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Water is poured into a tank containing a 650 kg aluminum sphere (sp gr 2.7) until the cube is half covered. A.) What is the normal force on the cube at that point. ...

Question

Water is poured into a tank containing a 650 kg aluminum sphere (sp gr 2.7) until the cube is half covered. A.) What is the normal force on the cube at that point. Then a 2nd liquid (sp.gr 8.6) is poured into the tank until the normal force goes to zero. b.) What is the height of the 2nd liquid layer at that point?

Water is poured into a tank containing a 650 kg aluminum sphere (sp gr 2.7) until the cube is half covered. A.) What is the normal force on the cube at that point. Then a 2nd liquid (sp.gr 8.6) is poured into the tank until the normal force goes to zero. b.) What is the height of the 2nd liquid layer at that point?



Answers

A cube of wood having a side dimension of $20.0 \mathrm{~cm}$ and a density of $650 \mathrm{~kg} / \mathrm{m}^{3}$ floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) How much lead weight must be placed on top of the cube so that its top is just level with the water?

Discussion covers the concept of the buoyancy and the band's forces given by that entity of the liquid, into the volume of the less liquid into the gravity's and exploration cheap. So first minute to draw the free body diagram of the queue and the forces acting on the cube is the weight of the cube, and that is a cube well into the mass of the cube into the gravitational acceleration. And the radical force acting on the cube is the band's force. Okay? And we can break. The mass of the cube is equivalent to the identity of the cube, into the volume of the cube, into the gravitational axillary artery. The identity and the volume of the cube is the mass of the group. Okay. And uh the volume of the replaced water is equivalent to the cross sectional area of the cube. That is a square into the the height by which the cube is submerged in water and that is a minus H. So the volume of water replaces a Squire into a minus H. So from Newton's second law of motion, since the cuba's uh equilibrium, we can write the submission of all forces along the wind direction is zero. Or from this, the balance force is equivalent to the weight at smg. Or we can write the band's forces. The identity of the water into the volume of the replaced water. That is a square into a minus H. Times G is recovering too the mass of the cube. And that is uh the identity of the cube into the volume of the cube and the volume of the cube is a cube into the gravitational acceleration G. Now we can cancel our G on both sides and a square on both sides. So by this we get uh the value of edge is the cuba. Into the silence of the cube, into the density of water. Minister density of cube upon the density of water, african salt, The height by which the top of the Cubism of the waters officers is 20 centimetres into the density of water. That is 1000 kg per meter cubed and density of cube. That is 6 50 kg Permitir que upon the density of water. That is 1000 kilogram poor meter cube. Our NHS equivalent to seven centimetre. So this is the solution of party. I'm in the 2nd part B. The mass of the lead that is Emel uh will be given by the mass of the replaced water. And that is a cuba into the intensity of the water, into the volume of the replaced water by because of the Masaoka. And that is equivalent to a square times H. So the mass of the ladies equivalent to 1000 1000 kg for me to do into a square. And that is uh, 0.2 meters squared Into the height, and the height is 0.07 notice the mass of the lead is equivalent to 2.8 kg.

