Answers: 1. Wooden block tips down. 2. 20π m longer. Solutions by Extreme Cases: 1. Imagine that instead of a wooden block and a lead block, you have nothing on one side of the scale, and a balloon full of helium attached to the other side. With air in the chamber, the side with the helium balloon attached will rise as the helium balloon pulls up on it - this is because the air is providing a buoyant force on the balloon. Once all the air is removed from the chamber, the side with the helium balloon falls - the lack of buoyant force from air makes it evident that the side with the helium balloon has more mass than the side with nothing. 2. Imagine that the Earth was a single point with a radius of 0m, and therefore a circumference of 0m. With the telephone poles, the circumference would then be 2*10m*π = 20π m, and so the difference is always 20π m. Solutions by Principles: 1. Here are the facts: a) Air is a fluid, and so we can apply Archimede's principle to this situation of blocks in a chamber full of air. b) Archimede's principle states that the buoyant force applied on an object in a fluid is equal to the weight of the volume of fluid displaced by that object. c) We can conclude that the wooden block, having a greater volume, displaces more air than the lead block, and so there is a greater buoyant force on the wooden block pushing it upwards. d) When the air is removed from the chamber, there is no longer that greater buoyant force pushing the wooden block upwards and so the wooden block tips down the scale. 2. If we let r = radius of the Earth, we can solve this by simple math: Difference in circumference = Circumference of telephone wires - Circumference of Earth = 2π(r+10) - 2πr = 2πr + 20π - 2πr = 20π Therefore, the telephone wires will have a circumference that is 20π meters longer than the circumference of the Earth.
+Rizwan Awan (TheOnlyRizzy) You have the best explanation. Your comparison with a helium balloon gives the best idea. The balloon wants to go up because the weight in that volume is less than that from air. Remove the air and the balloon will fall down. The wooden block has a larger volume. Remove the air and the wooden block looses the advantage of the larger volume.
+Rizwan Awan (TheOnlyRizzy) really a great explanation. Very interesting to see the difference between (i) extreme and (ii) analytical approach in action.
+Rizwan Awan (TheOnlyRizzy) alternative scenario, even easier: imagine the scales have one identical box standing on each arm. the wooden block completely fills box 1. the lead block doesn't fill box 2, so box 2 has the lead block + some air in it. both boxes and their contents cause the scale to balance out. (here we can see that actually, the blocks themselves do not have the same mass to begin with, the wooden block is heavier!...and the point is that you are weighing more air on one side) when you evacuate, you remove that air from box 2, but you remove nothing from box 1. so you took mass away from box 2.
+Science4Fun Thanks! I think this was "Physics Girls" whole point - to show how both approaches come to the same conclusion and compliment one another :)
OK. Here's another one: Two guys are in a boat with three cigars. No matches, how will they smoke their cigars? Answer: They throw one cigar overboard, that makes the boat a cigar lighter. Simple....
But then their boat is now a cigar lighter which, as you can imagine, is not sufficient to support the weight of 2 men, and they are now floundering in the open water.
If the rock is more dense then water then the water level will go down because the displacement is pure the volume of the rock not the volume equivalent in what to the weight of the rock
+mike more What about if the object is very heavy but less dense than water? wouldn't it still make the water decrease after than when the object was on the boat? I think so because the weight of the object makes the boat decrease which replaces the molecules and then makes the water level higher. when the object is in the water, the weight on the boat decreases which makes the water level decrease.
AcEDndJhard if the object is heavy and still less dense than the water, than it must have a bigger volume to compensate (remember, the density is the ratio between the volume and the weight). Which defeats the purpose of the extreme-case chosen. And, it will make the water decrease when it left the boat, but it would make the water higher again when it would enter the water and float.
My guess is it depends on the density of the rock. When it's in the boat, it displaces water based on its weight. When it's under water, it displaces water based on its volume.
You already know the density, she said it will sink! This means the lowest density the object can have, is high enough for this to work. As long the density is higher than that of water, the weight will displace more than the size.
A floating object displaces it's mass in water, a sunken object displaces it's volume of water. Since the rock ways more then its volume in water, when it is on the boat it displaces more water then it would if it had sunk. So the water level will decrease. Edit: Huzzah! I was correct, however, you cannot make this argument for any rock that starts in the boat. There bound to be rocks that are porous and less dense then water.
but the water level will never rise because the rock can only ever displace the same amount of water whether in the boat or tossed into the water, if it is less than the density of water. your first sentence still holds true 100% of the time, because if you throw a rock over board that sinks, as the riddle suggests, then that rock was denser than what you were floating in
Devin Royce You are correct in the essence of that once the system has settled that will be the outcome. However a rock that weighs less then its volume in water will still 'sink' temporarily if it is thrown into the water, during that time before the system reaches equilibrium the water level will have risen. yes there would be some force required to keep the rock at the bottom, but the original question was just a friendly way out asking whether it displaces more on the boat, or submerged.
3:33 either they have the same mass or they are balanced. They cant be both at the same time due to the air pushing them up with different forces. If they have the same mass the gravitational force is the same, but because the wooden block suppresses more air then the lead one, it got a bigger force pointing upwards- the lead block sinks. If they are balanced they still feel the same upward-force as above. For the resulting forces to be equal, the wooden block needs to have a higher mass then the lead one. In air they cant have the same mass and be balanced (at different sizes) I know what you meant but it is a wrong formulation.
If you forget that statement about them having the same mass and just focus on them being in balance. then when air is removed the balance tips towards wooden block (because it has to be heavier to counteract bigger buyoant force).
Leeengold exactly , the mass balance doesn't actually equates the masses it equates the moment of gravitational forces so the wooden block has to be of greater mass initially to be balanced . when the air is sucked out wooden block will fall because there is no buyoncy and it has greater mass then iron block . which implies she told us wrong question .
The question is impossible to answer based on a flaw in the wording of the question itself. You will never remove all of the air. Perfect vacuum is not obtainable. The highest vacuum man has achieved I believe is around 1E-12 Pascal or 7E-15 Torr. Even space is not a perfect vacuum. That is not to say the outcome will not be the same theoretically, just pointing it out, it's a bad question to begin with the way it's asked.
+Tymon0000 Yeah, that gives the exact answer. I thought of this more from an engineering perspective, where you could argue that the radius of the planet is much, much larger than 10m. So a more accurate extreme case is that the planet's radius is practically infinite compared to the poles, so the circumference of this wire circle essentially 2*PI*R, where R is the radius of the planet. So the answer to the question, "how much longer is the circumference of the wire than the circumference of the planet", is 0.
+Brandon Bocklund It's constant... If you add 10m poles, you end up with a circumference that has 10 extra m of radius. So if your circle (earth of any other circle) has radius R, and the wire has radius R+10, then circumference is (2*Pi*Radius), so the difference will be: Diff = 2 * Pi * ( R+10m ) - 2 * Pi * R Diff = 2 * Pi * 10m + 2 * Pi * R - 2 * Pi * R Diff = 2 * Pi * 10m Diff = 62.83m So, the circumference of the wire circle will be 62.83 m longer than the circumference of the earth.
Right, but lets say we're looking at Pluto, which isn't even a planet anymore. Pluto has a radius of ~1 million meters. So the wire has a circumference of 1 million meters + 62 meters which is pretty much just 1 million meters, so the difference between them is zero in a practical sense.
+Brandon Bocklund So how long would the poles need to be, for you to say, that the circumference would be considered longer? Let's consider the Pluto's radius of ~1 million meters.
Problem 1: Wood down, lead up. This one is tricky and I haven't seen a good explanation in the comments yet, so I'll see if I can get it. First of all we have to assume that the wood does not outgas at all when the atmosphere is removed, if id did this would mean that the mass of the wood would change. But since this video is talking about buoyancy I think that's more what the question is driving at. So, the block should tip towards the wood. Wood down, lead up. The reason for this is buoyancy. The atmosphere is pressing in on the wood over a greater surface are and is thus creating an upward force that is larger than the upward force on the lead block, but of course everything is balanced. When the air is removed it can no longer provide this force and the small difference in mass becomes apparent. Problem 2: 20Pi meters longer. But others have mentioned this already.
+axelasdf If that were true a helium filled balloon would fall like any other object. Why does a helium balloon get pushed up? The helium balloon has the same amount of pressure on all sides, and yet it floats up. The thing is that while the forces are balanced for each side of an object, the surface area for the balloon is huge compared to it's weight this means it would have to move many more particles of air out of the way below it to fall down than you or a block of wood. Since gravity is acting on all the air and on the balloon the weight of the air (pressure from the atmosphere) pushes up against anything moving within it. Only objects that have a high density (small surface area) can overcome this force and fall through the air. If I'm not making sense, try veritasium's video about buoyancy.
+YouCanScienceIt The forces only cancel out sideways. But more force is pushing up than down, because the air pressure decreases when you are higher. Even the few cm of the height of the cubes makes a difference. If you do the math, it only comes down to the density of the fluid (air/water/...) and the volume, in other words, the displaced weight. (Weight assumes gravity; alternatively you could say mass + gravity, which is the same thing as weight.)
If the density of the object thrown out is greater than the density of the fluid that the boat is floating in, then the total volume of displaced water is greater when the object is in the water. Otherwise, the total volume of displaced fluid remains the same. That is, an object that floats make no change to the water level when it's in the fluid; the level rises when an object that sinks is put into the fluid. This is a lot easier to see with a few simple density and volume equations.
This was way more didactic than I thought it would be, thanks a lot for that! My little sister could understand everything just fine, and got really excited about those problems at the end.
+Physics Girl i just tried this and you are wrong in the video. if the rock is very light then the water stays about the same when its in the boat and it rises slightly when you throw it in the water. i think you went the wrong way with your extreme.if the rock is so light that it is almost bouyant, then the water is slightly higher when the rock is submerged than when it is in the boat, so, you are wrong. it's very upsetting that you teach people incorrect things.
+Killer Apple You should be much more careful about what you post (i.e. don't outright say that she teaches incorrect things, and say that's she's wrong). Instead, explain how you conducted your experiment (i.e. posting a video about it would be great and informative), and ask for an explanation. Especially since you saw in the video that she did the experiment as well. You should also clarify that the water level you're referring to is with respect to the boat or to the container. Based on your logic "if the rock is very light then the water level stays about the same", then removing the rock would also have no impact (see phrontdoor's explanation). Her extreme example was well done. I.e. imagine you're in a small boat that has some concrete blocks in it (the blocks are much denser than water). If you take it out, then the water level will drop since there's less water being displaced. Putting the block back in the water only pushes up slightly. The water level drop is greater than water level rises. With regards to your second comment: "if the rock is so light that it is almost boyant, then the water is slightly higher when the rock is submerged than when it is in the boat," Let's assume that by almost buoyant, you mean that the rock has the density of water (and is almost completely submerged when dropped in a body of water). When the rock is on the boat, the rock displaces its own weight in the water. Then when the rock is temporarily lifted, the water level drops, equal to the weight of the rock being lifted. When the rock is put back in the water, then the rock again has to the displace the its own weight in the water. This rise and fall is nearly equal. Plus, If Physics girl was incorrect, I highly doubt nobody would have picked up on it already. (whether from the comments below or from the producers of the show)
Best one since the Chladni plates video. w.r.t the circumference problem, I found it easier to do 2π(r+10) - 2πr than think through the extreme case. But, thinking of the extreme case gave me a different perspective on the problem.
+Anirudh Surendranath You can do that too! But then take the extreme case to be r = 0 (is that what you did?) And the answer becomes immediately clear. :)
+Anirudh Surendranath I even did it with "l" as an arbitrary length instead of 10 m. I somehow have an aversion against doing formula manipulation with values in it and always try to do the general case first. That's what classical science education does to you. Sometimes when I'm stuck I try to imagine stuff with actual numbers, but it may occasionally lead you astray, just as measuring things in drawings can.
+Physics Girl Unfofrtunately for some reason, extreme cases make physics much more difficult for me to grasp than regular formulas and explanations (not so much true for math for some reason), so really seeing "2pi*(r+10)- 2pi*r" just somehow makes more sense to me than "imagine r=0".... or maybe they feel like more of a proof to me? I just don't see how r=0 makes it immediately clear what the answer is. Then again physics were always somehow hard for me to grasp (probably because of the way it's explained?) despite the fact I'm pretty good at math and study chemistry.
pokestep I feel the same way most of the time. And unless you've done the formula, how can you be sure that the answer for r=0 is the general answer for all cases?
pinkdispatcher Exactly! Mathematical proof is the way to go for me. Except for, if you think of r=0 as the extreme case, then it's "easy" to see the difference (but that's again the mathematical proof so I'm not sure if that's what is supposed to make it "immediately clear") because you automatically see the difference is 20pi... I dunno. In general I don't like how physicists explain things and usually act condescending to others (at least the ones I encountered) so physics really isn't the way to go for me... We did have a great physics teacher at uni tho.
I think that pumice stone sinks as well; if not you can substitute it with a marble stone with many holes. I think that defining density is definitely a mandatory issue before you can solve the riddle
For Problem 1, I guess you can assume the wooden block is pushed up by air so when the air is gone, it'll push down more. So the wooden block is heavier.
You forget that air is also a fluid. If the wooden block is less dense it has bigger volume so the boyant force acting on it, is greater. When the air is removed there is no boyancy. That means that the scale will tilt towards the wooden block and the led will appear to be less heavy. And that is the truth. Mwooden> Mled . Sorry for my bad English.
