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Uniform Circular Motion | 27 mins | 0 completed | Learn |

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Example #1

**Transcript**

Hey guys. So they don't to introduce a very classic problem in gravitation. We're going to have an object's being pulled by other objects by multiple objects in a two dimensional plane. So let me show you how it works. All right so here I have four small spheres they're all one hundred kilograms. So I can say I can just say these are all m's, m's everywhere and m's equals 100 kilograms there are small which means we can disregard the size of the radius right. The range is shown forming square of sides 10 centimeters square so all the sides have a length of L everywhere. I mean just do this L L L L and L equals point 10 meters and I want to know, I want to calculate the magnitude and direction of the net force acting on the sphere on the bottom left corner. So I want to know what is the net force here. So remember the ideas are masses according to the universal law of gravitation attract each other. So this guy is going to get pulled up by this force. It's going to get pulled to the right. I'm sorry by this mass it's going to get pulled to the right by this other mass and it's going to get pulled in this direction by this other mass over here. I'm going to call this one two three four so I can call this Fone of Ftwo and Fthree. Remember that these these forces are mutual forces. So you know this guy is also being pulled down this way. But now we're focusing on an object four the one on the bottom right there. OK. So the net force acting on four is just the sum of all forces on four. So it's just Fone plus Ftwo plus Fthree. But remember forces are vectors so this is a vector addition. Which means you can't simply calculate those numbers and add them up as you would at regular scaler addition. You have to do vector addition here. So you I need to remember is that every vector is made up of components and the magnitude of every vector vector is given by the Pythagorean theorem of its components. So Fnet4 is the square root of I'm going to call this F4x plus F4y. K so it's two components. And that's because you can end up with something like this Fnet on 4x will be this way Fnet on 4y will be this way. And then that will, the two will combine to form a Fnet on object 4. So this is just take us back to vector addition and the angle of the net force on 4 is going to call that theta 4 is the arc tangents the arctangent of y over x. So 4y, 4x. So these are the two things I want. So to get these two variables all you have to do is find F4x, F4y and then we'll be able to plug you back into this equation. All right. So let's do that. How do we find F4x. Well if Ftotal is F1 of F2 of F3 then F4x. Is F1x, F2x and F3x for y and follows the same pattern F1y, F2y, F3y. So I have to get these six numbers and then I'll be able to plug it all in.

