now, this is between m one and M two, so I'm gonna call this F 12 and then this is F one and three. So I'm gonna call this F 13 okay? And so how is what is the Net gravitational force gonna be? Well, I can't just add these things together, so I have to use Vector Edition. If I add these two vectors tip to tail. What I'm gonna get is I'm gonna get a net gravitational force that points in that direction. Right? So these things pointed this way. The net gravitational force has to be between them. So this is the net gravitational force, and I'm to figure out the magnitude of that. So how do I figure out the magnitude. Well, remember that we're figuring that the magnitude of a force and we have the sort of components like the X and Y components. We do that using Pythagorean serum. So we've got F 13 is the ex components. We've got a square that and then we've got f 12 That's the white component. And then we've got a square that so this is like the high pot news of this triangle. Right? So we're trying to figure out what this distance is right here. Um, that we have to figure out what sort of like the have partners of this triangle that it makes right? Okay, So now we actually have to figure out what these two forces are. What is F 13 and what is F one to? Well, I can figure out f 13 just by using Newton's Law of gravitation for point masses. So I've got big G times mass one mass three divided by the distance between our one and three, or and one and three squared. So that distance between and one and then three is equal to 0.1, and then I have both of those masses so I could go ahead and just figure that out. We've got 6.67 times 10 to the minus 11 than the product of the two masses 25 20 divided by 0.1 squared. So when you get that is, you get to their sorry 3. 34 times 10 to the minus six. So I got 10 of minus six there. Alright? And if I do the same exact thing for F one and two, I've got G m one m two divided by the distance between one and two squared, which is at 0.15. So just got the same same set up here 6.67 times, 10 to the minus 11. Then I've got 25 now, 30 now divided by 0.15 squared. So what I get is 2. 23 times 10 to the minus six. And both of these things are in students. All right, so now that I actually have the components of these in the actual forces now, I could go ahead and just figure out the net gravitational forces So the magnitude of F net is going to be square roots. And then I've got just basically plugging these values. It got 3 34 times, 10 to the minus six squared, plus 2.23 times 10 to the minus six squared. If you go ahead and plug that in, you should get a net force of 4.2 times 10 to the minus six. And that's the Newton's. How about the direction? How do I figure out what the direction is? Remember that that direction is gonna be this angle theta right and theta. How do we relate data from the tangent? So we really the tangent data is equal to the y component divided by the ex components. So in other words, it is the, uh uh f 12 divided by F 13 So we'll go ahead and figure out this angle here. I've just got to take the inverse tangent, so that's gonna be the inverse tangents of the F one to which is going to be that 2.23 times 10 to the minus six, divided by 3.34 times 10 to the minus six. If you do that on your calculator, what you should get is 33.7 degrees. All right, guys, that is the magnitude and the direction of this net gravitational force. And let's keep going on. Let me know if you have any questions.

2

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Using Symmetry in 2D Gravitation