The buoyant force acting at the bottom of both plates are equal and when we take out the air, they will have no impact on the balance. However the weight on the video right plate is the weight of the smaller block plus the weight of the air colum above this block plus the weight of the air collums around the block and in the plate. For the left plate, it's the same situation, but the air collums are smaller, since the volume of the wooden block is bigger, leaving less space for as around and above it. Then, with air, the weight of the wooden block plus the weight of the smaller collums is equal to the weight of the other block plus the weight of the bigger collums. So, the wooden block is heavier and will bring the plate down without the air to balance it. We can also use the extreme case, where the wooden block is the size of the plate and extremely high and the other block is extremely small. The solution is the same.
***** No... do you even physics... the boat floats because of the shap and material. The boat floats because it has the round shape, and it's made of fiber glass which evenly displaces the water so that it doesn't tip. Even if the rock was perfectly round it wouldn't float.
Kibbis Kibertus A boat floats because the overall density of the air and the body of the boat is lower than the density of water. In this case buoyant force compensates the weight so it remains on the surface of the water rather than going up or down. FYI, the rock boat would float if the body was thin enough.
Using extreme cases is extremely useful. In almost every case ever, in every field, everywhere. Lots of people seem to really not like it though. When they propose an idea and you immediately consider how it would work in the extreme case, or ask a question about the extreme case, they get displeased. It often leads to, at least, nailing down a problem very precisely, and they seem to not want to be so 'strict' about things usually. It prevents them from thinking effectively in a lot of cases. I'm not sure if it's because it often reveals much more complexity to an issue than they presume it to have or if they're not confident in their ability to nail things down like that or what, but maybe if the benefits of 'thinking in the extreme' were more often talked about it would help. Thanks for your video!
Great video. One comment: If the rock in the boat had the mass of a black hole, then the boat would sink to the bottom, which means that we have to modify the original statement that the boat is "floating" in the water.
+Eugene Khutoryansky There are also tiny black holes; they don't have to be huge. Tiny ones don't live long (look up Hawking radiation), but as tiny a mass as that of a single proton could theoretically become a black hole.
+Eugene Khutoryansky The real problem is that the black hole rock would be so dense that, unless the boat is infinitely strong, it would punch right through it, probably causing the boat to sink.
pinkdispatcher Yeah. For the purposes of an extreme case thought experiment, it works fine. But I found some amusement in working out the consequences of actually running such an experiment. :)
I have yet to watch the answer, but I've thought this through and perhaps my explanation is different than the Physics Girl's and may help some people understand this problem better. So my thinking is that when you throw the rock off, the water it displaces should be equal to V, the volume. This would make it seem the water level would go up, however you must remember that the boat now weighs less by mg (m is the mass of the rock). This means the buoyant force must be mg less than it was before and the boat will rise, displacing mg less water. So then the water will rise if the amount of water the rock displaces is more than the amount of water replaced by the boat rising--meaning V>mg. Otherwise it will stay the same if V=mg or sink if V 1 (simplified to density*g) then the water will sink. As long as the rock sinks, you know the density is greater than 1 and the water level will sink. EDIT looks like the answer is right, but I feel like something is wrong with my work. maybe it should just be m/V > 1, if anyone spots a mistake plz let me know!
The answer to the planet-pole puzzle: Let radius of planet = r To find : outer circumference - circumference of planet = 2pi(r+10) - 2pi(r) = 2pi(r+10 - r) = 2pi(10) = 20pi
It was on the boat, is my first clue, so now it's in the water: water level stays the same, displacement. Water cannot rise since the weight of the rock on the boat was already in effect. It only moved from the boat into the water. Resume play.
1) the problem is impossible given that the two blocks are of equal mass. Ignoring this assumption, and instead assuming that the masses are balanced under pressure, then evacuating the air will cause the wooden block to move down. The wooden block must have a greater mass in order for its weight to counteract the greater buoyant force it experiences. 2) the circumference will increase by 20pi. 2pi(r + 10) - 2pi(r) = 20pi.
For all of u wondering what would happen if rock was bigger/smaller with greater/less mass i have writed the equations and the rise or fall of water level depends ONLY on its mass and density
Here is how it looks like: (Mk * R - Mk * r)/R*r where Mk is mass of rock R its density and r density of water. So as long as density of rock is bigger then waters the water lvl will drop
if the rck i less dense than water, it wont sink, so wouldnt it be incapable of displacing more water than it had while it was in the boat? actually the rock would now be another boat correct?
But what if the rock is ultra light, but very voluminous, like a plastic hollow ball? Wouldn't the water level rise? So ultimately the question depends on the density of the rock.
+poketopa1234 Yes this is 100% true, but she said in the question that the rock sinks, so you know its density is > water which means the water level must lower.
+poketopa1234 I thought about the exact same thing right after her explaination. I've been looking but still can't find comments that can explain how I am wrong. I understand the sinking/floating thing but i'm not talk about that. I'm still talking about a sinking rock. She used the extreme case of tiny/high density rock. What about a huge/less dense (not than water) rock, with a very buoyant boat? Say the rock is the size of the bath tub, but the boat is buoyant enough to carry it right over the water. If the rock is dropped into the water, obviously the water level will increase. Someone please explain to me, thanks.
w I l l I a m I think she just omitted that senario. The general rule she made only applied to the first question. If you want more info, just look up archimedes principle.
Two guys are in a boat with three cigarettes. They had no fire, so they threw one of the cigarettes overboard, and the entire boat became a cigarette lighter.
because if diameter increases in 20 m, radius increases in 10m, and each new meter will multiply the other two variables on the equation C= 2 x pi x r(C beeing circumference and r beeing radius) thus the increase in the sise will be 10x pi x2 or 20 pi (pi = 3,1415)
No...the value of the 'rock' was 'already there' as a displacement-factor....changing the position of the rock will not alter it's original 'displacement-value' in relationship to the water.This question is a 'closed-loop' scenario, and all the factors were present BEFORE the rock was tossed over!It does mean the boat will 'rise' in exactly co-equal value to the no longer present 'rock' IN the boat itself...yet the water displacement remains at the same value, neither rising nor falling.This is an exercise in 'logic and proportion' rather than 'shifting paradigm' values of the body of water itself.
You forget that the rock is denser than the water. So in order to support the rock's weight far more water needs to be displaced than if the rock were in the water. That's why the levels go down.
TheMegoShow Read what I wrote again. It's the entirety of the displacement under consideration...not 'weight'. The volume of displacement would still be the same, regardless of the density of the matter displacing a given volume of water. A 'ping-pong' ball would displace the same amount as a ball of steel of the same dimensions.
you can do this from first principles by balancing the forces. Since in both scenarios (1. rock in the boat 2. rock in the water's bottom) nothing is accelerating, the forces of gravity and forces of buoyancy must balance. "W" is weight, "V" is volume, "D" is density, "F" is force. Assume D_rock > D_water = 1 (kg/L) Scenario 1: Eq 1: W_boat_with_rock = W_displaced_water ....... so-> W_boat + D_rock * V_rock = D_water * V_displaced_water_of_boat_with_rock Scenario 2 gives us two equations, since the boat and the rock have to be stationary but here they are separated: Eq 2: W_boat = D_water * V_displaced_water_boat_alone Eq 3: D_rock * V_rock = D_water * V_rock + F_acting_on_rock_by_bottom_of_pool Now, a few easy manipulations of these three equations (I'll skip) and you get: Eq 4: V_displaced_water_of_boat_with_rock - (F_acting_on_rock_by_bottom_of_pool / D_water) = V_displaced_water_boat_alone + V_rock ... and because the (FORCE_acting_on_rock... / D_water) must be a positive quantity we modify Eq 4 to an inequality: V_displaced_water_of_boat_with_rock > V_displaced_water_boat_alone + V_rock So water level goes down in scenario 2
Hey +Physics Girl, Your Problem 1 (3:30) is confusing. It does not, of course, follow from the fact that these blocks are balanced that they have the same mass. That's whole point of the puzzle, right? If they do have the same mass, then, when measured in a vacuum, they should be balanced _assuming_ that both arms of the scale are of equal length. However, you imply that they are balanced when measured in air. So, given that they are balanced when measured in air and that they have the same mass, we can only conclude that the arms are not of equal length. You never mentioned that they were of equal length. The arm holding the wooden block must be longer to account for its greater buoyancy. Remove the air and the buoyancy is lost so the longer arm exerts more torque thus the wooden block drops. I don't suppose that that was the intended solution. Scales aren't meant to be defective. However, the information given forces this conclusion. I guess you'd intended to have decent scales with arms of equal length. If the arms are of equal length and the scale is balanced in air, the wooden block must have more mass to counter the difference in buoyancy. Again, the wooden block will drop once the air is removed.
I don't think it'll involve arms of the balance. The wooden block will have air in its pores in normal conditions, and when all air is sucked out, its mass will become less.
Ahmed Nauman Tariq It _shouldn't_ have involved them and wouldn't have if the problem had been worded better. We should have been given that the arms were the same length rather than the blocks being the same mass. Imagine deforming the metal block into the shape of one great big pore. What difference would you now expect to observe? Yeah, nah, it's not about pores but buoyancy. Another way to think about the problem is to consider the difference in air pressure between the top and bottom of the block. There is greater pressure at the bottom than at the top. The pressure at the bottom pushes up and that at the top pushes down. The consequent difference in force is proportional to volume. Thus the wooden block is more buoyant. Taking away the air nullifies this.
actually since the iron block barycenter is closer to the the earth than the wooden cube, wouldn't the gravitational force be higher on the iron block ? which means that it weighs more: (F = m * a) if m is the same and a is bigger, F is bigger. which means the balance should tip to the iron side, provided we are in a perfect vacuum with no friction etc etc EDIT: ah, she actually says that the cubes "are balance so they have the same mass" that's a contraddiction. I had missed that
Remember: both cubes are _inside_ the medium (air). Buoyancy doesn't apply here, since air pressure happens at all sides: up, down, and sideways; they effectively negate each other. In the end, the scale will stay balanced.
I thought about it another way. when the rock is on the boat, water exerts enough force to keep it from sinking. when you throw it in the water water fails to keep the rock over sea level, thus water is producing less force than before and that can only happen if the sea level goes down (archimedes)
problem 1: imagine a really small and dense block on the right side and a balloon filled with air on the other. now for the scale to stay balanced the balloon itself (without the air inside it) should weigh as much as the block on the right side, because the weight of the air inside it is held up by the pressure of the air around the balloon. now if you create a vacuum, the left side will tip down because now the weight of the air inside the balloon is added to it. you could also solve this problem using the case of an extremely dense air (or even water) problem 2: imagine a tiny planet the size of a grain of sand. the circumference of the circle that the telephone poles make is Radius*2*3.14=62.8 meters. i'm a master of extreme cases XD remember if you make the problem harder, then do the exact opposite. for example i had a problem with the planet being almost infinitely large. then the surface of the planet will be flat and therefore circumference=infinite so i imagined an infinitely small planet instead.
But what about 2 dense objects? I thought that the bigger block would have more air pressure due to it's larger surface. Once the air pressure is gone it will tip to the right. At least that was my reasoning.
zeroZyra yes, both ways the bigger block will tip down after you create a vacuum. the good thing about using extreme cases is that there are several factors that you can change and take to an extreme to find your answer
+MilesCrlsn No. In the problem stated, it is _not_ given that either block is porous or that it contains its own internal volume of air, simply that they are of equal mass but unequal density. Since they are of equal mass, _theoretically_, removing the air from the chamber will have no effect. Why? Because the air pressure is uniform (if you ignore gravity) on a stationary object. Since they're of equal mass, there is no net force and thus no net acceleration (assuming starting from stationary), and thus the air pressure is uniform over the surface of the two objects (neither has a net influence caused by the air). Now in _reality_, because the objects have weight and are not in freefall, gravity is relevant here. Specifically, Newton's law of universal gravitation that states that (paraphrased slightly for simplicity's sake) gravity is proportional to mass, and inversely proportional to the square of the distance between two bodies. Therefore, the air itself will have a very slightly lower density along the top of the wooden block than along the bottom, giving a net upwards force causing it to ride ever so slightly higher (practically probably hundredths of femtometres). When the air is evacuated, that effect no longer exists and thus the larger object will fall a practically immeasurable, imperceptible distance.
Kill3rCat being porous doesn't change anything. (as long as the air inside them is considered a part of them. i didn't remove the air from inside the balloon in my example.)the bigger object is affected more by the pressure of the air around it and the smaller object is less afected. so after you remove the air from around the objects the bigger one measures heavier. also as i mentioned, using the case of 2 objects under water (extremely dense air) it's obvious that the bigger objects with the same weight is going to possibly float (negative weight) and after removing the water, both objects are going to be the same weight.
MilesCrlsn I derped. I said 'porous' when I meant 'hollow'. And in the situation, it is not given that either object has an internal volume of air. There is only 'air inside them' in your example, which is wholly irrelevant to the situation described in the video.
+Physics Girl I like it actually because they're always so stumped :P its good to know smart people can learn something every day too!! PS. i LOVE all your videos :) ive showed them to so many people and they do too :D
There are 3 types of rock, rocks that float, rocks with density equal to the water and rocks with density greater than water. Case 1, When the density of the rock is greater than water. The rock sinks and the floor supports the weight of the rock. The water level sinks because the delta volume displaced by the boat is greater than the volume of the rock. Case 2, the density of the rock is equal or less than the density of water. The water level stays the same because the delta volume displaced by the boat is equal to the volume of the portion of the rock that stays underwater.If the rock makes it to the floor, the floor will not provide any support of the rocks weight.
pausing at 1:08 it could be either way, depending on what you mean by "water level". If you're talking about the level of the water vs the boat . . . then it would lower; that is to say: the level of the water would be lower on the boat, as there is less weight in the boat & the boat would have less mass beneath the surface (thus lower -- a.k.a. more boat above). If you're talking about the level of the water after throwing it in, vs before throwing it in the water level would be higher. . . as you're not removing mass from the water, so that lump on the bottom displaces the water that was there & that water would need to rise up . . . so the water to the edge of the tank would rise -- if it were not for the same displacement on the boat reducing mass & increasing displacement proportionately to the boat's total mass.
yeah, video is a bit misleading because it does not ask about the waterline on the boat, it asks about the water level, but then proceeds to "prove" the water level goes down (on the boat) which is not the water level in the tank.
