Alright, guys. We're going to work this one out together as well. So, we have the gravitational force between a rod of mass \( m \) and length \( l \), and this mass over here. We're solving for the gravitational force, but this is a mass distribution, remember, because this is non-spherical. So, we're just going to go ahead and stick to the steps. The first thing that we're going to have to do, and by the way, the first part of this problem is just set up the integral expression. The first thing we do is just write out the formula for \( f \). Right? So we've got \( f \) is equal to, we've got \( g \times m \), and then it's going to be the integral of \( \frac{{dm}}{{r^2}} \). So boom, that's the first step. You always start with this equation.

Now for the second step, it says we're supposed to pick 2 \( DMs \) along our sort of shape or object, and so we're going to pick this \( DM \) over here. So this is going to be \( DM \), and we're going to get this one over here. This is also going to be \( DM \). Cool. So we need to figure out an expression for little \( r \), which is the center mass distance. So between this \( dm \) and this mass over here, this is \( r \). And between this mass and this mass, this is \( r \). So clearly, we can see that \( r \) is actually changing throughout the shape. Right? So, let's come up with an expression that relates the variables that are given to us in the problem. That's \( l \) and \( d \). So if you think about this, this whole entire distance over here is actually \( l + d \). But the thing is, we're not taking all of that distance, remember? Because we're actually not including this little piece right here. So, that means we're not actually including sort of that piece. And if we're looking at this \( DM \) over here, then this distance over here sort of ignores all of this distance, that I've highlighted in blue. So, it's clear that we're not actually going all the way from \( l + d \). We're going to subtract something. And because this sort of distance here is variable that we're not using, I'm going to assign it a variable, and we're going to be in the \( x \) direction. So, I'm just going to say that this is \( x \). So that means that this little \( r \) distance here is actually \( l - x \). It's the whole entire thing, minus this little piece right here.

For step 3, it says we're going to break \( DF \) into its \( x \) and \( y \) components. So that we only do that if the \( DXs \) or \( DFs \) actually have components to them. But if you take a look here, this mass is always just going to be pulled straight to the left no matter what. Right? Because everything is sort of on the axis of the \( x \) axis. Right? So this \( DF \) doesn't have any components, so that means we can kind of just get rid of this whole step here. We don't really have to do it because there are no components. And we just get \( f \) is equal to, and then \( g \times m \), and then integral of \( \frac{{dm}}{{r^2}} \). So it's just going to be the exact same thing.

Now, step 4, we're going to expand into the \( x \) and \( y \) components, but we don't have any, so we don't have to do that. Right? So now we're actually going to skip steps 3 and 4, and we're just going to go straight into step 5, which is we're actually going to plug in our expression for \( R \) into the integral, and then we're going to take out all the constants. Right? So that means that we're going to do \( f \) equals, and now we've got \( g \times m \) integral of \( \frac{{dm}}{{(l - x)^2}} \). So can we pull out any constants? Well, remember that \( l \) and \( d \) are constants. Right? These are actually just lengths that are given to us. The problem is we can't pull them out because all of these constants here are sort of enveloped in this term that involves \( x \). So this guy right here is actually the variable. And so because these two distance here are attached to this variable, I can't remove them from the integral. Right? So basically, what happens is I'm already done with that step. So there are no constants to be pulled out of the integral.

So now let's take a look at step 6. Right? Step 6 has 2 different possibilities. If we're just left with the integral of \( DM \), we're going to replace it. But as we've seen, we're not going to pull out any constants, and so we're not just left with the integral of \( DM \). We actually have this term here on the bottom. Which means that we're not going to go with 6a, we're actually going to go with 6b. Which is, otherwise, we're going to change the differential of \( DM \), right, this guy, to match the changing variable. So this right here is our differential. And because it's \( DM \), and I have a variable that's changing that's \( x \), these two things don't talk to each other. They're not the same thing. So what I have to do is, unlike the unlike we did for the ring, I actually have to match this changing variable, using the density right here. So this is something we haven't talked about just yet.

So, what happens is we're going to get this \( DM \), and we can relate this \( DM \) to the density, which is \( \lambda \), and we can write a different expression for that involving the mass and the length of the object. And by the way, you're going to see all of these expressions. This is only for 1-dimensional shapes, whereas we're going to use these equations for 2-dimensional shapes, and then these for 3-dimensional shapes. So this would be like a flat sheet or a surface or something like that. And this would actually be a three-dimensional object like a cube or a sphere or something like that. So we're dealing with just a horizontal rod, which is just one dimensional. We can imagine it's very thin. So we're actually going to use the one-dimensional expressions. So, what I need to do is I need to relate \( DM \), to the mass and the length that are given to me. And so that's actually going to be this expression right here. So \( DM \) is equal to \( \frac{{m}}{{l}} \times dx \), and then I just substitute it in for that expression.

