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<Homework 11Alternative Exercise 14.114980-kg cylindrica can buoy floats vertically in salt water: The diameter of the buoy 0.800 m...

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<Homework 11Alternative Exercise 14.114980-kg cylindrica can buoy floats vertically in salt water: The diameter of the buoy 0.800 m

<Homework 11 Alternative Exercise 14.114 980-kg cylindrica can buoy floats vertically in salt water: The diameter of the buoy 0.800 m



Answers

A 950 -kg cylindrical can buoy floats vertically in salt
water. The diameter of the buoy is 0.900 $\mathrm{m} .$ Calculate the additional
distance the buoy will sink when a 70.0 -kg man stands on top of it.

Everyone, It's Alex. Be here with a question on buoyancy from chapter 12 of university physics. So the question tells us that we have a cylindrical boy which is floating in seawater at some depth, which we're gonna call H. And it tells us that a man comes along and he stands on top and it asked us, How much further will the boy sink? Well, you know how much further is going to sing? First, we're gonna have to figure out how far it's thinks originally. So we know that they're two forces acting on the boy. One is the force of gravity, which is pulling it down. But we know that the boy isn't going anywhere. So it must be another force which is pushing it back up. And in this case, that is the force of buoyancy we're gonna call FCB. Now we know from our committee's principal of the force of buoyancy is equal to the weight of the sea water displaced. So the mass of the seawater times t and we know that as usual, the force of gravity s a G is just a massive the boy. I'm gonna call capital end again times a gravitational constant g. Now I figure out the mass of the sea water displaced. We're gonna need to know the density of seawater, which he just did not regular into us by the book and the volume of sea water displaced. Now I know that the boys cylindrical, so it's gonna push out a cylindrical volume of water. And in particular, we know that the volume of a cylinder is equal to pi times its radius squared times its height h where in this case, H is the depth that the soldier is sinking into the sea and not the entire site of the cylinder. So putting that into our condition for the boy not going anywhere, which is a balance of forces. So we want the force of buoyancy to be equal to the force of gravity. Pointing this in, we should find that we have grow some seawater that ends. Do you see water times Pi r squared H, which is the volume of sea water displaced Times G is equal to capital M G and he received It teaches cancers up as often happens and we're left with an equation for H the height of the cylinder sinks in Hey, capital and the masses, the boy divided by the density of seawater times pi r squared. Now, this is the equation for the depth that the cylinder sinks in without anyone on top of it. But the question asked this What? How much further doesn't sink in when it 80 kilograms man stands on top? So we see that the only thing that actually changes in this equation over right here no way we can see. So before we had, the death is equal to the mass of the boy divided by the density of seawater times pi r squared. So we see that the only thing that's actually going to change here is mass because the masses, the boy is now gonna become heavier when a man stands on top of it and in particular, the change. It's just going to be equal to the mass of the man divided by the density of seawater Times pi r squared again. And so we actually don't even need to know the mass of the boy. We just need to know the mass of the man who stands on top of it, and that will tell us how much further is going to sink it now, playing in numbers that are given to us, you know that the mass of the man is equal to 80 kilograms. We know that the density of seawater given to us in the chapter is 1000 and 30 kilograms per cubic meter. We know that the radius is equal to 0.9 meters, which is tthe e diameter, given to us, divided by two. That's gonna give us your a point for five meters and pies, just a constant, of course, which we always know. So from this, we're going to calculate that don't h. It's about 0.1 two two meters here. I'm using three significant figures because the question gave us three significant figures for each of the constants and bring that to more normal units. That's about 12.2 centimeters. It's not very far, but that's to be expected, given the fact that the boy has quite a wide diameter and the man actually doesn't weigh that much. Thanks for listening

Okay in this question we have to find out the additional distance the boy will think when 80 kg main stand on it. So according to according to our committee's principal keep Mideast principle. Indigenous blind force. Mhm. Additional boy force is equal to additional wet Disney. And we also know that the volume is given us must buddy even density. And uh when volume is equal to area into distance and these here the is the additional distance the boy will think so With men and boy must displace additional 80 Kg butter. So so the volume is given is as Moss is given is 80 kg. So 80 kg divide by 1030 Kg per two big meter. So we get the volume is zero x 0707. You beat me there. And we also know that the volume is given us area into distance. So the distance is given as volume divide by the Yeah. So we get the additional distances Find 0707 divide by area is given to us by our scared. So the additional distance become buying 07 07 divide by 3.14 multiply by buying 450 Scared. So the additional distance becomes fine 12 two m.