But what if instead of a small but very heavy rock, you had a large but light rock. then the water level would go up. p.s - excuse me if i get this wrong i'm 12
To tack onto what rodrigo said, the issue here - which the graphic doesn't actually show correctly - is that displacement caused by a floating object is based on the mass, not volume, of the object. For anything denser than water (meaning it would sink), its mass (in kilograms) is larger than its volume (in cubic meters), and hence once it sinks it now only displaces its smaller volume, not its larger mass, in water. For anything where its displacement is less than its volume, it would have to be less dense than water, and hence float.
TheMagicViper - Let the radius of the earth be R. The circumference of the earth is 2×pi×R The circumference of the telephone line will be 2×pi×(R+10), because you increase the radius by 10 So the differece between them is 2*pi×(R+10)-2×pi×(R) = 2*pi×(R+10-R) = 2*pi*10 = ~62.8
Navitas Nexus Double fail. Neither is the base value D, nor is the difference 10m. Combined with your previous double fail (see reply to earlier comment) you now managed to perform a classy, unparalleled quadruple fail. What is actually you point in hanging around in the comment section, spamming it with pathetic nonsense, instead of finally LEARNING SOMETHING ...?!
Boggless You don't really need the π, there. The answer is (r+10)/r (or its opposite, depending on how you compare the two), but if you're going with algebra, the comparison will look like this. 2πr : 2π(r+10)
Solutions: A) the rock problem: if the rock will float (volume>mass), the water level will not change. If it sinks(mass>=volume), the water level will go down. B) the wire problem: start circumference is x. Radius is x/tau. New radius is x/tau+10. Multiply by tau to get circumference and you get that the difference is 10 tau. C) the balance problem: considering the weight and buoyant force of the blocks are equal, we know that the wooden one generates more buoyant force then the lead one, so to compensate it's mass will be bigger, then by removing the air the buoyant force will go away, and the side with the wooden block will go down due to it's greater mass.
No, the circumference of a circle is 2rπ, which means he's right. Funny, I did the exact same calculation in my head... If it still confuses you though: 2rπ is the radius of the planet, everything added to that is what we're looking for. so, 2π(r+10) = 2rπ + 20π 20π is the added lenght
That sounds frustrating. Have you tried exercises to help you improve your concentration levels? There are plenty on TH-cam... maybe try a few to see which work well for you.
That's good, in a sense. It's the same for me (for any complex science video) I might understand slightly more in the long run, but in the short term, I have to debunk quite a few things I thought I understood but was incorrect about so therefore it'll feel like you understand less, but in actuality I understand more, and the things I thought I understood that were incorrect are now gone.
It's because she says way too much at once, and her analogies and examples are really unclear. This is a pretty common problem among highly intelligent people. They try to say things in simple terms but something gets lost in translation.
You are 'adding value' judgments to the 'rock'...it was already displacing a given amount of water at the start of the examination, regardless of 'where' it was located.The real question is 'How much water will the boat displace when the rock is tossed overboard?' and the answer is...the boat will 'rise' a tiny fraction, but the 'displacement' of the given body-of-water' will remain at it's ab initio value regardless of 'where' the rock is...the 'rock' was 'already there' as a factor, so moving it is moot.
Ammar Ali Now YOU have missed the point about extreme cases. Throw a real huge pile of rocks in sea and you'll have a mountain with all the water in deep down. So the sea level will be even negative.
If you keep throwing rocks overboard, the sea level will drop slowly. Way before it gets to 0, your boat will run out of rocks. Get a bigger boat that can hold more rocks and the sea level will drop more as you unload, but it will be dropping from a higher level it rose to when you added that bigger, heavier boat. Get one bigger still and it means when you put it in the sea, the sea level will rise even more initially before it falls from that higher level back to normal as you unload. In the end, it will just return to what it always was before you put a boat with rocks into it.
About the aluminum foil boat: if you jump out of a boat and you float in the water the level of the water stays the same -- whether your boat is heavy or light. This is because the amount of water being displaced is the same. Great videos, well explained and informative!
Well but what if the rock is super light and barely pushes the boat down. If you then throw it in the water it's going to rise. You didn't explain that very well...
EZ, water level down because if the rock sinks that means it is now displacing less than it's weight in water. in the boat, the rock's weight displaces an amount of water equal to it's weight, which is a greater amount than that of the rock's volume.
It stays the same if you assume that the volume and density are equal. The weight of the rock is still pushing down on the water while inside the boat, causing the water level to rise equally in proportion to the mass of the rock. Subsequently- when tossed into the water, the volume of the rock raises the water level. But it really depends on the density of the rock Vs. the volume of the rock. During the short amount of time after you drop it but before it enters the water, the level will have decreased temporarily.
About the last 2 problems, here is my try at them. Correct me if I'm wrong: Problem 1: Let's get 2 blocks, one with density 1.5 and volume 2 and another with density 2 and volume 1. We now put them in a place with some fluid/gas with density of 1. Now, their relative densities are, respectively 0.5 and 1, making their relative mass 1 (0.5*2) and 1 (1*1), therefore, balancing the balance. When we take the fluid, their relative masses are no longer the same, so, the less dense object weights more. Problem 2: the circumference of the planet is r*π*2 . the poles are 10m, so, the radius of the planet + the poles is r+10, and the circumference is 2*π* (10+r), so, the difference is 20π m
There could be a alternate reasoning. 1. in the first case when the rock sinks. when the rock was in boat. The total force acting on earth was the weight of water + boat + rock = W (this was the weight supported by earth) But here the total weight acts through pressure exerted by water= rhoXgXh = weight of water + boat + rock. When the rock sinks on floor The weight supported by earth remains same. = weight of water + boat + rock. But this time the weight of rock acts directly on floor and not through pressure. therefore W=weight of rock + force exerted by liquid = weight of rock + rhoXgXh1 h1 will be less than h. 2. in the second case when body floats. total weight W will remains same as before. but now the weight acts through pressure in both the cases i.e the body was in boat or it was floating. therefore rhoXgxh will not change.
I used the opposite extreme, that of a very large but very light rock. That rock would do almost nothing when on the boat, but when in the water it would displace its entire volume. Then I thought the answer must depend on the density of the rock. Judging by the answers given/suggested in this video for the rock, person and ice cube, I would guess that the density required to change the answer is equal to that of the water itself, which of course the rock is assumed to be denser than. I learned more than I expected to. :)
I think that Archimedes is the best reasoning to use. It is surprising that people are still unable to work out what the consequences are of using it. This all used to be about (as I was first asked it anyway) an anchor in the boat rather than a rock but the same principles apply. In the boat the anchor causes the volume of water displaced to be equivalent to its mass. In the water the volume of water displaced is equivalent its volume instead. As the anchor is heavier than water (normally anyway :-) it means that the water level MUST go down as the volume of water displaced in the second case is smaller than the volume displaced in the first case. QED.
Additional side note: If you're extremely fit (denser than water) and treading water, then you give up and sink, the water level goes up because your body is displacing it's entire volume and not just body minus head.
yes it dose make the water rise in that when you throw the rock out of the boat the time that the water has to change the position on the boat is almost instant. And yes the water level relative to the boat is going to drop after the rock leaves your hand, but the new level of the water in the container is going to rise when the rock is put in to the water. so the answer to your question is that it rises and drops just relative to two different points and time perspectives of observations.
I like this problem, but calling the object a "rock" could be slightly confusing. There are definitely rocks that exist that are as dense or less dense than water, which is what the question is asking about. In the boat, the water is displaced by the rocks weight. In the water, the water is displaced by it's volume. An object with density equal to water (pouring a bottle of water into the water) would not change the water level, and an object less dense than water (assuming you somehow forced it to sink) would make the water level rise.
when the rock is in the boat it is if floating(boat is floating). so the weight of the rock equals to the buoyancy force.( floating objects have same weight and buoyancy.) when the rock goes in the water it sinks that means buoyancy is lesser than weight.buoyancy is lesser in second case which means less liquid is diaplaced. which means lesser water level rises.(correct me if I am wrong)
At 3:36 you said 'mass', when you meant to say 'weight'. Please add a correction. And now my own 2 problems for you guys. The first is simple, the second is tricky. Both are set in a normal room, full of air. Two syringes have only air inside, they are nor empty nor full, they are capped (no air goes out/in) and balance each other on a scale. Then I uncap the left syringe, empty it off, and cap it again. What will the scale do? Two rubber baloons are filled with air, and they balance each other on a scale. Then the left one gets popped. What will the scale do? (suppose no piece of rubber breaks off the baloon)
For Answer 1) To be extreme, My take on it would be to replace the wood with with something even less dense like foam or polystyrene. It's less dense because of the air it contains, and the air inside it makes up part of it's weight. So if you suck out all the air, the less denser of the two, the one with more air, will lose more of it's weight. And the more dense item, the lead, will not lose as much and retain more of it's weight. Therefore, the side with the lead will tip down.... in my opinion that is. Answer 2) I'm tired, so just going to take a lazy guess and say it's the length of the telephone polls, 10m, times 2, multiplied by Pi ( 3.14) .... so 62.8 m... ish .
Boat question: If the rock sinks, it is denser than water, then the water will go lower. If doesn't sink it stays the same but If you leave an helium balloon fly away the water goes up. Vacuum bell: *IF* Masses are equal, balance stays equal. Porous materials can behave in weird ways trapping air and or other volatile gasses. Wood can dessicate,sublimate solvents or boil off water for example, water going away reducing it's mass. Denser material doesn't mean non porous or non soluble... Why not just try it ? Planet with telephone poles: 2*10*pi meters more
I first heard the telephone pole and wire puzzle stated as a steel band around the equator. If one yard is added to the length of the band, how much would it rise up to be at the same height ALL AROUND the equator; OR how much larger would Earth need to get to fill the band again. It is tempting top think it would hardly make any difference, but... ... Answer, ABOUT a foot in diameter, or about 6 inches all around. (pi is the actual ratio). ... Circum = pi*Diameter ... Wire around the planet C1 = pi*D1 ... Add the poles: C2 = pi*(D1+2*10m) C2 = pi*(D1+20m) Distribute the pi C2 = pi*D1 + pi*20m ... pi*D1 = C1 so... C2 = C1 + pi*20m ... So the difference is pi * 20 m, or 62.83 meters longer regardless of planet size. -- Regards, ScienceAdvisorSteve
I needed help for the two problems at the end!!! problem 1: She said the wood and the lead block has the same mass when we balance them. I search all over the internet, they all say that a balance is a mass measuring device. (really?) So, sucking all the air out, you are still comparing two blocks of the same mass, therefore, they balance. So far so good. Here is my confusion… #1. How does a beam balance works. If I hang two identical object on the two sides of the balance. According to Newton, the force acting on them would be (each, ignoring the force between the objects themselves since the earth is way massive than these two objects) f=G.m1.m2/r^2, where m1 is the mass of the object, m2 is the mass of the earth, r is the distance between the object and the center of the earth. Now, if I push down on the one side and then let go, the object on this side now is closer to the center of the earth, r is smaller, thus, f becomes larger, therefore this side would not comes up again since this side has a bigger downward force than the other. See the problem? Confusion #2… If I use f=ma method (in vacuum) to find a wood and a lead block of the same mass. Certainly they would have different volume. In use, they will displace different amount of air, according to Archimedes principle, the up thrust they experience would be different. Thus their weight (downward force of the earth acting on them) would be different, will they still balance the scale? (they still have the same mass) problem 2: I’ve difficulty using extreme case method to figure out the extra length of the cable. The way I see it, we have to establish the extra length has nothing to do with the size of the planet, shrink the planet down to zero, the different in length would just be 2 pi 10. OK, so far so good, but how do you convince me about the relationship of the extra length and the size. You may say … look at the equation, the different is 2 pi (r+10) minus 2 pi r. Alright, if you present this to me, I would say you don’t need the extreme case at all, since the equation clearly shown the extra length was independent of radius. See the dilemma? No equation, can not establish relation, with equation, no need extreme case. Would someone demonstrate the relationship without the equation to throw light on the problem.
the rocks gathers far less volume than the boat. It displaces less water than the boat does. Boat is less dense than rock as it floats and the rock sinks. But when you put the rock on the boat , using the boat's volume rock can displace more water by adding (and uniformly distributing) the mass to the boat. So, level is high when rock is on the boat and it restores when the rock is removed.
The water level falls. You didn't define your frame of reference in your question, so I define it as the boats deck. When I throw the stone overboard the boat gets lighter. A lighter boat float higher in the water, making the water level drop from the reference frame of the boats deck.
BUT consider an extreme case for the second question. If the bottom of the aluminium foil boat is like an extremely large upside down obtuse triangular pyramid, the volume of water for the mass of your body now have less space around the boat to fill when you are in the boat. Then the water level rises more. And when you are floating in the water, the water have more space around your body to fill since very small volume of the boat is in the water. Then the water level rises less. The rise in water level is then differently proportional to the mass the water displaces. Therefore, for displacing the same mass, the water level rises more when you are in the boat.
As a Software Developer of 20+ years & fan of physics I realized (after a while of head scratching) that using EXTREME CASES allowed me to make problem solving much easier.
Instead of the extreme case being a small rock with a high mass, why can't it be a large rock with a small mass (but greater than the mass of water so it sinks), so then the rock would not push the boat down far making the water level rise slightly, but then once you throw it into the water it would raise the water level because of its size.