So now, what happens is this \( f \) equation becomes \( g \times m \), and now I've got the integral of \( \frac{{m}}{{l \times dx}} \), right, because that's this piece right here, divided by \( (l - x)^2 \). So now, if you've looked if you've taken a look what have happened here, we've gotten a \( dx \) and we have this \( x \) variable. So now these two things talk to each other. They're the same sort of variable, so I can actually go ahead and carry out the integration. The last thing we have to do is we have to pull out the constants. Right? We want to be pulling out constants from the integral. So that \( m \) and \( l \), because they're capital letters, can be pulled out. So we've not got that \( f \) is equal to, and we've got \( g \times m \times m \), over \( l \), and now we've got the integral of \( \frac{{dx}}{{(l + d - x)^2}} \).

Okay? So, that is step 6b, and we're done with this. So, now the last two things we need to do are we just need to determine the limits of integration. So, that's based on the actual shape of the objects. So what happens is as we're integrating along the \( x \) direction because we integrate this way, we need to go from some starting point to some final point. The starting point is actually going to be right here. So, what I'm going to do is I'm going to call that \( x = 0 \). Right? We can sort of choose that to be the 0 points. Now, we get to the end of the rod and that's going to be over here. Now, that distance is \( x = l \), so that's going to be right here. So that's actually where we're integrating from, \( 0 \) to \( l \). So those are going to be our limits of integration. So that means that \( f \) is equal to and now I'm just going to reverse \( m \) and \( m \). So I've got \( g \times m \times m \), over \( l \), and now the integral of \( 0 \) to \( l \), and we've got \( \frac{{dx}}{{ (l + d - x)^2 }} \).

Okay. So if your professor only wants you to set up the integrals, sometimes you might just have to only do this in problems, then we're done here. This is the answer. This is the integral expression with the correct limits and all of that good stuff. Right? So this is the expression. So if your professor actually wants you to carry out the integral, then stick around because we're actually going to do that in part b. But if you only are asked to set up the integral, then this is the answer.

So in part b now, we're actually going to evaluate the integral, which is step 8. So we're going to use integration techniques to integrate this expression. Now, one thing we can do is I've got an \( X \) over here in the bottom, and I've got a \( DX \) that's going to be on top, but I have this whole expression right here. So one thing I can do is I can just use a \( u \) substitution. So just think back to \( u \) substitutions. We're just going to call \( u \) this whole entire expression right here, \( l + d - x \). And so we take the derivative with respect to to to \( x \), and with these two things are constants, the \( l \) and the \( d \). So that means that those go away or they evaluate to 0. And when I take the derivative, I just get negative \( dx \). Right? It only affects this piece right here. So, I don't want \( du \). I actually want \( dx \) because I have this \( dx \) right here. So one of the things I can do is say that negative \( du \) is equal to \( dx \). So this is the reason this is important is because every time I see \( l - x \) in this expression, which, which is over here, I'm going to substitute with \( u \). Every time I see \( dx \), \( x \), I'm going to substitute that as well. So what happens is this piece right here gets substituted, and this piece right here gets replaced as well. So what our new integral becomes is our new integral becomes \( f = g \times m \times m \), over \( l \), and now we've got the integral of \( du \) divided by and now we've got oh, or sorry. Negative \( du \). Right? So don't forget the minus sign. And then we've got \( u^2 \). Okay? That's on the bottom right here. So, this integral can actually be written in a different way, because this integral looks a little nasty. So remember we can actually sort of evaluate this or we can rewrite this as, negative \( u \) to the minus 2 power times \( du \). Remember, the minus sign just means that it goes in the bottom exponents. Right? Or the the denominator flips.

So how do we solve these integrals? Well, remember, if this is just a normal integral, so I'm going to scroll down. The integral of \( u \) to the \( n \) power \( du \) is just equal to we raise the power by 1 and then divide by that new exponent. So that means that this integral here \( f \), which is \( g \times m \times m \), over \( l \), is going to be, let's see. So when we evaluate this integral, we're going to raise the power by 1, so it's going to be negative \( u \) to the negative one power. And now we have to divide by the new exponent, so we're going to divide by negative one. Well, hopefully, you guys should see these negatives cancel, so we're going to cancel out those, and we just get \( u \) to the negative one power. So in other words, what happens is we just end up with \( g \times m \times m \), over \( l \), times \( u \). The problem is we still have to evaluate this at the limits, but what we have to do is we have to change this. So we have to turn this back into \( x \). So, let's see. We what we actually get is we get \( g \), big \( m \), little \( m \), over \( l \), and then we get \( l + d - x \). And there's only one power. And now we actually evaluated the limits, which is \( x = 0 \) to \( l \). So for our final answer, sort of right down here, what we get is we get \( g \times m \times m \), over \( l \), and now parenthesis, one over when we evaluate this top limit over here, we're going to get \( l + d - l \). And then we're going to get minus one over \( l + d - 0 \). Alright. So what happens is these \( l's \) will cancel out over here, and then we're just left with this expression. So that means that for your final answer, your force is equal to \( G \times M \times M \), over \( L \), and now we have one over \( d \) minus one over \( l + d \). And that's actually the whole entire integral. So that's that part b. Alright, guys. That's how you solve these problems. Let me know if you guys have any questions.