University of North Carolina at Chapel Hill
Free Body Diagrams - Example 2


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Video Transcript

Welcome to our second example video on free body diagrams. This video will continue. Thio consider the problem of drawing free body diagrams for multiple objects. So in this problem, we're going to go back to a similar problem that we considered in the initial video where we're looking at two boxes that are touching each other and we're going to apply some force. We'll just call it F one. So we're gonna put on push on box one with a forced F one and then box one will push on box, too. But we know that box too, will push back on box one because of Newton's third law, that for every action there is an equal and opposite reaction and we want to be able to draw a free body diagrams toe, figure out what's going on here and right out our equations of motion. So we have two boxes. We know that they both have some mass call this M one G and then forced to the floor on box one. And then we have box m two G and force of the floor unbox, too. So in the Y direction, they're not bouncing up and down on the floor. So we expect that force of the floor is gonna be equal to the weight in each situation. And then the box one has a force pushing it to the right that we're calling F one. Box two will be pushed to the right by a box one. So this will be the force of box one on box Tomb. But we know that as a reaction force box to will push back on box one. So we're going to need another vector. And this is the force of box, too. Unbox One. Okay, we know from Newton's third Law that F 21 is equal and opposite to F 12 Uh, then we can ride out our equations of motion and why it's just gonna look simply like of the floor. One minus m one G is going to be equal to zero. So this is the sum of forces equals mass times acceleration, which in the Y direction is zero. So these air wide directions, then we have, of course, the floor on box to minus the weight of box, too, is also going to be equal to zero. So both of these air zero Newton's, I should note. And then we think about the X direction. So in the X direction we have force F one minus force F 21 It's going to be equal to Mass one times acceleration and we have force of box went on Box two is equal to the mass of box two times itics its acceleration. We're going to assume that the boxes continue to touch, which means we have the condition that the accelerations will be the same. So now that we have these two equations here, we want to figure out to put them together. We see that we have f 21 and F 12 but we know that those air equal and opposite Now the big mistake that I C students make here is when they go to plug in f 12 they say, Oh, well, I need to multiply it by a negative and then plug it in here. But the way I've written this notice, I've put F one minus f 21 which means I've already chosen the direction to be to the left for F 21 which means all we have to plug in is the magnitude of F 21 In order to get this correct, we could have Britain is F one plus F 21 because this is a sum of forces in the X direction and then plugged it in with the negative sign. Personally, I prefer toe always put down the direction we'll talk about that more incoming videos. Because I've done that, we're gonna have a fairly simple problem here. We have F one minus force of box to on Box One sequel to em one. A force of box one on box two is equal to M to a So we're just going to take it and plug it in because we already have the direction listed in the negative. So it'll be minus M two a, which is equal to M one a, and we come up with that. F one is equal to mass two plus mass one times acceleration. Using the distributive property there Now notice. This makes a lot of sense. Essentially, it says we push on box one, and the reaction of the system is that it has an acceleration proportional to the some of the masses. So it's almost like we just have one great big box. That's the equivalent mass of box one plus box, too, and that makes a lot of conceptual sense. That's good to see that.

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