Okay, so we know that boy in force is equal to I really w intensive e g road avarice, the density of the water and these Ah, bottom off the object that was converse in the water G is the acceleration of gravity. So if we do some arrangement here we have these You could f boy and divide by wrote our future and b Here's the immerse warning. So since the question was asking us how much off the boy and was above the water so we no need it wanted it was above the water. That's a view buff is equal to the total bottom off the boy minus the immerse wouldn't be okay And we know the total volume on the boy. And can you go to the heights times the area on about. OK, so that's the ages the high ace area and minus a body which is buoyant, divided by world of your G. Okay. And we also know that the bottom area things since the boy is us ceiling sunder, it means the bottom area will be go two pi r square. And since the question didn't give us the radius but you give us the time and we know that radius is evil. The diameter by about two cities that I am under a safety gear. I too It's a radius. So the bottom area will be pi r squared, which is pie Test the over two square and m minus. Your boy in force. Okay. Do you buy by road apology it is Will give us age file turns be over to square minus. Remember, boy, enforced in these cases is equal to the gravity nobody boy. So therefore, when force can be subsidy as energy Okay And Mr Massive Boy Okay. And over throw wgi As you can tell GMG come because alot so therefore we have the water. There was a path The water is age by guys de over two square minus m over road Abu Okay, so there's find out the barrier for each variable, so we know the heights wouldn't means the right passage. So we know the high seas give mass. Um 2.1 meter. Okay. And diameters. Gimenez Joe, 0.33 meter masses Givens 1 20 kilograms. And we know the density of the water. Ow is 1000 kilogram per meter cubed Okay, So you re plugging these values back into the equation. Here we have the morning off the boil there was a body. Water is a save you. Buck is equal to, um, Lucy. Okay, so 0.1 meter times by times, Joe Boy, the reason we meet her do you want to announce square and then minus 1 20 kilograms over 1000 kilograms per meter, cubed good. And this will give us the bone is above the water is about Joe point through finance. Six meter cubed. Okay. So since the question was asking us the lane off the boy that was a bottled water. Oh, now that were you hide up the boy. That was about the war, about the water. So therefore we need to find the heights. It was above the water. So that's a angel. Buff is equal to the water. And there was a bubble water and anywhere by the bottom area or the everything about it. Which is, um they say hi times be over to square. Okay. And this will give us, um zero point zero by 96 mira que over hi and times 0.33 meter over to and then square. And this will give us the high all the land over the boy. That was above worries about, um, 0.70 meter. Okay. And this is the answer for this question. Thank you.

We want to calculate the height of the buoy above the surface of the water. We can say the buoyant force would be equaling the weight MG. And we can see the density of the seawater multiplied by the volume that is submerged, multiplied by G equals G. Of course, Gs cancel out. We can then uh substitute in for the volume of the bully that's submerged underneath the surface of the water. And we can say that. Then pie R squared, modeling essentially the louis as a cylinder multiplied by L plus R minus H Plus one half, multiplied by 4/3 pi R cubed Equalling then the volume submerged. And so the density of seawater multiplied by pi R squared L plus r minus age plus one half, 4/3 pi r cubed. So essentially this is the volume of a sphere but only half. And this is going to be equal to that. Of course I am. So after some manipulation, we can say that's the density of seawater multiplied by Then, In this. My apologies. That should be capital. Are just to keep with the with the notation. So this would then be equaling two pi r squared L plus r minus H. This becomes this becomes then plus two thirds pi r cubed. Equalling M, solving them for H. After some algebraic manipulation, we find that H is going to equal L plus r minus. Uh This would be then the mass over the density of seawater -2/3 pie R cubed over pie R squared. And we can simplify even further. And after further manipulation, H will be equal to l minus here. The mass of the buoy divided by pi r squared times the density of seawater. And from that this would be then Plus five are over three. So given this, we can actually solve and say H would be equaling .600 m- than The mass of 75.0 kg, divided by pi, multiplied by point 200 m, quantity squared multiplied by the density of seawater 1027 kg per cubic meter. and this would be plus five, multiplied by .200 m, and this would be then divided by three. And we can find that then the height that is submerged, The height of the buoy that is submerged underneath the surface of the water Is equaling .352 m. That is the end of the solution. Thank you for watching.


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