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Hey, guys, So often problems that you'll see in two D gravitation. We'll have a lot of big numbers and a lot of steps to follow. So this video, I want to give you a shortcut in order to minimize the amount of work that you have to dio. So let's check it out. So let's say I was trying to figure out what the gravitational forces on this top mass right here. So I wanna find F net. Now, How do I do that? Well, first I have to go ahead and label my forces. So there's two masses on this triangle and they're both exerting gravitational forces on this top mass right here. So we've got a gravitational force from this guy that points in that direction and then f g over here. Okay, so now that I've labeled the forces, I have to calculate them. But notice how if these masses are the same, the little EMS are the same, and I haven't given numbers for those. I've just I've just said that they were both little EMS, and the distances between them are the same. So this is the distance between them, then these things have to have the same gravitational force. These have to have the same f g. Remember that F g is just big G times the mass of both objects divided by their distance square. So if you have the same EMS and the same ours, gravitational force is the same. So now that we've calculated with the gravitational force ours, If I actually had numbers for this, I'd have to decompose the vectors into their its components Now, just to just to sort of refresh that the way I would do that is, if you have to factor that points off in this direction, then you could always split off into its components. So I have an X component, which is F X, and this is a white components F y. If this is a force or vector like F right and we do that using TRIG, right, If long as I have the angle theta relative to the X axis, I could get its components. So the next step for this is now that I have the forces Aiken split off into its components. So if this is the angle relative to the X axis, that means it's gonna have a X components here, and that X component is gonna be related to the coastline of data. Whereas the why component down here that points in this direction is gonna be related to the sign of data. All right, But I have the same exact forces. And if these things have the same angles and the same forces, then you're gonna end up with the same exact components in the X and Y directions. So the tip is if you end up with two components in the in the other direction that are equal, but opposite you can always cancel them if they're opposite. So this is another way where you can use symmetry to reduce the amount of work that you've done. So let me show you. So I have the X and Y components. But this force over here on the left is gonna break up into the same exact components F x and F y, except the white components are gonna add together here, whereas the X components are equal, but they point in opposite directions. So we don't even have to do this in our calculators. We won't have to worry about it because the Net force is gonna end up canceling both of those out. And so instead, what's gonna happen is that the f Y components are gonna add together and your net force is gonna be pointing down here. And that net force, which I wanted is just gonna be equal to times f Why? So this is one way that symmetry can reduce the amount of work that you have to dio. So I wanna attach some numbers to this stuff. So let's go ahead and work out this quick example. So go ahead and pause the video after looking this figure and see if you can figure out what the net gravitational force is on this little M from these two forces to these two masses over here on the rights. Okay, cool. So let's get into it. The first thing we have to do is we have to label our forces and calculate them. So I've got a force that points off in this direction. And I'm told that that F G is equal to five Newtons and I have another force that points off in this direction so that F G is equal to five Newtons and I'm told it feels a five Newton force from each em on the right. And that makes sense because we have the same exact are and same m for both of them. So we know that those forces are equal. So now we just have to split it up into its components. So I've got this data angle here that's relative to the X axis. So that means that this component here is gonna be FX and then I've got f y. That points off over here, So I've got f y. But this angle right here relative to the X axis is going to be the same as this angle right here because I had the same distances involved are, and we can assume that these masses right here are sort of like connected by that vertical line. So that means that the angles for both of them are the same. Which means that the why components are gonna be pointing in opposite directions, but they're going to be canceling each other out so that I can cancel out my f y with this F y right here. And the FX components are actually going to add together. So that means that the net gravitational force is going to be pointing off in this direction. So what are the X components for? Both of these things will remember that the X components if I look at my vector edition is just f times cosine theta. So I actually can put I could put numbers to that because I know what the force is. I know that these five Newtons and as for the angle I'm given that the angle is 53. degrees. So that means that each FX component is three Newton's. And as for the Net force these things we're gonna add together and they're gonna be perfectly equal to each other in terms of the components. So that means that the F Net is going to be two times FX. So that means that the net gravitational force is two times three, which is just six Newton's. Alright, guys, let me know if you have any questions with stuff

3

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Finding Net Forces in 2D Gravitation