I don't know if it has any significance, but when you partially fill a glass with ice and water, the water level goes down as the ice floats up (and before it gets to the top). I have no idea if it's right, but it almost seems like since buoyant force is stronger the deeper in water you go, that it might be causing more water to be displaced. If that is true then depth of water is also a factor to consider.
Change in Perimeter = 2πa Here a is, height of the pole. So in this case the answer will be approximately 60 meter. ∆P = 2πa ; this formula is independent from radius of our Earth. This is really awesome.
problem 2: telephone pole... planet circumference is 2 * pi * x; adding a 10 mtr pole is 2 * pi * (x + 10); let radius of planet be 10 meter or 100 or 1000.. results are all going to differ by 20 * pi (like using radius of planet, which is x, set to 1000 mtr. = planet circum is 2000 mtr * pi, and with the addition of a 10 mtr pole it would be 2020 mtr * pi) the difference in that example is 20 mtr * pi... any radius of planet would be the same at 20 mtr * pi...... 20 mtr * pi = 62.83 mtrs.. yeah! I be smart! I be smart!
stays the same SO LONG as the boat also remains in the water. if you mean "the default" water level (IE without your boat) then the rock raises the water level. SO rock in water. raises water level but IT ALREADY raised the water level buy being in the boat while that boat is in the water. SO the water level WITH boat stays the same but will be higher once that boat leaves the water since the rock remains.
Problem 1: Wood is much more porous than metal of any kind. Even iron, which is porous for metal is not as porous as wood. So in wood, more of its mass is water and air. Since we are assuming no water in either the wood or the metal and sucking air out, the balance is going to shift towards the metal because now the masses are no longer equal. The wood is now lighter than the metal. Problem 2: Circumference is 2πr so if you had 10 ft telephone poles all the way across the planet, you would be adding 10 ft to the planet's radius. That is tiny for planets. So starting with a unit circle and thus a circumference of 2π, if you add 10 to the radius, the circumference increases to 22π. This is a difference of 2r(2 times the added radius). So in this case, the amount you add to the circumference is 2πr. If the circle instead had a radius of 10 and you added 10 to the radius you would get 40π and the original circumference being 20π. So for any circle, even a planet, increasing the radius by 10 increases the circumference by 20π. And thus you get 2πr added to the circumference for all possible added radii, assuming of course that gravity does not affect this increase in circumference by putting an upper limit on the radius that can be added before it gets so massive that the added radius:added circumference ratio drifts wildly away from this 2πr.
Wooden block question... If I follow your suggestion of thinking in extremes. A grain of sand that weighs just as much as a plane and a huge plane and put them both on a scale balance them and put them in a vacuum chamber. When the air is sucked out, the air pressure on the larger object which would have been higher as it has more surface area would have been significantly reduced. Hence the plane would go up. Or rather the sand which had little air pressure acting on it would remained it's weight and hence, be relatively heavier, thus, would go down. Am I wrong?
+Kavin Mitran Barathidasan Yeah, the plane would go up. Although reason for that is not more surface area but rather more volume because more volume means bigger buoyancy, which is caused by pressure difference from the bottom to the top of the plane.
Yes--what about this, if density = mass/volume, then the volume of water displaced by the boat with concrete block inside is much more than that displaced by the empty boat plus the concrete block at the bottom of the water, since the boat initially supports and evenly distributes the mass of the block, allowing the block less total density in relation to the water. When the block is dropped overboard, it resumes its own density in relation to the water (its mass is taking up a lot less space), and the water level drops. We tried it with a rock, a toy boat, and a measuring cup.
If the rock is denser than water the water level lowers. If the rock is the same density as water it stays the same. If the rock is less dense that water the water level rises (but of course the rock floats). Pumice floats and I have a piece of sandstone that floats for about 30 seconds until it soaks up the water.
I'm using this video to demonstrate to people that considering extreme cases is a good way of memorizing physics equations. F=ma. How do those 3 variables relate to each other? Consider extreme cases and you don't have to memorize the formulas. If you apply a force to an aircraft carrier then apply the same force to a pebble, the pebble will move way faster than the aircraft carrier. Small mass means big acceleration and vice versa.
The question at the end was bugging me with the wood and lead block. What makes sense for me is solving an opposite problem and then assuming the opposite results. Please tell me if this is wrong, but this is what makes sense to me. Imagine that instead of air, you had some liquid of an unknown density (this would represent the air). Now begin with an empty container where you have a boat on the scale, and a rock. It's safe to assume that the boat would go down if there was no liquid in the experiment. Now consider that the liquid was water. After inserting the water, the boat would go upwards, and the stone would sink. Based on the same principles, it seems like the wood would be more prone to rise due to the air being in the container, so then it would follow that the wood should fall without it.
I paused then I thought the water would stay the same. I thought it depends on the weight per volume ratio of the rock to the weight per volume or density of the water. So assuming the they're equal, the water should stay the same height because when the rock is with you on the boat, it had weight to displace equivalent water and when you throw it to the water off your boat, it displaced same volume of water as your boat did when you had it with you. Now Im resuming the video and see if my hypo is somewhat relevant. :D
0:16 I believe it's the same as when the stone was in our hand or in the boat, it's weight had pushed the boat as much inside the water as when the stone is inside the water. Watched the video... Well, I was wrong. didn't regard the mass density.
Soooo for problem 1, I guess it would tip toward the wooden block, as it has a greater volume, and it has more buoyant force due to it would displacing a greater amount of air than its lead counterpart. That means when there's no air, the buoyant force would disappear, tipping the wooden block downwards. That's how I interpreted it. Now to problem 2. We'll have to add 10 -- the length of the telephone poles -- to the radius of the planet, which shall remain a forever unknown 'r'. We are asked to find how much longer the length of the wire is compared to the circumference of the said planet, so L - C, where 'L' is the length of the wire -- AKA the circumference of the planet including the telephone poles. Because circumference = 2πr, and we gotta add 10 to the radius, the equation will look rather like this: 2π (10 + r) and due to the fact that we're asked to find the difference in circumferences (?), 2π (10 + r) - 2πr [which is the original circumference of the planet] so, 2πr + 20π - 2πr = 20π The circumference of the planet after adding the telephone poles is 20π greater compared to its original circumference. [P.S.: Sorry if this was too long; I tried to make it as brief as possible]
I thought of the opposite extreme case, what if you had a so heavy boat it was standing on the bottom, and then you throw the rock in the water, the boat would still take up as much space as before, but the rock would now too, so the water level rises. I feel that this depends on the weight and buoyancy of the boat and rock.
Ok I paused the video to keep it fair and real. On the boat the rock is displacing its weight in water. In the water the rock is displacing its volume. Rock is denser than water (it sinks so you know that’s true, no need to look it up). That means the water its weight displaces is greater than the water its volume displaces. Therefore the water level drops when the rock is thrown overboard. Imagining the rock being made of a super dense substance like osmium helped me intuit this. Such a substance would “push down” on the boat far harder than the small amount of water its volume displaces. (Edit: just watched the rest of the video - same mental exercise)
Answer 1: Since the masses of the blocks are same, in order to make the 2 sides balanced, both upthrusts (U = vdg) should be equal. So when the air is removed, the upthrusts will be not there anymore. But the balance is maintained sinces the masses are equal. So neither wooden nor metal block tips down. That is beginners' physics. Answer 2: She asks how longer would the wire's length would be THAN the planet's circumference. Not how many TIMES. Simply enough!! Extra length = C(planet) - L(wire) = 2piR(planet) - 2pi[R(planet) + 10) = 2pi10 (if pi = 3.14) = 2×3.14×10 = 62.8m
+Chirath Madurapperuma I really want to agree with you on problem 1. This is the first thing that came into my mind. But after reading all these comments I'm starting to believe that the buoyancy from the air would slightly affect the larger shaped object.
+Cory In the video she says that the wooden block is larger ryt?? I didn't hear at the 1st time. With same MASSES?? Yeah now that's a problem to me as well. Certainly the buoyant force is more on the larger block (wooden). So how can 2 equal weighted blocks be balanced by 2 unequal buoyant forces? That's a big problem!!
+Cory Yeah.. Wooden block = W; Metal block = M; Mass of W = Mass of M = m The equation should be as follow, Weight of W - upthrust on W = Weight of M - upthrust on M mg - Volume of W × air density × g = mg - Volume of M × air density × g Volume of W = Volume of M This can't be true because W should be always larger than M: Therefore the fact that these two are having EQUAL masses can't be justified.
Depends upon the density of the rock. And possibly its porosity. And what shape is the rock. (concrete hull boats do float.) Are there rocks less dense than water? Porosity should be discounted as included air would be replaced by water. In theory.Does the rock absorb water, as opposed to trapped air being replaced? Is the water pure? Salt water is more dense. How dense can "water" be? (refer to density of rocks, above) Is the water in solid or liquid state? Just havin fun
So at 1:14 she say's Archimedes principle is "excessive" and so I thought wow! gonna learn something new today. Yet she uses precisely Archimedes principle about 30 seconds later without calling it by that name. The only subtlety is when the "black hole rock" is in the boat the upward force can match the rock's weight because the boat is displacing the water.
it depends, You need to know the difference of the volume of water the rock displaces and the volume of water displaced by the weight of the rock. So If the area of the rock is 1 cubic foot of water, that would equal 7.48 gal. of water. If the weight of the rock displaces 2 cubic feet of water, that would be 14.96 gals of water. Toss the rock in the air the water level drops 14.96 gal. (-14.96) The rock enters the water , the level goes up 7.48 gal. (7.48) (-14.96gal)+(7.48gal)= -7.48gal in my example it drops.
A stone (and everything) on a floating boat is like an air pocket(neglecting the excess amount of boat's building stuff got inside) to the buoyancy. And a black hole wipes all water out.
but the size of the object matters and not its weight ... it could be as light as an exercise ball tied to a rock at the bottom of the water (extreme version). Archimedes maybe right about 2:55 but that is talking about weight and not size. a lighter object but rather huge can be tied to the bottom and it will displace water significantly and at the same time if its not tied to the bottom and allowed to float the amount of water it displaced will be significantly lower than when it is tied to the bottom. so now im confused.
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Answers:
1. Wooden block tips down.
2. 20π m longer.
Solutions by Extreme Cases:
1. Imagine that instead of a wooden block and a lead block, you have nothing on one side of the scale, and a balloon full of helium attached to the other side. With air in the chamber, the side with the helium balloon attached will rise as the helium balloon pulls up on it - this is because the air is providing a buoyant force on the balloon. Once all the air is removed from the chamber, the side with the helium balloon falls - the lack of buoyant force from air makes it evident that the side with the helium balloon has more mass than the side with nothing.
2. Imagine that the Earth was a single point with a radius of 0m, and therefore a circumference of 0m. With the telephone poles, the circumference would then be 2*10m*π = 20π m, and so the difference is always 20π m.
Solutions by Principles:
1. Here are the facts: a) Air is a fluid, and so we can apply Archimede's principle to this situation of blocks in a chamber full of air. b) Archimede's principle states that the buoyant force applied on an object in a fluid is equal to the weight of the volume of fluid displaced by that object. c) We can conclude that the wooden block, having a greater volume, displaces more air than the lead block, and so there is a greater buoyant force on the wooden block pushing it upwards. d) When the air is removed from the chamber, there is no longer that greater buoyant force pushing the wooden block upwards and so the wooden block tips down the scale.
2. If we let r = radius of the Earth, we can solve this by simple math:
Difference in circumference
= Circumference of telephone wires - Circumference of Earth
= 2π(r+10) - 2πr
= 2πr + 20π - 2πr
= 20π
Therefore, the telephone wires will have a circumference that is 20π meters longer than the circumference of the Earth.
+Rizwan Awan (TheOnlyRizzy) You have the best explanation. Your comparison with a helium balloon gives the best idea. The balloon wants to go up because the weight in that volume is less than that from air. Remove the air and the balloon will fall down. The wooden block has a larger volume. Remove the air and the wooden block looses the advantage of the larger volume.
+Rizwan Awan (TheOnlyRizzy) really a great explanation. Very interesting to see the difference between (i) extreme and (ii) analytical approach in action.
+Rizwan Awan (TheOnlyRizzy)
alternative scenario, even easier:
imagine the scales have one identical box standing on each arm. the wooden block completely fills box 1. the lead block doesn't fill box 2, so box 2 has the lead block + some air in it. both boxes and their contents cause the scale to balance out. (here we can see that actually, the blocks themselves do not have the same mass to begin with, the wooden block is heavier!...and the point is that you are weighing more air on one side)
when you evacuate, you remove that air from box 2, but you remove nothing from box 1. so you took mass away from box 2.
+Science4Fun Thanks! I think this was "Physics Girls" whole point - to show how both approaches come to the same conclusion and compliment one another :)
+Nico van Os Thanks! I was quite pleased when it clicked in my mind, and thought I'd share :P
Ha ha, I got asked this question during a chemical engineering job interview.
Omg you are here
OK. Here's another one: Two guys are in a boat with three cigars. No matches, how will they smoke their cigars? Answer: They throw one cigar overboard, that makes the boat a cigar lighter. Simple....
old post, but hilarious
...If you start this knowing that there is a type of boat called a lighter it's just that much better.
That joke is so old I fell off my dinosaur, laughing, the first time I heard it.
But it’s not a physics riddle.!??
But then their boat is now a cigar lighter which, as you can imagine, is not sufficient to support the weight of 2 men, and they are now floundering in the open water.
everytime she says "easy!" my self steem lowers a little bit...
@Imagine No Religion What, it is just logical thinking. The answear is obvious, no?
