Dear Mr. Walter Lewin, your lectures are such a pricess gift to me during this hard time. I had always struggled with course name fluid mechanics because the teachers in my school did not explained and demonstrated as clear as you did. It help me regain my motivation to strive to become an engineer. After all, thanks you a lot!!!!.
Why am I watching this? Actually I'm a lecturer in medical school and I had a lecture on blood flow in blood vessels which is based on fluid physics and Bernoulli's rule I couldn't think of anything better to prepare for my lecture than watching your lecture on fluid physics it's a good thing that even doctors are watching your lectures Lot of respect professor From Iraq
Thank you very much Prof. Lewin, I have to take my final oral test in physics for engineering next Monday and after all this years of your lecture being around, you are saving my life as those of many other students ! These should be saved as hsitory of science for years to come !
THATS HOW PHYSICS IS TAUGHT.....I AM COMPLETELY AMAZED...... EARLIER I FOUND FLUIDS VERY DIFFICULT TO UNDERSTAND BUT NOW I ENJOYED LEARNING IT IN THIS LECTURE......THE PRACTICAL DEMONSTRATION WERE JUST TOO GOOD.....
Not only hydrostatics.... Well it's physics which I seem to feel like.... Hope I could ever meet you.... Wanna study in MIT but you don't teach nowadays... Still your lectures are powerful..
@@lecturesbywalterlewin.they9259 sir, at 20:02 the balloon will rise considering there is an atmosphere, because of helium you mentioned, there is a buoyant force even when there is no gravitational acceleration, because of difference in densities.
You are the best teacher ever ! I really love your teaching ! I wanna wish you a long and happy life !! You are someone irreplaceable in my heart ...I wanna meet you so badly and say how much your teachings helped me ! Also I really loved your birthday series 💕 Thank you so much again professor Lewis!
Now i am preparing for my neet exam i have many boubts in fluid mechanics after this class i feel comfortable to solve problems of fluid mechanics thank you professor 😊for wonderful class love from India ❤💓
Thanks again sir... For the first time I have felt Fluids... Before this I had been thinking that in fluid part there is nothing but puzzles... Now I feel very comfortable in fluids.... Can't pay you against this but infinite respect will always be for you...
I think that to understand the pool level problem, it helps being exaggerated. Consider a super dense object, 1 ton with the size of a coin. Now imagine we drop it in a boat that could carry it. Intuitively, the water level would rise significantly to counter the added weight. Now if we drop it in the water, there will be a massive weight relief in the boat and the buoyant force required to keep it floating will be therefore much less. The object has the size of a coin, so the volume of water displaced by it as it sinks is negligible and so is it's buoyancy. The total buoyant force that the water produces will be way less, and so the pool level will drop. As long as the boulder's density is greater than that of water, the same reasoning could be applied.
In the answer to your waterline of swimming pool question, I think it will go down. When we seperate the stone from the boat, the waterline goes down more than the volume of the stone since the density of the boat is lower than the stone. So when the stone sinks the net effect is a lower waterline.
I have an answer to the swimming pool question posed at 11:20, but I also have two additional questions regarding the same. The answer to your question: The water level goes down, because if we assume that the density of the stone is greater than the density of water (a reasonable and intuitive assumption), then the stone will sink to the bottom of the pool and settle down. When that happens, the buoyant force that was originally holding up the stone before it was thrown into the water will now be shared between the upward normal force at the bottom of the pool AND the new buoyant force together. Therefore, the new buoyant force is only a fraction of the original buoyant force. Since lower buoyant force displaces lesser water than higher buoyant force, the water level in the pool goes down. (In the event that this is a bottomless pool, then the stone will keep accelerating downwards under its own weight as there is a net force pulling the stone downwards. This net force can only exist if the new buoyant force is lower than its weight, and hence the new buoyant force is lower than the original buoyant force. Thus, the pool level drops). However, I have two variations on the stone that was thrown out: Q 1. What if, instead of a stone, a fish of the same volume as the stone but of the same density as water is thrown out of the boat and into the pool such that the fish goes below the surface of the water but doesn't sink? Would the water level in the pool go up, stay the same, or drop? My (probably wrong) answer: The water level would stay the same. Since the buoyant force is neither being shared by any normal force at the bottom of the tank (as the fish isn't sinking to the pool floor) and nor is the fish accelerating downwards, there is no change in the total buoyant force, which means there is no change in the displaced water. Hence, the water level stays the same. In fact, when the fish is thrown out of the boat, for the brief moment that it is in the air, the water level drops (as the buoyant force drops), but once the fish is under the surface of the water, the water rises again by the same amount that it had dropped, thus equalizing the level. Q 2. What if, instead of a fish, a wooden block of the same volume as the fish but of lower density than water is thrown into the pool such that it floats? Would the water level in the pool go up, stay the same, or drop? My (definitely) wrong answer: The water level would stay the same. There is no change in the total buoyant force here either, as the wooden block isn't sinking to the floor. But I still feel that my answer to this last question is wrong because my inductive reasoning would have led me to believe that if a higher-density object (stone) lowers the water level in the pool, and an equal-density object (fish) keeps the water level the same, then a lower-density object (wood) would have raised the water level. But that is not what the equilibrium equations tell me. Please correct me if I am wrong in any of the above. And yes, your lectures DO make me ♥ physics :)
professor i am from i india, i love to watch your lectures , you explain practically everything. i am in class 10 but i understand everything. it creates me a passion to get phd in physics .thank you professor
Whenever a body is partially or completely submerged in a fluid,it will experience an up thrust which is equal to weight of the fluid which has been displaced.
