Finding Net Forces in 2D Gravitation

by Patrick Ford
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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