Welcome back, guys. Let's try to solve this one together. So we've got these three point charges arranged in this triangle right here, and we're trying to find out what the net force on this three Coulomb charge is. So let's go ahead and first draw. What? The forces that are acting on this three Coulomb charge. So we got We're focusing on this guy right here. And the first is that he feels an attractive force from this negative two Coulomb charge. Why is it attractive? Because these things are unlike charges, one's positive, one's negative. So this is going to point in this direction, and I'm going to call this F2 because it comes from the two Coulomb charge.

Now, there's also this one Coulomb charge here on the bottom left corner, that is some distance R away from this other charge, and it's going to exert a repulsive force on the three Coulomb charge because these things are like charges. They're both positive. I'm going to call that F1, and what we're supposed to do is take these forces, which is the only force acting on it, and figure out what the net force is on this object. So in order to do that, we have these forces there pointing in different directions. We have to use vector addition. So with vector addition, we always have these steps right here. First, we label and calculate these forces, which we've labeled already. So I'm going to take that over here, and then we have to decompose them, pick our directions, and add them in this all for the net force.

Alright, so let's go ahead and calculate the forces. So I got F2 and I'm going to be using Coulomb's Law. Coulomb's Law is just k times the Q1 Q2 divided by the distance between them squared. So that means that F2, which is the force between the negative two Coulombs and the three Coulombs, is going to be k which is that constant 8.99 × 10⁹. And then I've got the two charges involved. So I've got to two and three divided by the distance between them square 0.6 square. Now, some of you might be wondering why I haven't chosen that negative sign, and it's because remember, whenever we're finding the Coulomb force, we're always just going to plug in positive numbers. We're going to worry about the direction later. So if you work this out, you should get the F2 is equal to 1.5 × 10¹³. And that's in Newtons right here. Right?

So you don't need the units for that if we try to work out F1. So in order to work out F1, we need 8.99 × 10⁹. We've got the two charges, right? You've got this three Coulomb and the one Coulomb. but we don't have the center of mass or sorry. We don't have the distance between them. That little R. So hopefully you guys realize that this is a 6-8-10 triangle. Right? So this should be 10. But if you didn't, you could always just use the Pythagorean theorem. So you've got 6 squared plus 8 squared, and that equals 10 centimeters. Alright, so we want this in SI this actually should be 0.1 m. So first we have to go. We have to do three Coulombs, one Coulomb, and then divided by 0.1 square. And if you work this out, you should get 2.7 × 10¹². Alright, I'm trying, right? That's small enough just so we don't run into this little thing right here. But anyways, now we've labeled and calculated both of these forces. So we're done with that step that first step. Now we just have to go ahead and decompose this thing. Right?

So you've got these forces that are pointing in opposite directions at once pointing to the left one is pointing upwards, so we got to decompose it. So I've got to decompose this F1 vector by basically moving it and projecting it down to the X and Y axis. And I do this by needing by using my trigonometry and my cosines and things like that. Right? So, in other words, I need my F1 X components and my F1. Why components. Now I've got this. Um, I know if I wanted to break up this F1 into its X and Y components, I know I relate that using the sine and cosine. So F1 is going to be cosine F1 X is going to be F1 cosine of theta and F1. Why is going to be F1 sine of theta? So I know what these forces are all I have to do is just figure out how to get Thetas. So I use that by going ahead, looking at my triangle, and usually I'm told what the angles of these things are so I can go ahead and figure what this angle is. Or I could just use the relationship of sine and cosine in the triangle itself. What I mean by that is that I've got this angle theta right here, which, by the way, is the same as this angle theta right. These things are parallel lines. This thing is a parallel horizontal lines, so could basically complete this little triangle. And I know that this is 8 centimeters. So I've got 6-8 and I know the hypotenuse of this triangle is 10. So that means Well, let's use our sine and cosine rules. Right? The cosine of anything is just adjacent over hypotenuse. So, in other words, cosine of this angle right here is the adjacent side over the hypotenuse. That's just 6/10. And if I did the same thing for sine using so gotta I've got opposite over hypotenuse. So I've got 8/10. So, basically, instead of having to figure out the angle theta and then having to use that, which is normally what we do here, all we have to do is just replace these cosines and sines with these fractions right here that we figured out it's the same exact thing. So let's keep going. We've got F1 X is just going to be F1, which is 2.7 × 10¹², and I have to multiply by cosine of theta, which is just 6/10. So if you do that and you work it out, you should get Let's see, I've got 1.62 × 10¹². Now, the Y component is going to be very similar to 0.7 × 10 of the 12 Now, instead of 6/10 you do 8/10 for a sign. So if you work this out, you're going to get 2.16 × 10¹².

Alright, so now you've got the components. So now you have to basically just pick our X and Y sorry Are positive or negative directions. So we've been working with up into the right. Usually, that's what we picked to be positive. So let's go ahead and just stick with that up into the right is gonna be positive. Which means left and downwards gonna be negative. And now all we have to do is at all of our components together. So I'm gonna go down here and make some room for myself. So I got Let's see, that should be enough room, That little box right here. I'm going to have the F2 and F1, and I add them together to figure out what the Net Force is these are my X and Y components. Now, if you remember, F2 pointed purely along the X direction and to the left. So that means that X component is going to be negative. So that's negative 1.5 × 10¹³ and a Y component of zero. And now the F1 pointed in this direction, which means both of the components gonna be positive up into the right. So we got positive components for that. We've got 1.62 × 10¹² and then 2.16 × 10¹². So, again, this is why we don't wanna concern ourselves with the negative directions by plugging in the charges first. Because all these things are gonna change anyways, So we're gonna go ahead and pick our directions after all of that stuff. That's why this is that process is really important. And if you go ahead and add everything straight down, we're gonna get that the F net components, there's gonna be negative 1.34 × 10¹³ because it's actually one power higher. Let me just go ahead and make sure you can, As you can see, that that's going to be 13 and then this is just gonna be 2.16 × 10¹². So if I actually go back into the diagram, I can actually draw out what that vector is going to look like. It's going to look like this is going to have a high negative x component, and then it's going to have a positive Why components gonna look something like that? Because this F2 is actually really, really powerful compared to the white components or start compared to the X component of this guy. This one's actually relatively weak. Um, okay, so I've got my components all added up, So that means that the magnitude of the net Force is just going to be using the Pythagorean theorem. So I've got negative 1.34 × 10 to the squared. Plus, Now I've got 2.16 × 10¹², and then I've got to square that Let me just make sure that's written properly. So 10 of the 12 square that when you get the magnitude of F net is equal to what I got was 1.36 × 10¹³ because again, this is like one power higher. So it's actually really strong. And that's Newton's. Now the question didn't ask us to find out what the direction is, but when you know, you could do that using tangents and things like that. All right, so that's how you find the net force of these kinds of problems. Let me know if you guys have any questions