At 38:55, syphen experiment. when we suck the water at other end, we created a differential pressure at top and because of this water is raised to max height of tube. once it crosses the max height, gravitational P.E is converted to K.E which resulted into water fall.
Um, King Hieron II was called a "virtuous man" by Machiavelli and he had a long and seemingly friendly relationship with Archimedes, and his father (a court astronomer) so he'd hardly "kill" him if he got it wrong. :P Great lecture.
Hello Dr. Lewin, this is what I think about the question about the water level changing when you throw the rock overboard. I am a little confused on what you mean by will the water line remain change. Relative to the boat or relative to the walls of the pool? Initially, the water is at a height y_0. The volume of the fluid V_water is given by x*y_0*z where x is width and z is the thickness. For the boat to float, the F_b must equal the weight of the boat and rock. So F_B_0 = (M+m)g (1) where m is the mass of the rock and M is the mass of the boat. Now analyzing the situation when you throw the rock overboard. The mass of the entire boat-rock system must change because we no longer have m. So, in order to float, F_B =Mg. (2) So applying Archimedes Principle to (1.1) V_fluid_0*p_water = V_(M+m)*p_(M+m) and to (2.2) V_fluid*p_water = V_M*p_M . F_B_0 > F_B because of (1) and (2). Therefore. V_fluid_0*p_water > V_fluid*p_water. p_water = M_water/V_water = M_water/x*y_0*z so we have it that: V_fluid_0*M_water_0/xy_oz > V_fluid*M_water/xyz We know that the density of water must remain constant so as we throw the rock overboard, the mass of the fluid M_water increases. So if M_water > M_water_0 and the densities are equivalent, y_0 > y to keep a constant density so the water level will sink.
+Dr. Science Sc.D please summarize your conclusion yes the water level will go down. You did way more work than was needed. Try this. If the volume of the rock is V when the rock is in the boat, more water is displaced than V (Archimedes). When the rock is at the bottom the water displacement is V. Conclusion ==> the water level goes down when you throw the rock over board. .
+Dr. Science Sc.D In these problem the water level is taken relative to the walls of the pool. If it was to be taken in relation to the boat (and then the question would be if the boat sinks in a little bit or a little less, respectively the water level in relation to the boat would rise or would lower). Coincidently, I think the answer of both of the problems is that the water goes down.
In this case, the helium balloon is already in motion due to its upward buoyancy force, which is caused by the difference in density between the helium inside the balloon and the air outside. When the container is accelerated forward, the balloon, being part of the container, also experiences the same forward acceleration. Since the air in the container is also accelerated forward, there is no relative motion between the air and the helium balloon. Therefore, the buoyancy force acting on the balloon is not affected, and it will continue to move forward with the container. I think this make the concept INTUITIVE
I think the water will lower slightly. The stone will displace its volume of water when tossed in and therefore sink due to density. At this point I believe that the boat will float higher in the water and displace less. The boat was displacing water equal to the weight of the stone which is more water than the volume of the stone. If the stone was same density as water , the level stays the same.
I guess the reason for the last problem being the following; when he turns the glass upside down a tiny amount of liquid runs through the microscopic gap between the glass and the cardboard with a high speed, causing so much low pressure thus the Mg of the liquid is supported. Please correct if I'm wrong.