What, it is just logical thinking. The answear is obvious, no?
My self confidence goes up... simply based on the fact that I’m attracted to her intelligence.
If the rock is more dense then water then the water level will go down because the displacement is pure the volume of the rock not the volume equivalent in what to the weight of the rock
+mike more I think it's not clear enough in the video that the result depends on the density in both cases...
the problems say "if the stone sinks" and "if you float" as parts of the problems. so this is given.
+mike more What about if the object is very heavy but less dense than water? wouldn't it still make the water decrease after than when the object was on the boat? I think so because the weight of the object makes the boat decrease which replaces the molecules and then makes the water level higher. when the object is in the water, the weight on the boat decreases which makes the water level decrease.
AcEDndJhard if the object is heavy and still less dense than the water, than it must have a bigger volume to compensate (remember, the density is the ratio between the volume and the weight). Which defeats the purpose of the extreme-case chosen.
And, it will make the water decrease when it left the boat, but it would make the water higher again when it would enter the water and float.
+nilsvids I love that the video is so certain of the answer when it's really "it depends on the rock."
My guess is it depends on the density of the rock. When it's in the boat, it displaces water based on its weight. When it's under water, it displaces water based on its volume.
You already know the density, she said it will sink! This means the lowest density the object can have, is high enough for this to work. As long the density is higher than that of water, the weight will displace more than the size.
@@handlotion8244“the weight will displace more than the size” WHEN ITS IN THE BOAT.
Ohmygod this sentence. The way you framed it made it suddenly so clear all of a sudden!!! Thank you big time.
I failed an engineering job interview on this question 14 years ago.
A floating object displaces it's mass in water, a sunken object displaces it's volume of water.
Since the rock ways more then its volume in water, when it is on the boat it displaces more water then it would if it had sunk. So the water level will decrease.
Edit: Huzzah! I was correct, however, you cannot make this argument for any rock that starts in the boat. There bound to be rocks that are porous and less dense then water.
but the water level will never rise because the rock can only ever displace the same amount of water whether in the boat or tossed into the water, if it is less than the density of water. your first sentence still holds true 100% of the time, because if you throw a rock over board that sinks, as the riddle suggests, then that rock was denser than what you were floating in
Devin Royce You are correct in the essence of that once the system has settled that will be the outcome. However a rock that weighs less then its volume in water will still 'sink' temporarily if it is thrown into the water, during that time before the system reaches equilibrium the water level will have risen.
yes there would be some force required to keep the rock at the bottom, but the original question was just a friendly way out asking whether it displaces more on the boat, or submerged.
Someone who spells "weighs" as ways, shouldn't even be near a computer
this is maths Jack hoarses four coreses mate
I'd like to whey in on this spelling debate.
3:33 either they have the same mass or they are balanced. They cant be both at the same time due to the air pushing them up with different forces. If they have the same mass the gravitational force is the same, but because the wooden block suppresses more air then the lead one, it got a bigger force pointing upwards- the lead block sinks.
If they are balanced they still feel the same upward-force as above. For the resulting forces to be equal, the wooden block needs to have a higher mass then the lead one. In air they cant have the same mass and be balanced (at different sizes)
I know what you meant but it is a wrong formulation.
@Leeengold ur statment is 100% correct!
Leeengold I feel smart because I actually came up with the lead box sinks on my own, with the same explanation u gave
If you forget that statement about them having the same mass and just focus on them being in balance. then when air is removed the balance tips towards wooden block (because it has to be heavier to counteract bigger buyoant force).
Leeengold exactly , the mass balance doesn't actually equates the masses it equates the moment of gravitational forces so the wooden block has to be of greater mass initially to be balanced . when the air is sucked out wooden block will fall because there is no buyoncy and it has greater mass then iron block . which implies she told us wrong question .
The question is impossible to answer based on a flaw in the wording of the question itself. You will never remove all of the air. Perfect vacuum is not obtainable. The highest vacuum man has achieved I believe is around 1E-12 Pascal or 7E-15 Torr. Even space is not a perfect vacuum. That is not to say the outcome will not be the same theoretically, just pointing it out, it's a bad question to begin with the way it's asked.
answer to problem with planet and wire:
Let's use the extreme method. The planet is a point, then we add 10m poles, R= 10, circumference 2 * PI * R.
+Tymon0000 whats the problem
+Tymon0000 Yeah, that gives the exact answer.
I thought of this more from an engineering perspective, where you could argue that the radius of the planet is much, much larger than 10m. So a more accurate extreme case is that the planet's radius is practically infinite compared to the poles, so the circumference of this wire circle essentially 2*PI*R, where R is the radius of the planet. So the answer to the question, "how much longer is the circumference of the wire than the circumference of the planet", is 0.
+Brandon Bocklund It's constant... If you add 10m poles, you end up with a circumference that has 10 extra m of radius. So if your circle (earth of any other circle) has radius R, and the wire has radius R+10, then circumference is (2*Pi*Radius), so the difference will be:
Diff = 2 * Pi * ( R+10m ) - 2 * Pi * R
Diff = 2 * Pi * 10m + 2 * Pi * R - 2 * Pi * R
Diff = 2 * Pi * 10m
Diff = 62.83m
So, the circumference of the wire circle will be 62.83 m longer than the circumference of the earth.
Right, but lets say we're looking at Pluto, which isn't even a planet anymore. Pluto has a radius of ~1 million meters. So the wire has a circumference of 1 million meters + 62 meters which is pretty much just 1 million meters, so the difference between them is zero in a practical sense.
+Brandon Bocklund So how long would the poles need to be, for you to say, that the circumference would be considered longer? Let's consider the Pluto's radius of ~1 million meters.
Problem 1: Wood down, lead up.
This one is tricky and I haven't seen a good explanation in the comments yet, so I'll see if I can get it. First of all we have to assume that the wood does not outgas at all when the atmosphere is removed, if id did this would mean that the mass of the wood would change. But since this video is talking about buoyancy I think that's more what the question is driving at.
So, the block should tip towards the wood. Wood down, lead up.
The reason for this is buoyancy. The atmosphere is pressing in on the wood over a greater surface are and is thus creating an upward force that is larger than the upward force on the lead block, but of course everything is balanced. When the air is removed it can no longer provide this force and the small difference in mass becomes apparent.
Problem 2: 20Pi meters longer. But others have mentioned this already.
the air exerts forces on all sides cancelling out...
the air exerts forces on all sides cancelling out...
+axelasdf If that were true a helium filled balloon would fall like any other object. Why does a helium balloon get pushed up? The helium balloon has the same amount of pressure on all sides, and yet it floats up.
The thing is that while the forces are balanced for each side of an object, the surface area for the balloon is huge compared to it's weight this means it would have to move many more particles of air out of the way below it to fall down than you or a block of wood. Since gravity is acting on all the air and on the balloon the weight of the air (pressure from the atmosphere) pushes up against anything moving within it. Only objects that have a high density (small surface area) can overcome this force and fall through the air.
If I'm not making sense, try veritasium's video about buoyancy.
+YouCanScienceIt The forces only cancel out sideways. But more force is pushing up than down, because the air pressure decreases when you are higher. Even the few cm of the height of the cubes makes a difference. If you do the math, it only comes down to the density of the fluid (air/water/...) and the volume, in other words, the displaced weight. (Weight assumes gravity; alternatively you could say mass + gravity, which is the same thing as weight.)
Agreed.
If the density of the object thrown out is greater than the density of the fluid that the boat is floating in, then the total volume of displaced water is greater when the object is in the water. Otherwise, the total volume of displaced fluid remains the same. That is, an object that floats make no change to the water level when it's in the fluid; the level rises when an object that sinks is put into the fluid. This is a lot easier to see with a few simple density and volume equations.
This was way more didactic than I thought it would be, thanks a lot for that! My little sister could understand everything just fine, and got really excited about those problems at the end.
+brianpso Glad to hear that! How old is your little sister?
lol she's 18, but she's not that into science and stuff, so by making this very friendly she didn't go away after 5 seconds =)
+brianpso When I read your comment I thought she was like 12, you made me laugh x)
+Physics Girl i just tried this and you are wrong in the video. if the rock is very light then the water stays about the same when its in the boat and it rises slightly when you throw it in the water. i think you went the wrong way with your extreme.if the rock is so light that it is almost bouyant, then the water is slightly higher when the rock is submerged than when it is in the boat, so, you are wrong. it's very upsetting that you teach people incorrect things.
+Killer Apple You should be much more careful about what you post (i.e. don't outright say that she teaches incorrect things, and say that's she's wrong). Instead, explain how you conducted your experiment (i.e. posting a video about it would be great and informative), and ask for an explanation. Especially since you saw in the video that she did the experiment as well.
You should also clarify that the water level you're referring to is with respect to the boat or to the container.
Based on your logic "if the rock is very light then the water level stays about the same", then removing the rock would also have no impact (see phrontdoor's explanation). Her extreme example was well done. I.e. imagine you're in a small boat that has some concrete blocks in it (the blocks are much denser than water). If you take it out, then the water level will drop since there's less water being displaced. Putting the block back in the water only pushes up slightly. The water level drop is greater than water level rises.
With regards to your second comment: "if the rock is so light that it is almost boyant, then the water is
slightly higher when the rock is submerged than when it is in the boat," Let's assume that by almost buoyant, you mean that the rock has the density of water (and is almost completely submerged when dropped in a body of water). When the rock is on the boat, the rock displaces its own weight in the water. Then when the rock is temporarily lifted, the water level drops, equal to the weight of the rock being lifted. When the rock is put back in the water, then the rock again has to the displace the its own weight in the water. This rise and fall is nearly equal.
Plus, If Physics girl was incorrect, I highly doubt nobody would have picked up on it already. (whether from the comments below or from the producers of the show)
Best one since the Chladni plates video.
w.r.t the circumference problem, I found it easier to do 2π(r+10) - 2πr than think through the extreme case. But, thinking of the extreme case gave me a different perspective on the problem.
+Anirudh Surendranath You can do that too! But then take the extreme case to be r = 0 (is that what you did?) And the answer becomes immediately clear. :)
+Anirudh Surendranath I even did it with "l" as an arbitrary length instead of 10 m. I somehow have an aversion against doing formula manipulation with values in it and always try to do the general case first. That's what classical science education does to you. Sometimes when I'm stuck I try to imagine stuff with actual numbers, but it may occasionally lead you astray, just as measuring things in drawings can.
+Physics Girl Unfofrtunately for some reason, extreme cases make physics much more difficult for me to grasp than regular formulas and explanations (not so much true for math for some reason), so really seeing "2pi*(r+10)- 2pi*r" just somehow makes more sense to me than "imagine r=0".... or maybe they feel like more of a proof to me? I just don't see how r=0 makes it immediately clear what the answer is.
Then again physics were always somehow hard for me to grasp (probably because of the way it's explained?) despite the fact I'm pretty good at math and study chemistry.
pokestep I feel the same way most of the time. And unless you've done the formula, how can you be sure that the answer for r=0 is the general answer for all cases?
pinkdispatcher Exactly! Mathematical proof is the way to go for me. Except for, if you think of r=0 as the extreme case, then it's "easy" to see the difference (but that's again the mathematical proof so I'm not sure if that's what is supposed to make it "immediately clear") because you automatically see the difference is 20pi...
I dunno. In general I don't like how physicists explain things and usually act condescending to others (at least the ones I encountered) so physics really isn't the way to go for me... We did have a great physics teacher at uni tho.
But doesn't it depend on the density of the rock and its size?
Kevin Yeoh
Doesn't depend on size only density
And I guess that its pretty easy ro say the rock s' density is more than water
Unless it’s pumice stone.😎
Bill Bill the riddle states that it sinks.... therefore it is denser than water
Maxime Boissonneault very true.
I think that pumice stone sinks as well; if not you can substitute it with a marble stone with many holes. I think that defining density is definitely a mandatory issue before you can solve the riddle
For Problem 1, I guess you can assume the wooden block is pushed up by air so when the air is gone, it'll push down more. So the wooden block is heavier.
For Problem 2: Difference = 2*Pi*(R + 10) - 2*Pi*R where R is radius of the planet. So, it is 2*Pi*10 .
+TheSlimyDog the blocks are equally pushed by air in every direction, so it'll stay the same
+TheSlimyDog I agree, thus she unfortunately misspoke at 3:35 when she said that the two blocks have the same mass.
You forget that air is also a fluid. If the wooden block is less dense it has bigger volume so the boyant force acting on it, is greater. When the air is removed there is no boyancy. That means that the scale will tilt towards the wooden block and the led will appear to be less heavy. And that is the truth. Mwooden> Mled . Sorry for my bad English.
The buoyant force acting at the bottom of both plates are equal and when we take out the air, they will have no impact on the balance. However the weight on the video right plate is the weight of the smaller block plus the weight of the air colum above this block plus the weight of the air collums around the block and in the plate. For the left plate, it's the same situation, but the air collums are smaller, since the volume of the wooden block is bigger, leaving less space for as around and above it. Then, with air, the weight of the wooden block plus the weight of the smaller collums is equal to the weight of the other block plus the weight of the bigger collums. So, the wooden block is heavier and will bring the plate down without the air to balance it. We can also use the extreme case, where the wooden block is the size of the plate and extremely high and the other block is extremely small. The solution is the same.
well it depends on the size of the boat, and the density of the rock...
+Kibbis “Kibnutz” Kibertus The density of the rock must be greater than the density of the water since it sinks; on the other hand, the boat does not.
***** No... do you even physics... the boat floats because of the shap and material. The boat floats because it has the round shape, and it's made of fiber glass which evenly displaces the water so that it doesn't tip. Even if the rock was perfectly round it wouldn't float.
