The point masses m and 2m lie along the x-axis, with m at the ori...

Question

The point masses m and 2m lie along the x-axis, with m at the origin and 2m at x = L. A third point mass M is moved along the x-axis. (a) At what point is the net gravitational force on M due to the other two masses equal to zero?

Accepted Answer

Welcome back everybody. We are taking a look at a couple different particles that are all sitting on the y axis at the origin. Here, we actually have a particle with a mass of two little M. Then at some distance above, we're just going to call that distance why we have a particle that is three little M. Mass right now. We are also told that somewhere in the middle here we're going to have another particle that lies between them. Now we don't know how far away it is from either side. We just know that it has a mass of big M. Right? And we need to find the coordinate of this point in terms of why? Well, first, let me just name a variable for that distance. So I'm gonna say that from the origin, it is just a distance of D away, meaning that this distance between the three M particle and the big M particle is y minus D. Now We are actually giving a condition in which we need to find the coordinate. In terms of why we are told that the gravitational force of the 3M particle uh minus the gravitational force of the two M particle acting on M is going to be equal zero minus because the two M. Is going to be pulling the big em down and positive for three M. Because the three M. Is going to be pulling the big M upwards. So we need to find D. When this equation is satisfied. Well, let me add this gravitational force of the two M particle to the other side. This gives us that we are going to find D. When the gravitational forces of each particle are in fact equal. So let's use kepler's third law here and plug in some values for these two sides of the equation. Kepler's third law states that the total gravitational force is going to be Newton's gravitational constant times the mass of the first particle in question times the mass of the second particle in question, all divided by the distance squared between those two. So looking at our left side of the equation here, we're looking at the three M. And M. Distance between them. We said is why minus squared. Great. Now looking at the right side of the equation, Same thing here, Newton's gravitational constant times the mass of our middle particle times the mass of our bottom particle all divided by the distance between them D squared. Remember we're trying to find D. In terms of why? So we need to solve all of this for D. First and foremost, we can divide by big G. Big M and little M on both sides. This gives us three over y minus D squared is equal to two over D squared. I'm gonna divide by two on both sides are multiplied by one half on both sides. And then I'm gonna multiply this little part up to the top. Right? That's tricky to fall along. This is what this is what it works out to. We have that three halves is equal to y minus D squared over D squared, meaning that it is the entire thing squared. So now let's take the square root of both sides. And we get that y minus D. Over D. Is equal to the square root of 3/2. A lot of this is just algebra. So let's just keep going here. I'm gonna multiply by D. On both sides these days are going to cancel out which gives us that y minus D. Is equal to the square root of 3/2 D. Right? Uh adding the D to both sides. We then get that Y. Is equal to D. Times the square root of 3/2 plus one. Now remember we're trying to sell for D. Here. So I'm actually gonna divide by this term E squared over two plus one on both sides. These terms are going to cancel out and we get that D. Is equal to y over the square root of 3/2 plus one. Which when you plug into your calculator you get that D. Is 0.45 Y corresponding to our final answer choice of B. Thank you all so much for watching. Hope this video helped. We will see you all in the next one

The point masses m and 2m lie along the x-axis, with m at the origin and 2m at x = L. A third point mass M is moved along the x-axis. (a) At what point is the net gravitational force on M due to the other two masses equal to zero?

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