Just one word. Wow sir. What an amazing lecture. Wish I could meet you someday soon but it happens as if time doesn't allow,, but I'll change it. 😀. For the love of physics -Ayushman(India🇮🇳)
Water line goes down. The rock is added weight which the water must then displace by allowing more of the boat in the water. If the boat and its contents are to be considered a system, then just pretend their density is shared, Having a rock in the boat effectively increases its density (its weight per volume). Due to this the boat will sink to a point where Vwater*Pwater*Gravity = Weight of boat. The boat weighs more with the rock, when the rock is thrown away, the volume of water required to meet this wight is lessened. The boat rises, the water line goes down on the side.
EUREKA is from REKA, rekao (sam) which on Serbian means Told (I told). Reka also menas river, flows of something, in this case words. It is similar as rhetoRICS, where Rika means roar (also talking meaning, but more in animal terms).
{39:00 -->} You said that when a hole is made in the vessel, water will flow with the same velocity as in the 'syphon case'. But what if both were done simultaneously ? The approximation that v2~0 would not hold good right ?
Professor in the syphon demonstration, the reason why the water runs against gravity is probably due to two reasons........ 1) you mentioned that area of the tube is much smaller than area of the vessel so this means adhesive force inside the tube will dominate over the weight of the juice. 2) if suppose after juice starts flowing and at some point the flow breaks then at that point there is vaccum while at the end of the tube dipped in juice pressure is 1atm.....so this will also drive the juice upwards.......
If the boat has a flat bottom,or otherwise, and is raised to the level of the water,its weight will remain the same,so, if now the stone is thrown into the water will flow overboard so the level of the water will go down,just as the water in Archimedes bath the water fell to the floor.
I do not thin Bernoulli's equation is that Bizarre when you think about it. Pressure is a static energy measure while flow is a kinetic energy measure so it stands to reason that they would have a inverse relationship. No different than Amps and Voltage. Any measure of energy in motion will have a inverse static measurement as well. It only stands to reason at least that has been my observation as a mechanic. Cheers!
{Cranberry Juice} I think the cardboard is held to the glass because, some of the area of the cardboard that is outside the glass is at a low pressure than the area that is in contact with the juice. So, according to Bernoulli's principle the cardboard stays with the glass.
I've always thought the Archimedes legend was a bit much and probably inflated, mostly because Archimedes was an incredibly brilliant mathematician and engineer. You'd think he'd be reasonable. So I couldn't imagine him reacting that absentmindedly to such a relatively basic discovery. But if u consider that he spent an immense amount of effort trying to work out the areas and volumes of weird shapes, having discovered the volume of only a few (The sphere for example), then it becomes much more believable that he'd become euphoric after finding a general way to measure the volume of ANY object using some previously unknown function of nature. Which in that era suggested there may be some relationship between nature and mathematics.
For the iceburg in salt water, you have not accounted for the lower concentration of salt in the ice vs concentration of salt in the water. The iceburg should float higher than the 92% formula indicates.
I think the level of water will go down because when the stone was in the boat , the boat had to replace an amount of water which was equal to the weight of the man , the boat and the stone itself but later when the stone was thrown , the stone didn't replace anymore water than its equal volume . I presumed that the density of the stone is greater than the density of water. So the water level goes down . I am not sure if this is an appropriate answer or not .
@@mdsakhawathossen7088 yes you are right. The nice point in this problem is that everything depends on the density difference between the object you throw and the liquid of the swimming pool. If you empty a bottle of liquid which has the same density of the liquid in the swimming pool, the level remains the same. If you throw a stone the level goes down. Nice problem prof. Lewin! Inspiring as usual
@Jeremy Nathan Well, I'm not so sure about that.. I mean, if you fill the boat with water, the water in the pool still has to displace the same volume that the filled water weitghts, making the boat to sink a lil bit more and therefore, making the water level to rise... When you remove the water inside the boat, the only volume to be displaced is the one from the weight of the boat, making the water level to go down. So I don't think that the density of the rock (or any object inside the boat) plays an important role (as long as it is higher than the density of the air).
Since the density of the rock is so high. it’s volume is quite small, for a given mass, therefore, the water displaced by the rock is much less. The water level will go down
For the upside down glass trick: Upward Force caused by atmospheric pressure exerted on the cardboard is higher than the downward force produced by weight of water (plus the small force caused by the trapped air on top of the fluid).
The waterline must go down as the buoyant force isn't changing as the boat displaces the volume, however, the weight of the system(man and boat and rock) is reducing so the upthrust is greater now and with respect to the boat, the man shall see the water going down.