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6m

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Hey, guys. So for this video, I wanna pull together everything that we've learned about gravitation in two D. So Newton's law of gravity vector components and using symmetry to solve a real problem that you guys might see. Let's check it out. So I've got 3 50 kg masses, I'm told with the masses, and the length of all of these things are in an equilateral triangle. And I'm supposed to be figuring out the magnitude and the direction of the net. Gravitational force is on the bottom mass. So we're gonna be taking a look at this guy right here. But we know the steps to solve any too deep gravitation problem. We've got a label, the forces, and then calculate them. Then we've got to decompose them into their components and then see if we could use some symmetry. And then finally, we're gonna add those components and figure out the Net forces. Let's go ahead and take a look. So first things first, I wanna I wanna label these forces. So on the bottom mass, I have a gravitational force that points in that direction. And because it's between M one and M three, I'm gonna call this F 13 And then I got another gravitational force that acts between M one and two. So I'm gonna call F 12 Alright, So I've labeled them. Now it's time to actually calculate them. How do I calculate with the gravitational force is I go back to my Newton's law of gravity. So the gravitational force between one and three is going to be. I'm gonna treat them as point masses. So I've got G and one and three divided by r squared due. I have everything I need. Well, I have the masses of both of the objects, and I have the center of mass distance between them. So I've got everything I need to go ahead and solve for this. So doing that, I get 6.67 times, 10 to the minus 11 then times 50 times 50 and then divided by 0.6 squared. And if you do that, you should get F 13 is equal to 4.63 times 10 to the minus seven Newton's. So we've labeled the forces. Um, Now we have to actually calculate the other force, right? You have to figure out what f 12 is. So what is F 12? We'll notice here I have the same exact masses involved. So I have the same masses, so same ems and the distance between them. So God, let's see, I've got the same exact masses right here, and I also have the same distances between their centers. So because I have the same EMS and the same ours here, then I know that F 12 is also just going to be 4. 63 times 10 to the minus seven. I don't have to do that over again because I've already done it. Okay, so now I've calculated the forces. Now it's time to decompose them and then use symmetry. So if I'm ever using vector decomposition, I need a coordinate system first. So let's go ahead and draw that. I've got a Y coordinate system right here. Why Axis And I've got an X axis right over here. In order to get components, I need an angle, but not just any angle. I need an angle with specifically with respect to the X axis. So I need that angle theta. How do I go about getting that angle what? We've looked at everything about this problem. We've got the masses and we've got the sides between them. But the one thing we don't know is the angle. And the one thing we haven't used yet is that equilateral triangles have 60 degree angles between their sides. So that means that these angles right here, 60 degrees. How can I go with how can I go from using this? This information here, the 60 degree angles to figure out what this thing is right here. There's a couple of different ways I could do it using geometry. Remember that if these air parallel lines so this bottom and top line your parallel lines, then these angles right here are called opposite or alternative interior angles, and they're the same. Another way you can think about this is you could sort of draw like a perpendicular line that goes down between both of these lines. Here. This is a 90 degree angle and this is also a 90 degree angle. So that means that this has to be 30 degrees because 16 30 makes 90. Which means that 30 and 90 means that this is a 60 degree angle. I know that's a whole lot of geometry stuff, but there's a couple of ways you could sort of convince yourself that the angle that we're working with is actually 60 degrees as well. So now that we have that angle, we can figure out what the components are. Weaken split this thing up into FX and F y. And we know that the relationship between F X and F Y is f cosign data and have signed data. So I've got F co signed data and then I've got F signed data right here. Okay, And now for the other angle right here for the other vector f one to I've got this angle. Now, what's that angle? If this was 60 degrees and these were 60 degrees, then that means that these two angles also have to be the same. So let's take a look here. We have the same exact force, and we have the same exact data between them. So that means that when we split this up into this components here, this FX is gonna be the same, and this f y is going to be the same. So the question is now coming. Use symmetry to eliminate those things. And yes, we can, because we have the same exact force and we have the same exact angle. So these X components are just going to go away. They're gonna cancel each other out. So I'm gonna write that here. So these things are going to cancel because we have the same F. G s. We have same F gs, same theta, but opposite direction. So we end up canceling those things out. All right, so now that we're done with the decomposition and using symmetry, it's time to basically just add up our final components and then figure out what the Net force is. We know the Net Force is gonna be pointing straight up, so that actually takes care of the direction of the gravitational force. Now, as for the magnitude, let me go ahead and write that down here. The magnitude of that net force is gonna be two times f y. Now, what is F y Well, remember that f y using our decomposition was f times. The sign of data I have with the force is that's the gravitational force that we figured out in step two and then I have the angle now that we've done the decomposition. So I've got that f sine theta is equal to 4. times 10 to the minus seven. Whoops. Send the minus seven times the sign of 60 degrees. And if you go ahead and do that, you should actually get four times 10 to the minus seven. That's the white components. But remember that this is just one of the f wise, right? This is just one of them, and the Net Force is gonna be two times f y. So we get two times this number right here four times 10 to the minus seven. So that means that the net Force is eight. Let me go and move their eight times 10 to the minus seven. And that's Newton's. So that's for the magnitude of the gravitational force. And as for the direction we know, it points in the plus y direction. Alright, guys, let me know if you have any questions with this stuff

4

Problem

Problem

Find the magnitude & direction of the net gravitational force on the center 5-kg mass in the rectangle below.

A

8.6×10^{-8} N

B

4.28×10^{-7} N

C

-1.7×10^{-7} N

D

6.5×10^{-7} N

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