But what is F1, F1 is the pull the attraction between F1 is right here. It's the attraction between the 1 and the 4. So let's do that. It's G m1m2 divided by the distance between 1 and 4 square. So G which is six point six seven times ten to the negative 11 mass 1 all the mass or 100 kilograms is going to put a hundred here and 100 here divided by the distance squared the distance is L right now right here between these two. So that's point one square. If you solve this whole thing I have it here if you solve this whole thing you get. You solve this whole thing got. Six point six seven times ten to the negative fifth ok times 10 to negative fifth. All right so now. What I need to do is I need to split this into. F1y and F1x. But if you look at F1, F1 is going straight in the y axis which means F1x is zero and F1y is just a whole number. So when I found F1 it was actually the same thing as my F1y. So it's six point six seven times ten to the negative fifth, just to be clear if you find a force this way. This is also the Fy and then Fx is just 0 because it has no component. OK so now let's find F2 actually 3h here, to have three first because it's going to the right over here F3 will be Gm3m4 again it's between 3 and 4 divided by the distance between three and four square. And all these numbers are exactly the same. You might notice. And these questions are kind of long so they do sometimes to make it a little bit more manageable. They'll give you all the numbers are the same or something like the distance to a point 1. That way it's going to if the numbers are the same it's easier to work out faster. That's the same exact number so you get the same exact number here. Six six seven times 10 to negative fifth. When I split this F3x and F3y notice that F3 is flat this way. So F3y would be 0 F3x would be 6 1 6 7 times 10 to the negative fifth. So it's the inverse of F1, F1 is in the y, F3 is flat in the x, 2 however is that an angle. And when I calculate F2 I'm going to have to decompose it. And then the numbers will actually look different. OK. So let's find F2 real quick, F2 is six point six seven times ten to the negative eleven the two masses again one hundred and 100. Now there's something that's different about F2 the distance between four and two is not L is not L ok. So I'm going to have to get it by right by drawing a little triangle here. So I want to know what is this distance. Let's call this X just for now and then I have this length here is L, this lengt here is L, by the way I hope you figure this out. This is a 45 degree angle because it's splitting the square the square right down the middle to find you can use Pythagorean theorem, X squared equals L squared plus squared so x squared equals 2l squared. So X is the square root of 2L squared. In other words X is the square of two L and this case L is point 1. So X ends up being. Point 141 meters. I know it looks long because it took a bunch of lines but that's because because I was restricted to a tiny little space there. So the idea is that this number here is actually point 141 squared. And just to be clear this comes from here. So it's no longer going to look exactly like this is going to be a little bit different. And I guess three point thirty four times ten to the fifth tenth of the negative. When you split this into F2x and F2y, F2x is F2 cosine of theta. And theta Here's 45 and F2y is F2 sine of 45. So just have to plug in these numbers. You might remember that the sine of 45 on the cosine sign of 45 are the same. So when you do this you should get the same answer. And that shouldn't be a surprise. This is two Thirty six times ten to the negative 5th, 236 times 10 to the negative fifth. Obviously all these things are measured in Newtons really long but once I found Fone, Ftwo and Fthree and I found the x and y components I am now ready to just piece everything together back in here. So we're going to work our way back and work on this piece and then come back and work here. So F1x this is now just plug it in F1x is 0, F2x F2x is over here is two point thirty six times ten to the negative fifth. Now as you do this remember you have to worry about positives and negatives. If you look at F2x, F2 is going this way F2x is to the right to F2x to the right. So it will be positive. So F2x is positive, F2y is positive as well. In fact all these numbers are positive. Fone over here y is a positive and F3x is a positive. All the members are up or to the right. So they're all going to be positive which is nice plus F3x, F3x is right here. So six point six seven times Ten to the negative. Fifth if you go to the y axis I have F1y, F1y right here. Six point six seven times ten to the negative fifth, F2y is this number. Two point thirty six times ten to the negative fifth and F3y is zero. And you might notice that these numbers were. They both have a zero right.

So there's a lot of symmetry here. All these numbers are the same. You going to get the same answer for both. And it's going to be completely symmetric. The answer will be for both of these guys. The answer will be let's make this a little bit different. Nine point zero three times ten to the negative fifth. Obviously Newton's this is 903 times 10 to negative fifth Newtons. OK. Now that I've got these sort of the preliminary answers I can just plug them in here so we can finally work all the way back here. And this is 903 times 10 to the fifth squared plus 903 times ten of the fifth Square. And when you do all this you get a twelve point eight times ten to the negative fifth you might have done this as one point twenty eight times ten to the negative fourth. But everything has been to the power of negative five. So I decided to keep it that way. But it doesn't really matter when you do this here you get to the arctangents of 903 times ten to negative fifth. I hope you see that this is just one. So it's the Arctangent of one which is 45 and it's 45 not because I have choose 45 but because this is fully symmetric that this is 45. But also these two here the X and Y have the same magnitude. So the net result of these two is right down the middle. Plus the guy gets down the middle as well. So it should make sense that this is a 45 pretty long question. But again it typically comes with a lot of symmetry to make it a little easier. It's a lot of work. It's annoying because of these big numbers and multiplication that you have to do. So you've got to practice's make sure that you can do it. In terms of process. I think it's fairly straightforward. This is basically good old vector addition. Except that accountants are being given these forces. You have to actually calculate them using this big ugly equation. Other than that it's just basic vector addition to the vector addition that we have done in the past. That said I want you to try the next one. It's similar but obviously a little bit different. There's still some there's still some symmetry there that's going to make things a little bit simpler and we want to know what is the net force on this guy over here. All right let's give that a try.

Practice: Three small 100-kg spheres are arranged as shown, forming a triangle of equal 10-cm sides. Calculate the magnitude and direction of the net force acting on the bottom sphere.

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