47:55 Sir please crt me if I'm wrong! In this case..when there is no cardboard covered upon the glass the pressure at any point on the surface of the juice must be 1 atm. After the cardboard has been kept on it and closed ,still it is 1 atm. When I think about what happens when we flip the glass which is closed with a cardboard the air molecules inside the glass goes up and juice comes down where the pressure of air inside the glass is not 1atm anymore and it would have become very low such that you can neglect the pressure of the air . Now if you look at the cardboard there are two pressures acting on it from both the sides,one is 1atm due to the atmosphere and the other one is, whatever the hpg of the juice is. And the hpg is way more less than 1 atm because it is not 10 metres ,if so, it would have pushed the cardboard and as result the cardboard is pushed towards the glass by 1atm pressure outside ! Correct me if I'm wrong sir!🤓
@@lecturesbywalterlewin.they9259 sir that was because you were just sucking out the air inside the tube and when you do that the atmospheric pressure pushes the fluid up you can do it until you are at 10 meters of height...the reason that you were not able to suck it above 1 metre in the previous demonstration was ... you sucked it at a single breath. you could have done that if you do that by taking a breath in between .otherwise you will not be sucking out all the air inside the tube which would actually resist the water coming up....the main reason behind that is... Your lung capacity..you may not be able to suck it more than 1metre at a time but you can do it more than 1 metre by breathing through your nose in between... correct me if I'm wrong sir
Abjo Das I’m unsure if this is correct but here’s my reasoning. When he inverts the cup, the empty portion above the juice is essentially a weak vacuum with very low pressure. By contrast, the piece of cardboard still feels the full atmospheric pressure pressing it against the cup. Interestingly, however, even though I haven’t done the maths for this claim - I don’t think this will work if the volume or density of liquid is too high. Essentially, you need a fine balance where Patmosphere > (Pcup + weight/area). Is this correct Dr. Lewin?
When the cup is turned upside down, the water wants to fall out. The air-filled cavity is therefore stretched a bit as the gravity pulls down the water. This reduces the air pressure inside the cup, since increasing volume reduces pressure. Eventually this lower pressure pulls upwards with the same force as the weight of the water pulls downwards. The water is now kept in place and the pressure inside is lower than atmospheric pressure outside.
Good evening professor. I've tried different experiments about the upended cup. First I assumed it's because of the shape of the cup because the base is narrow and the opening is wide and when upended the Pressure*Area inside is lower than the P*A outside. However, the opposite shape also worked... I tried different quantity of water inside. But it worked no matter it's fully filled or with just little water. It has nothing to do with the covering material whether it's water-absorbable such as cardboard, a piece of paper or a piece of plastic sheet. But when I made a hole at the bottom of the cup and upended it, it didn't work. So It's absolutely due to the air pressure difference between the inside and outside. Since 1/100 difference of 1 atm produces 10 gw/cm2, it's possible. I noticed when upended if some liquid was intentionally let out (just some drops on the paper) it had high success rate. So my conclusion is that the volume of air inside increases due to the loss of some liquid (leaked out or absorbed in the paper) which decreases the pressure of air so the pressure difference supports the paper. Could you verify this ? Thanks a lot !
Professor, your lectures are thought-provoking. I want to share with you an application of this phenomenon in my field. I'm a dentist and when we fabricate a upper complete denture we have to make sure the borders are in close contact with the soft tissue so the saliva can adhere in between. We call this "border sealing" and when the patient put on the denture they will squeeze out some liquid or air inside and good border seal produces negative pressure therefore the denture will be sucked tight on the hard palate (just like the cardboard). There's a syndrome called xerostomia which means lack of saliva. Edentulous patients with this syndrome suffer from denture dislodgment when chewing because there'e no saliva to provide good adhesion.
Hello. If you have a leak out of fluid it could not work any more I think. And if you have a Piece of plastic, water can not be absorbed. In my opinion the Explanation is another one: if cardboard Surface is 20cm^2 than from atmospheric pressure you get 20Kg (200N) pushing on the paper. Instead, inside the galss, the weight of the water can be for example 0.5Kg. So you have 20Kg pushing against 0.5 Kg for example. The air inside the glass is pushing on the water with atmospheric pressure but the total weight that you get pushing down on the paper is always the weight of the water! What do you think of this solution?
