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Hint #1:Sum moments about the point the cabinet will tip.Hint #2:The cabinet tips when the moment from P is greater than the moment from the weight_...

Question

Hint #1:Sum moments about the point the cabinet will tip.Hint #2:The cabinet tips when the moment from P is greater than the moment from the weight_

Hint #1: Sum moments about the point the cabinet will tip. Hint #2: The cabinet tips when the moment from P is greater than the moment from the weight_



Answers

A $20-\mathrm{kg}$ cabinet is mounted on casters that allow it to move freely $(\mu=0)$ on the floor. If a $100-\mathrm{N}$ force is applied as shown, determine $(a)$ the acceleration of the cabinet, $(b)$ the range of values of $h$ for which the cabinet will not tip.

So in this wrong we have a block with myself from 1970 to kill a grants and it is moving to the left. The initial velocity hold 6.1 m per second and he has a force that has been apply on an angle of two degrees with the horizontal force Even make this equation see Depends on time. 20 plus giving bombs force. I want to find this below the blogger God fight sense now Does it always on moving in the horizontal axis? Only horizontal component off this force is going to provide Any impulse will have to. When we do the impulse integral we multiply body close enough to sandal Tina 32 degrees We got the vertical. What it does not provide any policy doesn't delectably changing amendments. So this matter. So we have the internal 0 to 5 off this when this form of f on. Now that here is an image conversion the equation for the foreskin, an important force here, just staters holding you nose time. Second, these are impulse intended using the implementation. We have bias and my initial velocity V minus because it's probably left plus the impulse. It's a classic. A simple going right is going to be the final moment. Reviews about are absolutely forward final velocities and you know they imposed a mass on the nation's lots of the and we find on the most papers 73 m percent.

Let's to consider the cabinet that we were calling in a previous problem. Problem was 50. Um, so we have this box here and we're worried about it tipping over. So in that case, we're just gonna have a force from the floor is just gonna be exerted at point? A. So just as this thing lifts off the ground, we only have reaction force there on DSO. We'll take moments about that. So I'm not worried about what these forces are, which I so I haven't drawn them. That call that clutter up the figure. Um, we are told in this case that h two were playing the force at the very cop here and in the horizontal direction. So data is zero and H two is Ellen, this coop, everything else is the same. And, um well, no f one don't know. In this case, we're trying to find it so we can take moments about this point and we get our moment equation here and again. This is just once of being one, and so f one. It's just the weight of the cabinet times the width over to and, uh, divided by the height of the cabinet. And that winds up being 0.3 times the weight of the Cabinet or 120 news. Now, let's see here. Well, we had the weight of the cabinet was given before it's 100 Newtons, and now we're asked to Let's see here. What's the national? What is the minimum coefficient of friction? Static friction for the cabinet not to slide with application of this magnitude. Okay, so we just no need to know that. So the force here is just the just gonna react with the weight because this forces horizontal. So the friction force is going to be the static coefficient of static friction times the weight. Um um And it has to be equal to F one, because again, we're in static equilibrium. So that tells us that if we take, we find that us just needs to be. Basically, this coefficient here is 0.3. Um, for this Thio, Uh, if it's if this, um, if this forces any greater than it will start if if this is the coefficient of friction, then if this forces any greater than it will start to slip, uh, eso again, depending on how we're doing this If we're looking for the for the coefficient of friction are looking for the force, which we called the slip, given the coefficient of friction. Now, we then asked what would be the let's see here, find the magnitude and direction of the force. Um, required to tip the cabinet is the point of application, and we chose it anywhere on the cabinet. Well, what we can see here is that we basically that this is going to stay the same and we have a force, and we want to basically give it, get as far away from point A as possible, but also perpendicular to direct the arm from point A. So the distance where we can go for maximum distance in the cabinet from point A is this opposite of this opposite corner up here. And then we want the force to be perpendicular because we want again. Therefore, we want to find the minimum force. Um, s O that would make the, you know, the projection on the projection perpendicular to the moment. Arm the maximum. Yes. So, in that case, we have this. Um So the force here, the moment from this force was going to be minus F to L to where I to find l two is this distance here which we can again figure out because just using provide agreement here and then the moment from the weight again, taking moments about this point and so we can see that. You know, F two is just mg times the with over two divided by l two, and we're gonna get to find l two and plug everything in, and we did 103 Newton's.

So to find out how long it will take those object to hit the ground, we're letting P F T equals zero, and then we're solving this equation. So let's add 16. T squared the ball signs, get this new equation, and then we'll divide both sides by 16. And yet T squared equals 70.31 to 5, and then we take the square root herbal sides. You actually get a positive and a negative value for this, but we'll just disregard the negative value. And we get about what we around into, uh, to the nearest second. So let's go about, ah, eight seconds.

Were given in this problem is ah, a couple box plots from the problem about weighing trucks. So here's the rough draft or her rough box plot of the weight in motion data. And then there's a rough plot of the static data, and we want to know, Is this sufficient to draw a conclusion about the accuracy of the weight in motions strategy of weighing trucks? So the key here is that these data are not paired up right? This maximum right here might not be the weight of the exact same truck as the maximum over here. We cannot tell from the box plot. Same with the minimum right. This lightest weighing track from weight in motion might not be the same truck that's the lightest one for static. So the key, like I said, is that they're not paired up, so it's really hard to draw conclusions here. We want to use paired data when we're trying to draw a conclusion about which, like if it's accurate, because in this scenario it completely it has to be paired up your weighing the same truck in two different ways, and so if we don't pair them, we are getting all confused and not really able to pull anything significant from the data. So what you could say to answer this is, um, bees. Rocks plucks are not sufficient for drawing any conclusions out. Um, if the weight in motion strategy is consistent, this is because the data in A to box plots is not hair. This scenario calls for Rashed chairs because in each truck, his way, with both strategies both away on wearing strategies.


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