University of North Carolina at Chapel Hill
Free Body Diagrams - Example 4


Comments are currently disabled.

Video Transcript

welcome to our fourth and final example video on how to draw everybody diagrams. And this video will consider how two of our more complex situations work when they're combined to each other. Consider combining an inclined plane with two boxes sliding down it now from previous work that we've done. We expect that both boxes air gonna have an acceleration down the ramp of G sine theta, so we'll see if that's going to be a reality or not. So we're gonna have Box one in Box two. Let's go ahead and give these two boxes. Masses say we have M one is equal to 1 kg and m two is equal to 2 kg. It doesn't really matter, but we'll take a look at that. And now we're going to say, All right, well, I've got my picture here. I need to draw free body diagrams to see what the forces air doing. So coming down here for box one, I like to draw a little dash line through it to show the inclined plane so you can clearly see what's going on there. So we have force of gravity down. Then we have the force of the inclined plane both on box one and then for box to. We have a similar situation with the inclined plane and then force of gravity straight down and force of the incline out perpendicular to the surface. Now we know that box one is gonna push on box, too. So would push on box to we'd say, with F force of one on two, and then box to would push back with force of two on one. And now we want to be able to put all this together, have our equations of motion. So in order to do that, we say All right, I'm going to separate this in a separate F G one and F G two to be in terms of forces that are perpendicular and parallel to the inclined plane. So remember the trick to doing that? We re draw a little larger up here. The trick to doing that is we say Okay, well, I know that f g two I look at it compared to the inclined plane. I see that if this is data, this angle must also be data. And if this is f g, then here would be f g cosine theta, which is going to interact with the real be the force that causes the force of the inclined plane to push back perpendicular to the surface. And we're not moving perpendicular to the surface, so we're not worried about that. Then we have f g sine theta. Is that going to be the force pushing it down the ramp? So in order to write our equations of motion, we'll have f g one sine theta pushing, pushing box one down the ramp minus F of box to one box one, and that's got to be equal to mass box one times its acceleration. Meanwhile, we have F G two sine theta minus F of one on to its equal to massive box two times acceleration. Now we should also remember that forced due to gravity is simply mass of the object times acceleration due to gravity. So that's going to be M one g sine theta minus F 21 close M one A and M two g science theta minus F 12 ghouls em to a Now if we solve for F 12 on this side and remember that F 12 the magnitude of 12 is equal to the magnitude of 21 We've already handled direction with AR minus signs, so F one to then is equal to M two g sine theta minus m two a. I take that and plug it in. Then I have m one g sine theta minus m two g sign theta plus m two A. It's equal to mass one times acceleration. I've distributed the negative sign through. Hopefully you can see that. So now when we go to solve for acceleration, what we'll have is M one minus m two over m one minus m to all times G sign data. If you don't see the algebraic steps, I really recommend that you go through and make sure you can get to this result because what this says is we have MM one minus m to overcome one in minus m two. So this is one which means our acceleration in this situation is simply g sine theta, which is the result that we were hoping to get

Next Lectures