Regarding the Problem of the cranberry Juice!! If cardboard/paper Surface is 20cm^2 than from atmospheric pressure (100000 Pa) we get 20Kg (200N) pushing on the paper. Instead, inside the galss, the weight of the water can be for example 0.5Kg. So we have 20Kg pushing against 0.5 Kg for example. The air inside the glass is pushing on the water with atmospheric pressure but the total weight (air + juice) that is pushing down on the paper is the weight of the water: so much smaller that force produced by air pressure! Professor Lewin is this correct?
sir I think that the water level will go down because the weight of the boat decreases. and the density of the water becomes higher than that of the density of the boat
sir in your earlier fluid mechanics video in which 5 meter hose magic was shown could we have even generated 0 atm with continuous block and inblow method
@22.22 what happens when i suck out all the air inside (vacuumed) that room? because the balloon has some pressure inside, will it explode? just curious. Btw I always enjoy your lectures.
Cranberry juice :: Because of the difference in pressure of inside air, in glass and outside the water will hang on .Also adhesion is critical as I tried many times with different cardboard... Plz correct me professor If i am wrong
As the water will slip down a little bit the volume available for air will be increased and according to BOYLE'S LAW the pressure will decrease and the pressure difference will balance weight of water..Is this right professor ?
Lectures by Walter Lewin. They will make you ♥ Physics. Sir plzz tell me and I did a rigorous search on Google ,,just tell me why the pressure inside the glass decreases ..plzzzzzzz
*the water lavel will go down.* *At first the mass of the rock is displacing the water not the volume. As the rock is highly densed, when put into water it will displace the water same as it's volume which is lesser than when the water is displaced by it's mass.
Dear Prof. Walter Lewin, The Ping ball- Funnel experiment was awesome. In case of inverted position (blowing down), what happens when the velocity is kept on increasing ? Will the ball fall down, stick to the top or stabilizes at level below the initial level ?
Dear professor for me it was astonished, it's plead to u please let me know as we r always taking direction of acceleration due to gravity upwards or downwards only irrespective of what is the direction of external acceleration whether horizontal, vertical but g l never studied about horizontal direction of acceleration due to gravity as u mentioned in case of Apple and balloon. Please reply
The level of water will go down cause in the begging to counter total weight(man+stone+boat) greater amount of buoyant force will act and thus greater amount of water will be displaced and later weight(man+boat) decreases thus amt. Of water displaced also decreases and thus level decreases🤟
Can I check if this is correct for swimming pool question thats why less water is displaced when the rock is in the pool? Volume(rock) x Density(rock) x G > Volume(water displaced) x Density(water) x G
The demonstrations with angular momentum and torques were less of a shock to me than the helium balloon going against gravity. Never really gave any thought to why it goes to the perceived "up".By the way, if you took the balloon out of the case would it still go against the perceived gravity? If you accelerated it? Thank you for the lecture series.
>>>a shock to me than the helium balloon going against gravity.>>> does it also shock you that if you release a piece of wood at the bottom of a swimming pool that it will go up against gravity? It's the same physics.
True, but when I see that balloon going to the front when accelerated will always seem strange, even with me understanding the physics behind it. Also I think that if you took the balloon outside the case it would behave just like the apple, because there wouldn't be a pressure difference. Am I correct?
when the closed container is accelerated forward the air pressure will be larger in the back than in front. That's the key. when you release the He filled balloon in your room it goes up becoz the atmospheric pressure below the balloon is large than above.
Dear Mr. Walter Lewin, your lectures are such a pricess gift to me during this hard time. I had always struggled with course name fluid mechanics because the teachers in my school did not explained and demonstrated as clear as you did. It help me regain my motivation to strive to become an engineer. After all, thanks you a lot!!!!.
Why am I watching this?
Actually I'm a lecturer in medical school and I had a lecture on blood flow in blood vessels which is based on fluid physics and Bernoulli's rule
I couldn't think of anything better to prepare for my lecture than watching your lecture on fluid physics
it's a good thing that even doctors are watching your lectures
Lot of respect professor
From Iraq
Watching this great lectures series during corona quarantine to enchance my intellect and educate myself in the meantime more.
me tooo
from India ..
That's a very wise decision. Good luck!
Me too also
nile red is way better
Your service and dedication to teach every hungry mind is truly selfless.
Resonance??
Ok go
Thank you very much Prof. Lewin, I have to take my final oral test in physics for engineering next Monday and after all this years of your lecture being around, you are saving my life as those of many other students ! These should be saved as hsitory of science for years to come !
THATS HOW PHYSICS IS TAUGHT.....I AM COMPLETELY AMAZED......