Eso This problem is another cube floating in a water question. And here, um, dimensions of our cube is 20 centimeters by 20 centimeters by 20 centimeters. And the density of the cube is 650 kg for meters cube. And our question is, when it's floating in water, uh, once the height, here s o. How much is the top of the box? Um floating away from the water surface. So that's part A. So how do we figure that out? Um, well, first, we can start out by figuring out something very important, which is the buoyant force. Sorry. Hold on. Before that, um, it's it's important to draw the force diagram. So we have the buoyant force going upwards, and then we have our gravitational force going downwards. And since this system is at equilibrium, um are buoyant force should be equal to our gravitational force. And obviously our gravitational force is just going to be equal to the mass times G and waters air mass here, but we know the density, so it's just going to be the density times the total volume. Alright, the sub tee times G. Okay, now let's figure out the buoyant force and hopefully, somewhere in here, we're going to get a variable h so that we can solve for that variable. So what is your point? Force be is equal to road W times the volume submerged, uh, times the gravitational exploration. Well, what's our volume Submerged? Um, What you look at it? We're going to have, uh, 20 centimeters time, 20 centimeters times. Whatever. Uh, this submerged distances. What is that? Well, that is just 20 centimeters minus h eso If I called 20 to 27 years s, I'm just going to write 20 centimeters as s for the clean nous, Um, in the algebra RVs is going to be s squared times. I didn't need this parentheses right here. It's going to be s squared. Times s minus h. So that is the volume submerged, uh, times the gravitational acceleration. Um, and like we said before, um, we now have agent equation. So let's go back to a Newton's second law and be right all this again Entity of water times. The volume times G is equal. The density of the cube times, the total volume, um, and our total volume. Now we know that it's s cube times, Gene, So we can see that the S squared cancels with the S, uh, squared on this side and so and you can see that the GS cancel out as well. So we get W s minus h equals grow s. And let's just try to figure out a change. So we get row rowed w s then this means that h bring it on this side, subject this side. So we have a CI equals s minus row over rows of w times s or equals s one minus row row W So that is our equation. And if we plug in the numbers you get, uh, this is 20 centimeters, so it's 2.2 m. Times one minus row was 650 kilograms per meters cube divided by 1000 kg per meters cubed. And as you can see, these canceled out and we're left with the units meters, which is exactly what we want trying to find out the height and this is equal 2.7 m or seven centimeters in part B. Ask asks us Well, if we want to make sure that the box, um is actually level with the surface of the water. How much? Um, extra mass they have. They want to put lead. So let's call this m sub l How much extra mass do you want to put in order to have the box level with the surface of the water? Well, remember we we figured out, um, you already figured out this and we can start out with the exact same thing is just that instead of m g um, instead of Roe v t times G. We're adding some extra mass in here so you can start out with B equals mm total g again. But in this case, instead of just the mass of her cube, which is Roe v. P. Or figured out that it's Row X cubed is going to be plus a mass of the lead times Geep. And then for a buoyant force, we will have the exact same thing. Um, it's just that our age is going to be zero. That's not what we're gonna be figuring out. So instead of that, our entire volume is going to be submerged, so you'll have the same thing. Road W b submerged mhm times G. But in this case the submerged is going to be the total volume or just sq eso Now we just want wanna find out ml right here. Cross out, G um and then we have road W B C s equals robe This cube on. We just figured out that piece of s is equal s cute. So we now have this and now you can see Oops. Now you can see that ends of l is going to be equal toe s cube road w minus well, which is 0.2 m cube 1000 kg meters per meter, cubed minus 6 50 kg per meters cube and like before you can see that the units cancel out. There's a good trick to make sure that you're doing everything not totally wrong is by checking the units. When you have something like this, you can see that meters cube cancel out or left with units of kilogram, which is what we want. Uh, now, if you plug in the numbers, you should get 2.8 kilograms

The problem says we have a wooden block. This is a wooden block here that's Ah, immersed in water here, up to a height from the heir to the water level is etch from the trouble lent off. The cube is s. So the area here is a little square. We have a cute So here's the water. This is about a the one a block since until the Boise forced he calls the weight of the block. That is when the equilibrium is reached so we can write a buoyancy force isn't called too. Tends to your for water water times, Aggie. Down to the bone. You loser s minor stitch Times Square, which is equal to density. Offer would would times g times s cube, SUV's William Walked Cube, then s minus h will be equal to the density over wood would divide it by the density or for water times s So from here we'll find extra wretch will be equal to ask into one minus density over wood divided by dense to your for water. Uh so will subdue the values here. So 20 one minus 6 50 divided by 1000 his density of water will give us the height is seven centimeter, so this itch will be seven centimeters. But we when the upper surface off the block level, level with the water surface, the boys, the force will be full too. W g times we be excused. Acidity the blue turns g times this cube. So then bonus. He forces a call to the mask off the lead lead PB and she plus, um, moss off the block. Times G. Then we can write. MPB will be called to lead by the gene minus moss of the book. So said St Ng Values here tells off the tent city that is a tense T offer. Water minus does to your food his time's S Cube. So then mass off. The lead will be equal to 1000 minus 6 50 times here, a zero point to cube. This will give us a Marcel's let be 2.8 kilograms

So if this is ah Cube with side, then okay, and it thoughts like this on this model X we're trying to find. So in this guess what happens is the upward Brian Force balances the downward we're equals m g. So MGI it was Be now the mass is given by a cube. Does the volume density off? Would she That's called these density one g equals the volumes of Mars, which is a square is the cross sectional area and the Haideri submerges a minus x time to roll off water Guy's got go to G So from here we will have no one overrode do equals a minus x over a equals one minus excell What a So now we can solve these and we confined that X equals Ruoff Would Ruoff water? So that's roto minus row one over rule water. Oh, that will come out to be much goodbye A so that will come out to be one minus 0.65 divided by one times 20 centimeter, 20 centimeter. So x comes out to be seven centimeter now if you add one more mass So basically, if we're adding if we're adding another Mass m Then when it floats totally submerged, what happens is mg. Thus MGI equals volume of the water displaced, which is the total in this case, Ruoff water, which is too g. So from here we will have the mass required equals a cube times density of water minus density of wood. So that would be 20 centimeter cube times go to minus 01 is one minus 0.65 gram equals 2800 gram equals 2.8 kilogram.


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