Kibbis Kibertus A boat floats because the overall density of the air and the body of the boat is lower than the density of water. In this case buoyant force compensates the weight so it remains on the surface of the water rather than going up or down.
FYI, the rock boat would float if the body was thin enough.
***** well yeah because then it would be lighter than the water... duh... and whatever you want to think
+Kibbis “Kibnutz” Kibertus I'm no expert on physics but am an expert on boats... and that's not fiberglass
Using extreme cases is extremely useful. In almost every case ever, in every field, everywhere. Lots of people seem to really not like it though. When they propose an idea and you immediately consider how it would work in the extreme case, or ask a question about the extreme case, they get displeased. It often leads to, at least, nailing down a problem very precisely, and they seem to not want to be so 'strict' about things usually. It prevents them from thinking effectively in a lot of cases. I'm not sure if it's because it often reveals much more complexity to an issue than they presume it to have or if they're not confident in their ability to nail things down like that or what, but maybe if the benefits of 'thinking in the extreme' were more often talked about it would help. Thanks for your video!
Is that Jake from vsauce3?
Yes!
PG is smarter than vsauce. My life is complete. :D
i dont think you knoe the definition of the word smart
And who cares about that?
i thought you were going to say is that Jake from Statefarm LOL
Great video. One comment: If the rock in the boat had the mass of a black hole, then the boat would sink to the bottom, which means that we have to modify the original statement that the boat is "floating" in the water.
+Eugene Khutoryansky There are also tiny black holes; they don't have to be huge. Tiny ones don't live long (look up Hawking radiation), but as tiny a mass as that of a single proton could theoretically become a black hole.
+Eugene Khutoryansky The real problem is that the black hole rock would be so dense that, unless the boat is infinitely strong, it would punch right through it, probably causing the boat to sink.
cOmAtOrAn Perhaps. But when she said "a black hole rock" or something, that was just a figure of speech. Hyperbole.
pinkdispatcher
Yeah. For the purposes of an extreme case thought experiment, it works fine. But I found some amusement in working out the consequences of actually running such an experiment. :)
cOmAtOrAn
Sounds a bit like one of xkcd's "what if". :)
I have yet to watch the answer, but I've thought this through and perhaps my explanation is different than the Physics Girl's and may help some people understand this problem better.
So my thinking is that when you throw the rock off, the water it displaces should be equal to V, the volume. This would make it seem the water level would go up, however you must remember that the boat now weighs less by mg (m is the mass of the rock).
This means the buoyant force must be mg less than it was before and the boat will rise, displacing mg less water. So then the water will rise if the amount of water the rock displaces is more than the amount of water replaced by the boat rising--meaning V>mg. Otherwise it will stay the same if V=mg or sink if V 1 (simplified to density*g) then the water will sink.
As long as the rock sinks, you know the density is greater than 1 and the water level will sink.
EDIT looks like the answer is right, but I feel like something is wrong with my work. maybe it should just be m/V > 1, if anyone spots a mistake plz let me know!
The answer to the planet-pole puzzle:
Let radius of planet = r
To find : outer circumference - circumference of planet
= 2pi(r+10) - 2pi(r)
= 2pi(r+10 - r)
= 2pi(10)
= 20pi
She said you don't need to know the radius
Joe Milton Karan's answer doesn't need a value for r
you're right
WAS THAT JAKE FROM VSAUCE3???????
yes
there was also that guy from all chat
No, that was Jake from State Farm!
yes
where?
It was on the boat, is my first clue, so now it's in the water: water level stays the same, displacement. Water cannot rise since the weight of the rock on the boat was already in effect. It only moved from the boat into the water. Resume play.
argh! went against intuition!
Right according to Archimedes the water should be displaced according to the weight.
Like how the Monty Hall Problem makes more sense when you think of it with a million doors instead of 3.
Problem 2 : 10*2pi = 62,83 meters more than at the surface (extreme case scenario being a planet of radius 0)
1) the problem is impossible given that the two blocks are of equal mass. Ignoring this assumption, and instead assuming that the masses are balanced under pressure, then evacuating the air will cause the wooden block to move down. The wooden block must have a greater mass in order for its weight to counteract the greater buoyant force it experiences.
2) the circumference will increase by 20pi. 2pi(r + 10) - 2pi(r) = 20pi.
U nailed it. It's 11th grade physics
For all of u wondering what would happen if rock was bigger/smaller with greater/less mass i have writed the equations and the rise or fall of water level depends ONLY on its mass and density
Here is how it looks like: (Mk * R - Mk * r)/R*r where Mk is mass of rock R its density and r density of water.
So as long as density of rock is bigger then waters the water lvl will drop
+strahinja nikolic congrats, you tried too hard lol
Naah it took me 10 mins
Would that work if you had a rock the size of the moon thats density is only just hight than water?
if the rck i less dense than water, it wont sink, so wouldnt it be incapable of displacing more water than it had while it was in the boat? actually the rock would now be another boat correct?
But what if the rock is ultra light, but very voluminous, like a plastic hollow ball? Wouldn't the water level rise? So ultimately the question depends on the density of the rock.
+poketopa1234 something less dense or equally dense to water = stays the same
something denser = lower.
+poketopa1234 Yes this is 100% true, but she said in the question that the rock sinks, so you know its density is > water which means the water level must lower.
Ebonyl93 ok cool, thanks.
+poketopa1234 I thought about the exact same thing right after her explaination. I've been looking but still can't find comments that can explain how I am wrong.
I understand the sinking/floating thing but i'm not talk about that. I'm still talking about a sinking rock.
She used the extreme case of tiny/high density rock. What about a huge/less dense (not than water) rock, with a very buoyant boat?
Say the rock is the size of the bath tub, but the boat is buoyant enough to carry it right over the water. If the rock is dropped into the water, obviously the water level will increase.
Someone please explain to me, thanks.
w I l l I a m I think she just omitted that senario. The general rule she made only applied to the first question. If you want more info, just look up archimedes principle.
In an extreme case, it deppends on the rock temperature that may cause an increase of the water level
😂
Two guys are in a boat with three cigarettes. They had no fire, so they threw one of the cigarettes overboard, and the entire boat became a cigarette lighter.
is the answer for the second one 20 pi
because if diameter increases in 20 m, radius increases in 10m, and each new meter will multiply the other two variables on the equation C= 2 x pi x r(C beeing circumference and r beeing radius) thus the increase in the sise will be 10x pi x2 or 20 pi (pi = 3,1415)
No...the value of the 'rock' was 'already there' as a displacement-factor....changing the position of the rock will not alter it's original 'displacement-value' in relationship to the water.This question is a 'closed-loop' scenario, and all the factors were present BEFORE the rock was tossed over!It does mean the boat will 'rise' in exactly co-equal value to the no longer present 'rock' IN the boat itself...yet the water displacement remains at the same value, neither rising nor falling.This is an exercise in 'logic and proportion' rather than 'shifting paradigm' values of the body of water itself.
You forget that the rock is denser than the water. So in order to support the rock's weight far more water needs to be displaced than if the rock were in the water. That's why the levels go down.
TheMegoShow Read what I wrote again.
It's the entirety of the displacement under consideration...not 'weight'.
The volume of displacement would still be the same, regardless of the density of the matter displacing a given volume of water.
A 'ping-pong' ball would displace the same amount as a ball of steel of the same dimensions.
Poveys-perplexing problems.........
you can do this from first principles by balancing the forces. Since in both scenarios (1. rock in the boat 2. rock in the water's bottom) nothing is accelerating, the forces of gravity and forces of buoyancy must balance. "W" is weight, "V" is volume, "D" is density, "F" is force. Assume D_rock > D_water = 1 (kg/L)
Scenario 1:
Eq 1: W_boat_with_rock = W_displaced_water ....... so->
W_boat + D_rock * V_rock = D_water * V_displaced_water_of_boat_with_rock
Scenario 2 gives us two equations, since the boat and the rock have to be stationary but here they are separated:
Eq 2: W_boat = D_water * V_displaced_water_boat_alone
Eq 3: D_rock * V_rock = D_water * V_rock + F_acting_on_rock_by_bottom_of_pool
Now, a few easy manipulations of these three equations (I'll skip) and you get:
Eq 4: V_displaced_water_of_boat_with_rock - (F_acting_on_rock_by_bottom_of_pool / D_water) = V_displaced_water_boat_alone + V_rock
... and because the (FORCE_acting_on_rock... / D_water) must be a positive quantity we modify Eq 4 to an inequality:
V_displaced_water_of_boat_with_rock > V_displaced_water_boat_alone + V_rock
So water level goes down in scenario 2
Hey +Physics Girl,
Your Problem 1 (3:30) is confusing.
It does not, of course, follow from the fact that these blocks are balanced that they have the same mass. That's whole point of the puzzle, right? If they do have the same mass, then, when measured in a vacuum, they should be balanced _assuming_ that both arms of the scale are of equal length. However, you imply that they are balanced when measured in air.
So, given that they are balanced when measured in air and that they have the same mass, we can only conclude that the arms are not of equal length. You never mentioned that they were of equal length. The arm holding the wooden block must be longer to account for its greater buoyancy. Remove the air and the buoyancy is lost so the longer arm exerts more torque thus the wooden block drops.
I don't suppose that that was the intended solution. Scales aren't meant to be defective. However, the information given forces this conclusion. I guess you'd intended to have decent scales with arms of equal length. If the arms are of equal length and the scale is balanced in air, the wooden block must have more mass to counter the difference in buoyancy. Again, the wooden block will drop once the air is removed.
I don't think it'll involve arms of the balance.
The wooden block will have air in its pores in normal conditions, and when all air is sucked out, its mass will become less.
Ahmed Nauman Tariq It _shouldn't_ have involved them and wouldn't have if the problem had been worded better. We should have been given that the arms were the same length rather than the blocks being the same mass.
Imagine deforming the metal block into the shape of one great big pore. What difference would you now expect to observe?
Yeah, nah, it's not about pores but buoyancy. Another way to think about the problem is to consider the difference in air pressure between the top and bottom of the block. There is greater pressure at the bottom than at the top. The pressure at the bottom pushes up and that at the top pushes down. The consequent difference in force is proportional to volume. Thus the wooden block is more buoyant. Taking away the air nullifies this.
actually since the iron block barycenter is closer to the the earth than the wooden cube, wouldn't the gravitational force be higher on the iron block ? which means that it weighs more:
(F = m * a) if m is the same and a is bigger, F is bigger.
which means the balance should tip to the iron side, provided we are in a perfect vacuum with no friction etc etc
EDIT: ah, she actually says that the cubes "are balance so they have the same mass" that's a contraddiction. I had missed that
archimede's law is valid also in air :P
Remember: both cubes are _inside_ the medium (air). Buoyancy doesn't apply here, since air pressure happens at all sides: up, down, and sideways; they effectively negate each other.
In the end, the scale will stay balanced.
Actually, the water goes down while the rock is in the air, then back up when the rock enters the water.
I thought about it another way. when the rock is on the boat, water exerts enough force to keep it from sinking. when you throw it in the water water fails to keep the rock over sea level, thus water is producing less force than before and that can only happen if the sea level goes down (archimedes)
problem 1: imagine a really small and dense block on the right side and a balloon filled with air on the other. now for the scale to stay balanced the balloon itself (without the air inside it) should weigh as much as the block on the right side, because the weight of the air inside it is held up by the pressure of the air around the balloon. now if you create a vacuum, the left side will tip down because now the weight of the air inside the balloon is added to it.
you could also solve this problem using the case of an extremely dense air (or even water)
problem 2: imagine a tiny planet the size of a grain of sand. the circumference of the circle that the telephone poles make is Radius*2*3.14=62.8 meters.
i'm a master of extreme cases XD
remember if you make the problem harder, then do the exact opposite.
for example i had a problem with the planet being almost infinitely large. then the surface of the planet will be flat and therefore circumference=infinite
so i imagined an infinitely small planet instead.
But what about 2 dense objects? I thought that the bigger block would have more air pressure due to it's larger surface. Once the air pressure is gone it will tip to the right.
At least that was my reasoning.
zeroZyra yes, both ways the bigger block will tip down after you create a vacuum.
the good thing about using extreme cases is that there are several factors that you can change and take to an extreme to find your answer
+MilesCrlsn No. In the problem stated, it is _not_ given that either block is porous or that it contains its own internal volume of air, simply that they are of equal mass but unequal density.
Since they are of equal mass, _theoretically_, removing the air from the chamber will have no effect.
Why? Because the air pressure is uniform (if you ignore gravity) on a stationary object. Since they're of equal mass, there is no net force and thus no net acceleration (assuming starting from stationary), and thus the air pressure is uniform over the surface of the two objects (neither has a net influence caused by the air).
Now in _reality_, because the objects have weight and are not in freefall, gravity is relevant here. Specifically, Newton's law of universal gravitation that states that (paraphrased slightly for simplicity's sake) gravity is proportional to mass, and inversely proportional to the square of the distance between two bodies.
Therefore, the air itself will have a very slightly lower density along the top of the wooden block than along the bottom, giving a net upwards force causing it to ride ever so slightly higher (practically probably hundredths of femtometres). When the air is evacuated, that effect no longer exists and thus the larger object will fall a practically immeasurable, imperceptible distance.
Kill3rCat being porous doesn't change anything. (as long as the air inside them is considered a part of them. i didn't remove the air from inside the balloon in my example.)the bigger object is affected more by the pressure of the air around it and the smaller object is less afected. so after you remove the air from around the objects the bigger one measures heavier.
also as i mentioned, using the case of 2 objects under water (extremely dense air) it's obvious that the bigger objects with the same weight is going to possibly float (negative weight) and after removing the water, both objects are going to be the same weight.