EARLIER I FOUND FLUIDS VERY DIFFICULT TO UNDERSTAND BUT NOW I ENJOYED LEARNING IT IN THIS LECTURE......THE PRACTICAL DEMONSTRATION WERE JUST TOO GOOD.....
Glad to hear that
Just because of you....
Today I can feel hydrostatics practically....
Great thanks to W. Lewin sir...
Love from India 💕💖
Not only hydrostatics....
Well it's physics which I seem to feel like....
Hope I could ever meet you....
Wanna study in MIT but you don't teach nowadays...
Still your lectures are powerful..
Wonderful!
@@lecturesbywalterlewin.they9259 🙏🙏🙏
@@lecturesbywalterlewin.they9259 sir, is there any video of quantum mechanics you may have done
@@lecturesbywalterlewin.they9259 sir, at 20:02 the balloon will rise considering there is an atmosphere, because of helium you mentioned, there is a buoyant force even when there is no gravitational acceleration, because of difference in densities.
You are the best teacher ever ! I really love your teaching ! I wanna wish you a long and happy life !! You are someone irreplaceable in my heart ...I wanna meet you so badly and say how much your teachings helped me !
Also I really loved your birthday series 💕
Thank you so much again professor Lewis!
Now i am preparing for my neet exam i have many boubts in fluid mechanics after this class i feel comfortable to solve problems of fluid mechanics thank you professor 😊for wonderful class love from India ❤💓
Thanks again sir... For the first time I have felt Fluids... Before this I had been thinking that in fluid part there is nothing but puzzles... Now I feel very comfortable in fluids.... Can't pay you against this but infinite respect will always be for you...
I think that to understand the pool level problem, it helps being exaggerated. Consider a super dense object, 1 ton with the size of a coin. Now imagine we drop it in a boat that could carry it. Intuitively, the water level would rise significantly to counter the added weight. Now if we drop it in the water, there will be a massive weight relief in the boat and the buoyant force required to keep it floating will be therefore much less. The object has the size of a coin, so the volume of water displaced by it as it sinks is negligible and so is it's buoyancy. The total buoyant force that the water produces will be way less, and so the pool level will drop. As long as the boulder's density is greater than that of water, the same reasoning could be applied.
Your lectures contain all theory demonstration and application ,looking forward to binge watch all your content
Dr. Lewin's ability to describe and draw complex principles is amazing.
what a MIND BLOWING lecture
This is all nice and stuff but the mind blowing part is 0:49
This is the best lecture i have found on Archimedes' principle..just wonderful demonstration.
I am in love with physics just because of you. I left my job to teach physics..❤️
In the answer to your waterline of swimming pool question, I think it will go down. When we seperate the stone from the boat, the waterline goes down more than the volume of the stone since the density of the boat is lower than the stone. So when the stone sinks the net effect is a lower waterline.
The water level stays the same
He (Walter Levin) said it would change! Need I say he is right?
How can he make the dotted lines so effortless? Pure skill!
Honestly I haven't the faintest idea.
he uses the other, or wrong point, of the now angled piece of chalk with pressure and speed on the board to make it "skip" like a stone on water
47:28 We can all admire the greatness of the MIT chalks in this shot... no wonder why they sound so satisfying
Best physics teacher as well as the best physics UA-cam I've ever come across. No one else comes even close
:)
I have an answer to the swimming pool question posed at 11:20, but I also have two additional questions regarding the same.
The answer to your question: The water level goes down, because if we assume that the density of the stone is greater than the density of water (a reasonable and intuitive assumption), then the stone will sink to the bottom of the pool and settle down. When that happens, the buoyant force that was originally holding up the stone before it was thrown into the water will now be shared between the upward normal force at the bottom of the pool AND the new buoyant force together. Therefore, the new buoyant force is only a fraction of the original buoyant force. Since lower buoyant force displaces lesser water than higher buoyant force, the water level in the pool goes down.
(In the event that this is a bottomless pool, then the stone will keep accelerating downwards under its own weight as there is a net force pulling the stone downwards. This net force can only exist if the new buoyant force is lower than its weight, and hence the new buoyant force is lower than the original buoyant force. Thus, the pool level drops).
However, I have two variations on the stone that was thrown out:
Q 1. What if, instead of a stone, a fish of the same volume as the stone but of the same density as water is thrown out of the boat and into the pool such that the fish goes below the surface of the water but doesn't sink? Would the water level in the pool go up, stay the same, or drop?
