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Wind Force on a Dropped Box

Patrick Ford
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Hey, guys, let's take this one out. So you got this 2 kg box, right? So there's 2 kg box like this. We're on a rooftop or something like this. We're basically gonna drop it straight down, so there's basically just gonna be some weight force once you drop it. However, there's also going to be a horizontal wind force that pushes this thing horizontally like this. We know this wind force is going to be three Newtons and we want we want to do. We want to figure out the direction of the boxes acceleration. So I've got one force that acts horizontally. Another one acts vertically, and I want to figure out what is the acceleration of this box. Well, basically, if you think about it, you get these two forces one to the right, and one down is going to produce an acceleration that acts this way like this. And I want to figure out what is the direction of this data here or what is the direction of this A. All right, So what I wanna do is first I want to draw the free body diagram. So I got this like this. I've got my weight force. This is my mg. If I go ahead and figure out this is two times 9.8 and this is actually gonna be 19.6. We also have our wind force R F W, which is to the right, and that's three Newtons and that's basically it right. There's no tensions. And because this thing is traveling in the air, there's no normal force or friction. So the next thing I have to do is that one. I would have to decompose all my two dimensional forces, but I actually don't have any, so I can just skip that step. What I want to do is one of right f equals m A in the x and y axis. So if you want to figure out the acceleration here, the direction of this acceleration. So this is data A and the way I would do this is by getting the tangent inverse of the Y components over the X component. So in order to do this in order to get a Y and a X, I'm gonna have to write f equals M A in both X and y axis, right? So in my X axis I've got f equals M a X and the y axis. I'll have f equals m a y. So now, in the X axis, the only force I have is f w. So this is f w equals m a X. So, really, this is just three equals two times a X, so X is equal to 1.5 m per second squared. And that's one of the That's one of the numbers, right? That's one basically one half of the puzzle. Now I just need to figure out what a Y is equal to. So remember that this thing is going to be pushed in the horizontal and the vertical, so you can't assume that the acceleration zero on either axis. So what are my only forces that are acting in the Y axis? I basically just have my negative mg, and that's just because I'm going to use the convention that up into the right is positive, like I usually do, So I've got negative M G equals M A Y. You'll see that the M's cancel and basically you're a Y is equal to negative G, which is just negative 9.8 m per second square. This makes sense because of the only downward force that's acting on this object is the weight force. Then you're just gonna accelerate at 9.8. That's what everything does, right? So basically, I got those two numbers, So whoops. Then I've got my negative 9.8, and now I can just go ahead and sulfur data a sofa A He's just gonna be tangent. Inverse. Now I've got my A y component, which is negative 9.8 divided by my ex components, which is 1.5. If you go ahead and work this out, you're gonna get negative 81 degrees. This negative just means that it's below the horizontal. And that's what we should expect because it points downwards like this. So that's our answer. 81 degrees below the horizontal. Let me know if you guys have any questions