MilesCrlsn I derped. I said 'porous' when I meant 'hollow'.
And in the situation, it is not given that either object has an internal volume of air. There is only 'air inside them' in your example, which is wholly irrelevant to the situation described in the video.
getting a veritasium feel from this vid, i like it :)
+vic cobb I've "borrowed" his method of asking people questions. ;)
+Physics Girl I like it actually because they're always so stumped :P its good to know smart people can learn something every day too!!
PS. i LOVE all your videos :) ive showed them to so many people and they do too :D
There are 3 types of rock, rocks that float, rocks with density equal to the water and rocks with density greater than water.
Case 1, When the density of the rock is greater than water. The rock sinks and the floor supports the weight of the rock. The water level sinks because the delta volume displaced by the boat is greater than the volume of the rock.
Case 2, the density of the rock is equal or less than the density of water. The water level stays the same because the delta volume displaced by the boat is equal to the volume of the portion of the rock that stays underwater.If the rock makes it to the floor, the floor will not provide any support of the rocks weight.
Jimmy Wong! Oh, how I miss Video Game High School ~
omg i just understood one of her videos.
why is it less dense
Haha me too🤣
pausing at 1:08 it could be either way, depending on what you mean by "water level". If you're talking about the level of the water vs the boat . . . then it would lower; that is to say: the level of the water would be lower on the boat, as there is less weight in the boat & the boat would have less mass beneath the surface (thus lower -- a.k.a. more boat above). If you're talking about the level of the water after throwing it in, vs before throwing it in the water level would be higher. . . as you're not removing mass from the water, so that lump on the bottom displaces the water that was there & that water would need to rise up . . . so the water to the edge of the tank would rise -- if it were not for the same displacement on the boat reducing mass & increasing displacement proportionately to the boat's total mass.
yeah, video is a bit misleading because it does not ask about the waterline on the boat, it asks about the water level, but then proceeds to "prove" the water level goes down (on the boat) which is not the water level in the tank.
so.. what if the rock is larger than the boat?
Your boat will be at the bottom of the ocean at the first place, in case you mean heavier aswell
So much for the Crow and the Pitcher story
Lol
+Nill Hill poor crow
+Ravenous Crow you are way to into this
I can't believe I just had the right solution and the right thought on a Physics Girl video.
But what if instead of a small but very heavy rock, you had a large but light rock. then the water level would go up.
p.s - excuse me if i get this wrong i'm 12
If the rock was light it wouldn't sink. Any rock heavy enough to sink will lower the water level.
Then it wouldnt sink
+Warrior Son
Physics won't break your heart ^.^
To tack onto what rodrigo said, the issue here - which the graphic doesn't actually show correctly - is that displacement caused by a floating object is based on the mass, not volume, of the object. For anything denser than water (meaning it would sink), its mass (in kilograms) is larger than its volume (in cubic meters), and hence once it sinks it now only displaces its smaller volume, not its larger mass, in water. For anything where its displacement is less than its volume, it would have to be less dense than water, and hence float.
Yes That's what I would think. But I am only 13 so I have no clue tbh.
Someone please explain the last puzzle!!
TheMagicViper - Let the radius of the earth be R.
The circumference of the earth is 2×pi×R
The circumference of the telephone line will be 2×pi×(R+10), because you increase the radius by 10
So the differece between them is 2*pi×(R+10)-2×pi×(R) = 2*pi×(R+10-R) = 2*pi*10 = ~62.8
medanman62 Thank you!!
The answer is D+10m.
Navitas Nexus
Double fail. Neither is the base value D, nor is the difference 10m.
Combined with your previous double fail (see reply to earlier comment) you now managed to perform a classy, unparalleled quadruple fail.
What is actually you point in hanging around in the comment section, spamming it with pathetic nonsense, instead of finally LEARNING SOMETHING ...?!
Boggless
You don't really need the π, there.
The answer is (r+10)/r (or its opposite, depending on how you compare the two), but if you're going with algebra, the comparison will look like this.
2πr : 2π(r+10)
Solutions:
A) the rock problem: if the rock will float (volume>mass), the water level will not change. If it sinks(mass>=volume), the water level will go down.
B) the wire problem: start circumference is x. Radius is x/tau. New radius is x/tau+10. Multiply by tau to get circumference and you get that the difference is 10 tau.
C) the balance problem: considering the weight and buoyant force of the blocks are equal, we know that the wooden one generates more buoyant force then the lead one, so to compensate it's mass will be bigger, then by removing the air the buoyant force will go away, and the side with the wooden block will go down due to it's greater mass.
2Pi(r+10) ?
How much /longer/ than the original rope?
about 62.8m longer?
Wouldnt it be just π(r+10)?
No, the circumference of a circle is 2rπ, which means he's right.
Funny, I did the exact same calculation in my head...
If it still confuses you though: 2rπ is the radius of the planet, everything added to that is what we're looking for. so, 2π(r+10) = 2rπ + 20π
20π is the added lenght
it'll be 20π
i love extreme cases
no i love reasoning by exaggerating examples to the extreme.
I am with you Dude, the same here.
I love using extreme cases, it tends to be a quick illustration of principle. More so, since for many questions the answer is at either extreme.
every time I watch her I don't understand anything
That sounds frustrating. Have you tried exercises to help you improve your concentration levels? There are plenty on TH-cam... maybe try a few to see which work well for you.
That's good, in a sense. It's the same for me (for any complex science video) I might understand slightly more in the long run, but in the short term, I have to debunk quite a few things I thought I understood but was incorrect about so therefore it'll feel like you understand less, but in actuality I understand more, and the things I thought I understood that were incorrect are now gone.
It's because she says way too much at once, and her analogies and examples are really unclear.
This is a pretty common problem among highly intelligent people. They try to say things in simple terms but something gets lost in translation.
u rocked bro !!
You are 'adding value' judgments to the 'rock'...it was already displacing a given amount of water at the start of the examination, regardless of 'where' it was located.The real question is 'How much water will the boat displace when the rock is tossed overboard?' and the answer is...the boat will 'rise' a tiny fraction, but the 'displacement' of the given body-of-water' will remain at it's ab initio value regardless of 'where' the rock is...the 'rock' was 'already there' as a factor, so moving it is moot.
so you're telling me, if I throw enough rocks into water from a boat, I could make the sea lvl go zero?
Lux Dues In a univese where this kind of rocks existe , yes it is possible, due this video :b
kon aris no its not, it would only go back to the level it was before you entered the water
Ammar Ali
Now YOU have missed the point about extreme cases. Throw a real huge pile of rocks in sea and you'll have a mountain with all the water in deep down. So the sea level will be even negative.
If you keep throwing rocks overboard, the sea level will drop slowly. Way before it gets to 0, your boat will run out of rocks.
Get a bigger boat that can hold more rocks and the sea level will drop more as you unload, but it will be dropping from a higher level it rose to when you added that bigger, heavier boat. Get one bigger still and it means when you put it in the sea, the sea level will rise even more initially before it falls from that higher level back to normal as you unload. In the end, it will just return to what it always was before you put a boat with rocks into it.
About the aluminum foil boat: if you jump out of a boat and you float in the water the level of the water stays the same -- whether your boat is heavy or light. This is because the amount of water being displaced is the same. Great videos, well explained and informative!
Well but what if the rock is super light and barely pushes the boat down. If you then throw it in the water it's going to rise. You didn't explain that very well...
If the rock is super light it wont sink. If it sinks, water level goes down, if it can float, the water level stays the same.
She doesn't explain things very well.
EZ, water level down because if the rock sinks that means it is now displacing less than it's weight in water. in the boat, the rock's weight displaces an amount of water equal to it's weight, which is a greater amount than that of the rock's volume.
It stays the same if you assume that the volume and density are equal. The weight of the rock is still pushing down on the water while inside the boat, causing the water level to rise equally in proportion to the mass of the rock. Subsequently- when tossed into the water, the volume of the rock raises the water level. But it really depends on the density of the rock Vs. the volume of the rock.
During the short amount of time after you drop it but before it enters the water, the level will have decreased temporarily.
About the last 2 problems, here is my try at them. Correct me if I'm wrong:
Problem 1: Let's get 2 blocks, one with density 1.5 and volume 2 and another with density 2 and volume 1.
We now put them in a place with some fluid/gas with density of 1. Now, their relative densities are, respectively 0.5 and 1, making their relative mass 1 (0.5*2) and 1 (1*1), therefore, balancing the balance. When we take the fluid, their relative masses are no longer the same, so, the less dense object weights more.
Problem 2: the circumference of the planet is r*π*2 . the poles are 10m, so, the radius of the planet + the poles is r+10, and the circumference is 2*π* (10+r), so, the difference is 20π m
There could be a alternate reasoning.
1. in the first case when the rock sinks.
when the rock was in boat. The total force acting on earth was the weight of water + boat + rock = W (this was the weight supported by earth)
But here the total weight acts through pressure exerted by water= rhoXgXh = weight of water + boat + rock.
When the rock sinks on floor
The weight supported by earth remains same. = weight of water + boat + rock. But this time the weight of rock acts directly on floor and not through pressure.
therefore W=weight of rock + force exerted by liquid = weight of rock + rhoXgXh1
h1 will be less than h.
2. in the second case when body floats. total weight W will remains same as before. but now the weight acts through pressure in both the cases i.e the body was in boat or it was floating. therefore rhoXgxh will not change.
I used the opposite extreme, that of a very large but very light rock.
That rock would do almost nothing when on the boat, but when in the water it would displace its entire volume.
Then I thought the answer must depend on the density of the rock.
Judging by the answers given/suggested in this video for the rock, person and ice cube, I would guess that the density required to change the answer is equal to that of the water itself, which of course the rock is assumed to be denser than.
I learned more than I expected to. :)
I think that Archimedes is the best reasoning to use. It is surprising that people are still unable to work out what the consequences are of using it.
This all used to be about (as I was first asked it anyway) an anchor in the boat rather than a rock but the same principles apply.
In the boat the anchor causes the volume of water displaced to be equivalent to its mass. In the water the volume of water displaced is equivalent its volume instead. As the anchor is heavier than water (normally anyway :-) it means that the water level MUST go down as the volume of water displaced in the second case is smaller than the volume displaced in the first case. QED.
Additional side note: If you're extremely fit (denser than water) and treading water, then you give up and sink, the water level goes up because your body is displacing it's entire volume and not just body minus head.
yes it dose make the water rise in that when you throw the rock out of the boat the time that the water has to change the position on the boat is almost instant. And yes the water level relative to the boat is going to drop after the rock leaves your hand, but the new level of the water in the container is going to rise when the rock is put in to the water. so the answer to your question is that it rises and drops just relative to two different points and time perspectives of observations.
I like this problem, but calling the object a "rock" could be slightly confusing. There are definitely rocks that exist that are as dense or less dense than water, which is what the question is asking about. In the boat, the water is displaced by the rocks weight. In the water, the water is displaced by it's volume. An object with density equal to water (pouring a bottle of water into the water) would not change the water level, and an object less dense than water (assuming you somehow forced it to sink) would make the water level rise.
when the rock is in the boat it is if floating(boat is floating). so the weight of the rock equals to the buoyancy force.( floating objects have same weight and buoyancy.) when the rock goes in the water it sinks that means buoyancy is lesser than weight.buoyancy is lesser in second case which means less liquid is diaplaced. which means lesser water level rises.(correct me if I am wrong)
At 3:36 you said 'mass', when you meant to say 'weight'. Please add a correction.
And now my own 2 problems for you guys. The first is simple, the second is tricky. Both are set in a normal room, full of air.
Two syringes have only air inside, they are nor empty nor full, they are capped (no air goes out/in) and balance each other on a scale. Then I uncap the left syringe, empty it off, and cap it again. What will the scale do?
Two rubber baloons are filled with air, and they balance each other on a scale. Then the left one gets popped. What will the scale do? (suppose no piece of rubber breaks off the baloon)
For Answer 1) To be extreme, My take on it would be to replace the wood with with something even less dense like foam or polystyrene. It's less dense because of the air it contains, and the air inside it makes up part of it's weight. So if you suck out all the air, the less denser of the two, the one with more air, will lose more of it's weight. And the more dense item, the lead, will not lose as much and retain more of it's weight. Therefore, the side with the lead will tip down.... in my opinion that is.
Answer 2) I'm tired, so just going to take a lazy guess and say it's the length of the telephone polls, 10m, times 2, multiplied by Pi ( 3.14) .... so 62.8 m... ish .
Boat question: If the rock sinks, it is denser than water, then the water will go lower. If doesn't sink it stays the same but If you leave an helium balloon fly away the water goes up.
Vacuum bell: *IF* Masses are equal, balance stays equal. Porous materials can behave in weird ways trapping air and or other volatile gasses. Wood can dessicate,sublimate solvents or boil off water for example, water going away reducing it's mass. Denser material doesn't mean non porous or non soluble... Why not just try it ?
Planet with telephone poles: 2*10*pi meters more
I first heard the telephone pole and wire puzzle stated as a steel band around the equator. If one yard is added to the length of the band, how much would it rise up to be at the same height ALL AROUND the equator; OR how much larger would Earth need to get to fill the band again. It is tempting top think it would hardly make any difference, but...
...
Answer, ABOUT a foot in diameter, or about 6 inches all around. (pi is the actual ratio).
...
Circum = pi*Diameter
...
Wire around the planet
C1 = pi*D1
...
Add the poles:
C2 = pi*(D1+2*10m)
C2 = pi*(D1+20m)
Distribute the pi
C2 = pi*D1 + pi*20m
...
pi*D1 = C1 so...
C2 = C1 + pi*20m
...
So the difference is pi * 20 m, or 62.83 meters longer regardless of planet size.