My (probably wrong) answer: The water level would stay the same. Since the buoyant force is neither being shared by any normal force at the bottom of the tank (as the fish isn't sinking to the pool floor) and nor is the fish accelerating downwards, there is no change in the total buoyant force, which means there is no change in the displaced water. Hence, the water level stays the same.
In fact, when the fish is thrown out of the boat, for the brief moment that it is in the air, the water level drops (as the buoyant force drops), but once the fish is under the surface of the water, the water rises again by the same amount that it had dropped, thus equalizing the level.
Q 2. What if, instead of a fish, a wooden block of the same volume as the fish but of lower density than water is thrown into the pool such that it floats? Would the water level in the pool go up, stay the same, or drop?
My (definitely) wrong answer: The water level would stay the same. There is no change in the total buoyant force here either, as the wooden block isn't sinking to the floor.
But I still feel that my answer to this last question is wrong because my inductive reasoning would have led me to believe that if a higher-density object (stone) lowers the water level in the pool, and an equal-density object (fish) keeps the water level the same, then a lower-density object (wood) would have raised the water level. But that is not what the equilibrium equations tell me.
Please correct me if I am wrong in any of the above.
And yes, your lectures DO make me ♥ physics :)
Watch my lectures - Your answers are there!
Wow.i enjoyed every sec of this lecture.
You are the best.
44:55 "That's the reason she couldn't get it up. That's what Bernoulli does to you" - Lewis
professor i am from i india, i love to watch your lectures , you explain practically everything. i am in class 10 but i understand everything. it creates me a passion to get phd in physics .thank you professor
I wish I could actually sit there and learn these lectures from you
Whenever a body is partially or completely submerged in a fluid,it will experience an up thrust which is equal to weight of the fluid which has been displaced.
another CLASSIC Lecture by Dr Walter Lewin :D .. great Demonstrations, Excellent Chalk Board Graphics.... Thank you.
:)
At 38:55, syphen experiment.
when we suck the water at other end, we created a differential pressure at top and because of this water is raised to max height of tube. once it crosses the max height, gravitational P.E is converted to K.E which resulted into water fall.
Woww
YES! I studied these exact same concepts, only to better understand them here.
This lecture is awesome. Thank you ❤️
:)
Um, King Hieron II was called a "virtuous man" by Machiavelli and he had a long and seemingly friendly relationship with Archimedes, and his father (a court astronomer) so he'd hardly "kill" him if he got it wrong. :P Great lecture.
Hello Dr. Lewin, this is what I think about the question about the water level changing when you throw the rock overboard.
I am a little confused on what you mean by will the water line remain change. Relative to the boat or relative to the walls of the pool?
Initially, the water is at a height y_0. The volume of the fluid V_water is given by x*y_0*z where x is width and z is the thickness. For the boat to float, the F_b must equal the weight of the boat and rock. So F_B_0 = (M+m)g (1) where m is the mass of the rock and M is the mass of the boat.
Now analyzing the situation when you throw the rock overboard. The mass of the entire boat-rock system must change because we no longer have m. So, in order to float, F_B =Mg. (2)
So applying Archimedes Principle to (1.1) V_fluid_0*p_water = V_(M+m)*p_(M+m) and to (2.2) V_fluid*p_water = V_M*p_M . F_B_0 > F_B because of (1) and (2).
Therefore. V_fluid_0*p_water > V_fluid*p_water.
p_water = M_water/V_water = M_water/x*y_0*z so we have it that:
V_fluid_0*M_water_0/xy_oz > V_fluid*M_water/xyz
We know that the density of water must remain constant so as we throw the rock overboard, the mass of the fluid M_water increases. So if M_water > M_water_0 and the densities are equivalent, y_0 > y to keep a constant density so the water level will sink.
+Dr. Science Sc.D please summarize your conclusion yes the water level will go down. You did way more work than was needed. Try this. If the volume of the rock is V when the rock is in the boat, more water is displaced than V (Archimedes). When the rock is at the bottom the water displacement is V. Conclusion ==> the water level goes down when you throw the rock over board. .
+Dr. Science Sc.D In these problem the water level is taken relative to the walls of the pool. If it was to be taken in relation to the boat (and then the question would be if the boat sinks in a little bit or a little less, respectively the water level in relation to the boat would rise or would lower). Coincidently, I think the answer of both of the problems is that the water goes down.
