welcome to our first example. Video on how to draw free body diagrams in this video will consider a very common problem, which is that we have a cart which we're gonna hook up to. Usually it's a horse. I'm not very good at drawing horses, so we're just gonna We're gonna call it some generic beast of burden here. Okay, so we hook up our cart to some generic beast of burden and we're going to ask, Okay, Beast of burden, Would you please pull this cart forward and the beast of burden turns to you and says, You know what? That's not possible. It's not possible because if I pull on the cart with some force, the cards just gonna pull me back with the exact same force, which means I won't go anywhere. Now. You look at the beast of Burden and you know he's wrong, but you need to tell him why he's wrong. So that's what we're going to discuss right now. First of all, it will help if we draw a free body diagram for each of these. We have a free body diagram for the cart and we have a free body diagram for our generic beast of burden or Bob. Okay, so first of all, we know that Bob's gonna pull on the cart Force F B and the cards also gonna have forced down due to gravity and a force up due to the ground pushing on it. Meanwhile, Bob is gonna have those same why Direction forces adjusted for his own mass. And then we know that the card's gonna pull backwards on Bob. So this is force of Bob on CART and force of cart on Bob. We know those air going to be equal and opposite. Okay, So this being true, then we've already shown what the problem with Bob's logic is. Bob was thinking that the force of him on the cart and the force of the card on him would all be applied to him. But that's not true. That's not true. Because when he applies a force to the cart that appears in the free body diagram for the cart, when the cart applies a force on him, then that appears in his own free body diagram. Now it still doesn't look great because it looks like the cart will accelerate forward and he'll accelerate backwards and they'll just run into each other. Clearly, that's not what happens either, because there's a third player here. There's also the road, which Bob is pushing on in order to move forward, which means there's also a force forward due to the road. As Bob pushes backwards against it, the road will push him forward. So, uh, now we've got all of our forces worked out and we'll be able to set up our equations of motion, will be able to say F Road minus force of cart on Bob has to be equal to Bob's Mass. Times is acceleration. Meanwhile, over here, the force of Bob on the cart will result in the mass of the cart being accelerated. Now we're going to assume that they're going to accelerate at the same rate, because if they didn't, then they would move apart from each other. And that's not what we want to see in this scenario. So they both have the same acceleration, and now we want to find a way to put these together. Well, we know that the force of Bob on the cart is the same as the force of the car on Bob except for some difference in direction. Now notice. I haven't written these down in in terms of vectors. Um, now we know that this is all in the eye hat direction. Once we have to deal with forces that are dynamic moving in, the J hat will worry about that. But right now, everything that's that's moving is in the eye hat direction. And I've already worked out the direction by putting a negative sign in front of the force of the car and Bob. That means I've already accounted for the fact that it's pulling to the left. I don't need to put it in and against again a second time. So when I go to substitute, I'm going to simply say, Force of the road minus EMC Times A C is equal to M B times A B, and we can get ahead and get rid of the sub scripts on the accelerations because we know that those air the same acceleration and then what I find is that the force of the road pushing Bob forward has to be equal to the mass of the cart, plus the mass. The Bob Times acceleration. So this is the actual force that's accelerating Bob and the cart. It's the road that seems to be doing all the work. It's what's applying the force that accelerates both of them. So this was again the classic horse and CART problem where we see that in fact, in order to resolve that, we need to draw two different free body diagrams and carefully label what forces are being applied to, what object.

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## Video Transcript

welcome to our first example. Video on how to draw free body diagrams in this video will consider a very common problem, which is that we have a cart which we're gonna hook up to. Usually it's a horse. I'm not very good at drawing horses, so we're just gonna We're gonna call it some generic beast of burden here. Okay, so we hook up our cart to some generic beast of burden and we're going to ask, Okay, Beast of burden, Would you please pull this cart forward and the beast of burden turns to you and says, You know what? That's not possible. It's not possible because if I pull on the cart with some force, the cards just gonna pull me back with the exact same force, which means I won't go anywhere. Now. You look at the beast of Burden and you know he's wrong, but you need to tell him why he's wrong. So that's what we're going to discuss right now. First of all, it will help if we draw a free body diagram for each of these. We have a free body diagram for the cart and we have a free body diagram for our generic beast of burden or Bob. Okay, so first of all, we know that Bob's gonna pull on the cart Force F B and the cards also gonna have forced down due to gravity and a force up due to the ground pushing on it. Meanwhile, Bob is gonna have those same why Direction forces adjusted for his own mass. And then we know that the card's gonna pull backwards on Bob. So this is force of Bob on CART and force of cart on Bob. We know those air going to be equal and opposite. Okay, So this being true, then we've already shown what the problem with Bob's logic is. Bob was thinking that the force of him on the cart and the force of the card on him would all be applied to him. But that's not true. That's not true. Because when he applies a force to the cart that appears in the free body diagram for the cart, when the cart applies a force on him, then that appears in his own free body diagram. Now it still doesn't look great because it looks like the cart will accelerate forward and he'll accelerate backwards and they'll just run into each other. Clearly, that's not what happens either, because there's a third player here. There's also the road, which Bob is pushing on in order to move forward, which means there's also a force forward due to the road. As Bob pushes backwards against it, the road will push him forward. So, uh, now we've got all of our forces worked out and we'll be able to set up our equations of motion, will be able to say F Road minus force of cart on Bob has to be equal to Bob's Mass. Times is acceleration. Meanwhile, over here, the force of Bob on the cart will result in the mass of the cart being accelerated. Now we're going to assume that they're going to accelerate at the same rate, because if they didn't, then they would move apart from each other. And that's not what we want to see in this scenario. So they both have the same acceleration, and now we want to find a way to put these together. Well, we know that the force of Bob on the cart is the same as the force of the car on Bob except for some difference in direction. Now notice. I haven't written these down in in terms of vectors. Um, now we know that this is all in the eye hat direction. Once we have to deal with forces that are dynamic moving in, the J hat will worry about that. But right now, everything that's that's moving is in the eye hat direction. And I've already worked out the direction by putting a negative sign in front of the force of the car and Bob. That means I've already accounted for the fact that it's pulling to the left. I don't need to put it in and against again a second time. So when I go to substitute, I'm going to simply say, Force of the road minus EMC Times A C is equal to M B times A B, and we can get ahead and get rid of the sub scripts on the accelerations because we know that those air the same acceleration and then what I find is that the force of the road pushing Bob forward has to be equal to the mass of the cart, plus the mass. The Bob Times acceleration. So this is the actual force that's accelerating Bob and the cart. It's the road that seems to be doing all the work. It's what's applying the force that accelerates both of them. So this was again the classic horse and CART problem where we see that in fact, in order to resolve that, we need to draw two different free body diagrams and carefully label what forces are being applied to, what object.

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