8. Centripetal Forces & Gravitation

Newton's Law of Gravity

# Where is Net Force on middle object zero?

Patrick Ford

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Alright, guys, let's check out this problem. It is a classic one in gravitation, so we're gonna work it out together. So we've got two spheres and we're gonna position them along the line like this. So I've got two spheres like this. I know what the masses are, and I know the distance between them. So I'm gonna call the left Mass the 10 kg mass mass A on the 25 1 mass. B. And I know what the distance between them is. I'm told that the distance between them is 5 m. That's little are. So the whole point of this problem is that I'm gonna position a mass somewhere along the line between them, and I know what the I'm just gonna call that mass mass. See, I don't actually know what that Massey is, but I want to figure out where I have to place it so that the gravitational forces of the Net gravitational force is equal to zero. So it means I need to find out what this distance is. I'm gonna call that are a So that's my target. Variable. Really? That's what I'm trying to solve to solve four. So I've got are a Is my target variable and I want the Net. Gravitational force Thio equal zero What We know that Newton's Law says Newton's gravitational law says that there are forces between any two objects. So there is a force from this guy, which I'm gonna call F A. C. And there's a force from that guy over there on the right, and that's fbc. And I know with the I know how to express that as the equation. And so we want these things to basically cancel out, which means that the magnitude of F A F A C needs to be the magnitude of fbc, but they have to be equal and opposite in direction. So I'm gonna actually gonna write out the gravitational forces for those We're gonna start out with Newton's Law of Gravity. So I've got G m A M C divided by R A squared. And then over here you got G M B EMC divided by r b squared where that r B is just the distance right here between, um m a r m c and N b. So that's our be all right off course. We could take a look at this equation, we could actually cancel some terms out. I noticed the G pops up in both sides and then also, the emcee also pops up in both sides. So it's actually good. We didn't need to know that the mass of that center thing in there and then the Ara is my target. Variable. Now I know what Mass A and Mass B both are. So all I need to do is just figure out what this RB is. That's my only unknown variable not told what that distance is, so I actually have to go here and figure it out. So what is the distance between this thing and M B? Well, we don't know what the individual distances are between our A and R B. We don't know what those things individually are, but we do know that the whole entire distance between the two spheres is five. So we could actually come up with another equation for this. We know that our A plus RB is equal to five. So now what happens is in this equation where we've come up with two unknowns. We've got this other equation that we're gonna use to help solve it. So if I could figure out what RB is that I can plug it back into that equation. RB is this If I move this guy over, it's just gonna be five minus r a So I'm just gonna substitute this equation back into that guy right there and now we're only gonna have one variable. So now we've got mass A divided by r A squared equals mass B divided by five minus R a squared Notice how we've gone from two unknown variables to now on Lee one and that's the one that they need to find out. So because I need to figure out what that variable is, I want to start getting everything over to one side. So I'm gonna do is I'm gonna take this expression right here and move it up to the other side on the left on the M B is gonna come down and trade places with it. So when I do that, I get five minus r a squared over a squared equals mass be over mass a. Now I actually know what these two masses are, right. Uh, this mass B was equal to 25. 25. I've got this mass A was equal to 10. So how do we get rid of this whole like squared thing? Right? We've got this, like expression here that squared you might be tempted to, like, foil it out and start like, basically multiplying everything out. But that's actually gonna be way too complicated. Notice how both of these things on the top and bottom or both squared so we can actually take the square root of both sides. So don't go. Don't don't go ahead and foil it because it's actually gonna be, like, way more complicated. So instead, we could do is basically square roots both sides of this thing, and we do that. We get that the squares cancel five minus ra over Ara. And then this is just a number, right? The square root of 25 divided by 10 is just a number, and that number is 1.58 So cool. How do we get now? This are a by itself. I've got a subtraction in the numerator, so I can't just, like, go ahead and start splitting this thing off. But what I can do is I could bring the r a over to the other side. And when I do that, I guess it makes more room. So I've got five minus r A equals 1.58 r a. So now the last step one of the last steps is I've got to move this other are a to the other side So I've got five equals and I've got 1.5 are a plus R a. So this is actually sort of like a grace common factor. This are a between both of these terms, So I've got five equals R a and then I've got one plus 1.58 and that's just 2.58 Right? So if I do five divided by 2.58 and I basically move this thing down to the other side, then I've got our A and that is just gonna be 1.94 m and that is our final answer. So this thing needs to be 1.94 m away, So let's go back to the diagram away from the left mass in order Thio workout. So this has to be 1.94 so it actually has to be a little bit closer. Because if this is 1.94 then this is gonna be This is gonna be, like, three or something like that, right? It actually has to be closer to the left Mass because the mass on the right is a little bit bigger, so it's gonna pull on it harder. Our guys, let's say for this one, let me know if you have any.

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