--
Regards, ScienceAdvisorSteve
I needed help for the two problems at the end!!!
problem 1: She said the wood and the lead block has the same mass when we balance them. I search all over the internet, they all say that a balance is a mass measuring device. (really?) So, sucking all the air out, you are still comparing two blocks of the same mass, therefore, they balance. So far so good.
Here is my confusion… #1. How does a beam balance works. If I hang two identical object on the two sides of the balance. According to Newton, the force acting on them would be (each, ignoring the force between the objects themselves since the earth is way massive than these two objects) f=G.m1.m2/r^2, where m1 is the mass of the object, m2 is the mass of the earth, r is the distance between the object and the center of the earth. Now, if I push down on the one side and then let go, the object on this side now is closer to the center of the earth, r is smaller, thus, f becomes larger, therefore this side would not comes up again since this side has a bigger downward force than the other. See the problem? Confusion #2… If I use f=ma method (in vacuum) to find a wood and a lead block of the same mass. Certainly they would have different volume. In use, they will displace different amount of air, according to Archimedes principle, the up thrust they experience would be different. Thus their weight (downward force of the earth acting on them) would be different, will they still balance the scale? (they still have the same mass)
problem 2: I’ve difficulty using extreme case method to figure out the extra length of the cable. The way I see it, we have to establish the extra length has nothing to do with the size of the planet, shrink the planet down to zero, the different in length would just be 2 pi 10. OK, so far so good, but how do you convince me about the relationship of the extra length and the size. You may say … look at the equation, the different is 2 pi (r+10) minus 2 pi r. Alright, if you present this to me, I would say you don’t need the extreme case at all, since the equation clearly shown the extra length was independent of radius. See the dilemma? No equation, can not establish relation, with equation, no need extreme case. Would someone demonstrate the relationship without the equation to throw light on the problem.
+Physics Girl - For problem 1 is the mass the same or the balance perfectly balanced - it cant be both and you say it is at ~ 3:30
the rocks gathers far less volume than the boat.
It displaces less water than the boat does. Boat is less dense than rock as it floats and the rock sinks.
But when you put the rock on the boat , using the boat's volume rock can displace more water by adding (and uniformly distributing) the mass to the boat.
So, level is high when rock is on the boat and it restores when the rock is removed.
The water level falls.
You didn't define your frame of reference in your question, so I define it as the boats deck. When I throw the stone overboard the boat gets lighter. A lighter boat float higher in the water, making the water level drop from the reference frame of the boats deck.
BUT consider an extreme case for the second question. If the bottom of the aluminium foil boat is like an extremely large upside down obtuse triangular pyramid, the volume of water for the mass of your body now have less space around the boat to fill when you are in the boat. Then the water level rises more. And when you are floating in the water, the water have more space around your body to fill since very small volume of the boat is in the water. Then the water level rises less. The rise in water level is then differently proportional to the mass the water displaces. Therefore, for displacing the same mass, the water level rises more when you are in the boat.
As a Software Developer of 20+ years & fan of physics I realized (after a while of head scratching) that using EXTREME CASES allowed me to make problem solving much easier.
Instead of the extreme case being a small rock with a high mass, why can't it be a large rock with a small mass (but greater than the mass of water so it sinks), so then the rock would not push the boat down far making the water level rise slightly, but then once you throw it into the water it would raise the water level because of its size.
I don't know if it has any significance, but when you partially fill a glass with ice and water, the water level goes down as the ice floats up (and before it gets to the top). I have no idea if it's right, but it almost seems like since buoyant force is stronger the deeper in water you go, that it might be causing more water to be displaced. If that is true then depth of water is also a factor to consider.
Change in Perimeter = 2πa
Here a is, height of the pole.
So in this case the answer will be approximately 60 meter.
∆P = 2πa ; this formula is independent from radius of our Earth. This is really awesome.
problem 2: telephone pole... planet circumference is 2 * pi * x; adding a 10 mtr pole is 2 * pi * (x + 10); let radius of planet be 10 meter or 100 or 1000.. results are all going to differ by 20 * pi (like using radius of planet, which is x, set to 1000 mtr. = planet circum is 2000 mtr * pi, and with the addition of a 10 mtr pole it would be 2020 mtr * pi) the difference in that example is 20 mtr * pi... any radius of planet would be the same at 20 mtr * pi......
20 mtr * pi = 62.83 mtrs.. yeah! I be smart! I be smart!
stays the same SO LONG as the boat also remains in the water. if you mean "the default" water level (IE without your boat) then the rock raises the water level.
SO rock in water. raises water level but IT ALREADY raised the water level buy being in the boat while that boat is in the water.
SO the water level WITH boat stays the same but will be higher once that boat leaves the water since the rock remains.
Problem 1:
Wood is much more porous than metal of any kind. Even iron, which is porous for metal is not as porous as wood. So in wood, more of its mass is water and air. Since we are assuming no water in either the wood or the metal and sucking air out, the balance is going to shift towards the metal because now the masses are no longer equal. The wood is now lighter than the metal.
Problem 2:
Circumference is 2πr so if you had 10 ft telephone poles all the way across the planet, you would be adding 10 ft to the planet's radius. That is tiny for planets.
So starting with a unit circle and thus a circumference of 2π, if you add 10 to the radius, the circumference increases to 22π. This is a difference of 2r(2 times the added radius). So in this case, the amount you add to the circumference is 2πr.
If the circle instead had a radius of 10 and you added 10 to the radius you would get 40π and the original circumference being 20π.
So for any circle, even a planet, increasing the radius by 10 increases the circumference by 20π. And thus you get 2πr added to the circumference for all possible added radii, assuming of course that gravity does not affect this increase in circumference by putting an upper limit on the radius that can be added before it gets so massive that the added radius:added circumference ratio drifts wildly away from this 2πr.
Wooden block question...
If I follow your suggestion of thinking in extremes. A grain of sand that weighs just as much as a plane and a huge plane and put them both on a scale balance them and put them in a vacuum chamber. When the air is sucked out, the air pressure on the larger object which would have been higher as it has more surface area would have been significantly reduced. Hence the plane would go up. Or rather the sand which had little air pressure acting on it would remained it's weight and hence, be relatively heavier, thus, would go down. Am I wrong?
+Kavin Mitran Barathidasan Yeah, the plane would go up. Although reason for that is not more surface area but rather more volume because more volume means bigger buoyancy, which is caused by pressure difference from the bottom to the top of the plane.
Yes--what about this, if density = mass/volume, then the volume of water displaced by the boat with concrete block inside is much more than that displaced by the empty boat plus the concrete block at the bottom of the water, since the boat initially supports and evenly distributes the mass of the block, allowing the block less total density in relation to the water. When the block is dropped overboard, it resumes its own density in relation to the water (its mass is taking up a lot less space), and the water level drops. We tried it with a rock, a toy boat, and a measuring cup.
If the rock is denser than water the water level lowers. If the rock is the same density as water it stays the same. If the rock is less dense that water the water level rises (but of course the rock floats).
Pumice floats and I have a piece of sandstone that floats for about 30 seconds until it soaks up the water.
I'm using this video to demonstrate to people that considering extreme cases is a good way of memorizing physics equations. F=ma. How do those 3 variables relate to each other? Consider extreme cases and you don't have to memorize the formulas. If you apply a force to an aircraft carrier then apply the same force to a pebble, the pebble will move way faster than the aircraft carrier. Small mass means big acceleration and vice versa.
The question at the end was bugging me with the wood and lead block. What makes sense for me is solving an opposite problem and then assuming the opposite results. Please tell me if this is wrong, but this is what makes sense to me.
Imagine that instead of air, you had some liquid of an unknown density (this would represent the air). Now begin with an empty container where you have a boat on the scale, and a rock. It's safe to assume that the boat would go down if there was no liquid in the experiment. Now consider that the liquid was water. After inserting the water, the boat would go upwards, and the stone would sink. Based on the same principles, it seems like the wood would be more prone to rise due to the air being in the container, so then it would follow that the wood should fall without it.
I paused then I thought the water would stay the same. I thought it depends on the weight per volume ratio of the rock to the weight per volume or density of the water. So assuming the they're equal, the water should stay the same height because when the rock is with you on the boat, it had weight to displace equivalent water and when you throw it to the water off your boat, it displaced same volume of water as your boat did when you had it with you. Now Im resuming the video and see if my hypo is somewhat relevant. :D
0:16 I believe it's the same as when the stone was in our hand or in the boat, it's weight had pushed the boat as much inside the water as when the stone is inside the water.
Watched the video... Well, I was wrong. didn't regard the mass density.
Soooo for problem 1, I guess it would tip toward the wooden block, as it has a greater volume, and it has more buoyant force due to it would displacing a greater amount of air than its lead counterpart. That means when there's no air, the buoyant force would disappear, tipping the wooden block downwards. That's how I interpreted it.
Now to problem 2. We'll have to add 10 -- the length of the telephone poles -- to the radius of the planet, which shall remain a forever unknown 'r'. We are asked to find how much longer the length of the wire is compared to the circumference of the said planet, so L - C, where 'L' is the length of the wire -- AKA the circumference of the planet including the telephone poles. Because circumference = 2πr, and we gotta add 10 to the radius, the equation will look rather like this:
2π (10 + r)
and due to the fact that we're asked to find the difference in circumferences (?),
2π (10 + r) - 2πr [which is the original circumference of the planet]
so,
2πr + 20π - 2πr
= 20π
The circumference of the planet after adding the telephone poles is 20π greater compared to its original circumference.
[P.S.: Sorry if this was too long; I tried to make it as brief as possible]
I thought of the opposite extreme case, what if you had a so heavy boat it was standing on the bottom, and then you throw the rock in the water, the boat would still take up as much space as before, but the rock would now too, so the water level rises. I feel that this depends on the weight and buoyancy of the boat and rock.
Ok I paused the video to keep it fair and real. On the boat the rock is displacing its weight in water. In the water the rock is displacing its volume. Rock is denser than water (it sinks so you know that’s true, no need to look it up). That means the water its weight displaces is greater than the water its volume displaces. Therefore the water level drops when the rock is thrown overboard. Imagining the rock being made of a super dense substance like osmium helped me intuit this. Such a substance would “push down” on the boat far harder than the small amount of water its volume displaces. (Edit: just watched the rest of the video - same mental exercise)
Answer 1: Since the masses of the blocks are same, in order to make the 2 sides balanced, both upthrusts (U = vdg) should be equal. So when the air is removed, the upthrusts will be not there anymore. But the balance is maintained sinces the masses are equal. So neither wooden nor metal block tips down. That is beginners' physics.
Answer 2: She asks how longer would the wire's length would be THAN the planet's circumference. Not how many TIMES. Simply enough!!
Extra length = C(planet) - L(wire)
= 2piR(planet) - 2pi[R(planet) + 10)
= 2pi10 (if pi = 3.14) = 2×3.14×10 = 62.8m
+Chirath Madurapperuma I really want to agree with you on problem 1. This is the first thing that came into my mind. But after reading all these comments I'm starting to believe that the buoyancy from the air would slightly affect the larger shaped object.
+Cory In the video she says that the wooden block is larger ryt?? I didn't hear at the 1st time. With same MASSES?? Yeah now that's a problem to me as well. Certainly the buoyant force is more on the larger block (wooden). So how can 2 equal weighted blocks be balanced by 2 unequal buoyant forces? That's a big problem!!
+Chirath Madurapperuma she better post an answer soon. its driving me crazy thinking about it.
+Cory Yeah..
Wooden block = W; Metal block = M; Mass of W = Mass of M = m
The equation should be as follow,
Weight of W - upthrust on W = Weight of M - upthrust on M
mg - Volume of W × air density × g = mg - Volume of M × air density × g
Volume of W = Volume of M
This can't be true because W should be always larger than M: Therefore the fact that these two are having EQUAL masses can't be justified.
Depends upon the density of the rock. And possibly its porosity. And what shape is the rock. (concrete hull boats do float.)
Are there rocks less dense than water?
Porosity should be discounted as included air would be replaced by water. In theory.Does the rock absorb water, as opposed to trapped air being replaced?
Is the water pure? Salt water is more dense. How dense can "water" be? (refer to density of rocks, above)
Is the water in solid or liquid state?
Just havin fun
So at 1:14 she say's Archimedes principle is "excessive" and so I thought wow! gonna learn something new today. Yet she uses precisely Archimedes principle about 30 seconds later without calling it by that name. The only subtlety is when the "black hole rock" is in the boat the upward force can match the rock's weight because the boat is displacing the water.
it depends, You need to know the difference of the volume of water the rock displaces and the volume of water displaced by the weight of the rock.
So
If the area of the rock is 1 cubic foot of water, that would equal 7.48 gal. of water.
If the weight of the rock displaces 2 cubic feet of water, that would be 14.96 gals of water.
Toss the rock in the air the water level drops 14.96 gal. (-14.96)
The rock enters the water , the level goes up 7.48 gal. (7.48)
(-14.96gal)+(7.48gal)= -7.48gal
in my example it drops.
0:07 she must have rather stupid tossing that rock, knowing it in fact did not land in water.
A stone (and everything) on a floating boat is like an air pocket(neglecting the excess amount of boat's building stuff got inside) to the buoyancy. And a black hole wipes all water out.
but the size of the object matters and not its weight ... it could be as light as an exercise ball tied to a rock at the bottom of the water (extreme version). Archimedes maybe right about 2:55 but that is talking about weight and not size. a lighter object but rather huge can be tied to the bottom and it will displace water significantly and at the same time if its not tied to the bottom and allowed to float the amount of water it displaced will be significantly lower than when it is tied to the bottom. so now im confused.
Man, this Neet question was eating my head off.
Thanks Physics girl