In this case, the helium balloon is already in motion due to its upward buoyancy force, which is caused by the difference in density between the helium inside the balloon and the air outside. When the container is accelerated forward, the balloon, being part of the container, also experiences the same forward acceleration.
Since the air in the container is also accelerated forward, there is no relative motion between the air and the helium balloon. Therefore, the buoyancy force acting on the balloon is not affected, and it will continue to move forward with the container.
I think this make the concept INTUITIVE
I think the water will lower slightly. The stone will displace its volume of water when tossed in and therefore sink due to density. At this point I believe that the boat will float higher in the water and displace less. The boat was displacing water equal to the weight of the stone which is more water than the volume of the stone. If the stone was same density as water , the level stays the same.
Amazing lecture. And the rod he used in the water to explain stable and unstable equilibrium has colours of Indian flag. :)
After watching your lectures
I sometimes doubt my intelligence it seems like what the hell i have studied from past 2 years😂😂
I guess the reason for the last problem being the following; when he turns the glass upside down a tiny amount of liquid runs through the microscopic gap between the glass and the cardboard with a high speed, causing so much low pressure thus the Mg of the liquid is supported. Please correct if I'm wrong.
Here; the water level goes down because when the rock was in the the boat the displaced water, is greater then when t it through down.
what if the rock is floating on the water?
@@neillin8212 then it stays the same
Thank you so much for these meneer Lewin, ze zijn erg behulpzaam aan mijn understanding van physics
Just one word. Wow sir.
What an amazing lecture. Wish I could meet you someday soon but it happens as if time doesn't allow,, but I'll change it. 😀. For the love of physics -Ayushman(India🇮🇳)
Water line goes down. The rock is added weight which the water must then displace by allowing more of the boat in the water. If the boat and its contents are to be considered a system, then just pretend their density is shared, Having a rock in the boat effectively increases its density (its weight per volume). Due to this the boat will sink to a point where Vwater*Pwater*Gravity = Weight of boat. The boat weighs more with the rock, when the rock is thrown away, the volume of water required to meet this wight is lessened. The boat rises, the water line goes down on the side.
Binging bigtime on these lectures rn
EUREKA is from REKA, rekao (sam) which on Serbian means Told (I told). Reka also menas river, flows of something, in this case words.
It is similar as rhetoRICS, where Rika means roar (also talking meaning, but more in animal terms).
"they will get some of their 25000 dollars intuition back"
Thank you so much professor,...
Very easy to understand with your demonstration...
{39:00 -->}
You said that when a hole is made in the vessel, water will flow with the same velocity as in the 'syphon case'. But what if both were done simultaneously ?
The approximation that v2~0 would not hold good right ?
both can be used simultaneously. Each would work as if the other was not there.
Wow! This is very interesting lecture.
Can't describe the experience, amazing Thank You🙏🙏
Professor in the syphon demonstration, the reason why the water runs against gravity is probably due to two reasons........
1) you mentioned that area of the tube is much smaller than area of the vessel so this means adhesive force inside the tube will dominate over the weight of the juice.
2) if suppose after juice starts flowing and at some point the flow breaks then at that point there is vaccum while at the end of the tube dipped in juice pressure is 1atm.....so this will also drive the juice upwards.......
google syphon - it's all there
If the boat has a flat bottom,or otherwise, and is raised to the level of the water,its weight will remain the same,so, if now the stone is thrown into the water will flow overboard so the level of the water will go down,just as the water in Archimedes bath the water fell to the floor.
do you have any video on surface tension , capillarity and all that stuff
I may have covered some of it in 8.01. I do not remember.
I will become a physicist its my aim ...
Now i am a 5 semester
.. thanks sir from Pakistan
I remember this Archimedes problem from your video - Problem #29 (=
I do not thin Bernoulli's equation is that Bizarre when you think about it. Pressure is a static energy measure while flow is a kinetic energy measure so it stands to reason that they would have a inverse relationship. No different than Amps and Voltage. Any measure of energy in motion will have a inverse static measurement as well. It only stands to reason at least that has been my observation as a mechanic. Cheers!
{Cranberry Juice}
I think the cardboard is held to the glass because, some of the area of the cardboard that is outside the glass is at a low pressure than the area that is in contact with the juice. So, according to Bernoulli's principle the cardboard stays with the glass.
incorrect
My name is Emmanuel from Nigeria
I love the way you teach.
Please I want you to be my tutor in physics so that I can impact knowledge in my students
I've always thought the Archimedes legend was a bit much and probably inflated, mostly because Archimedes was an incredibly brilliant mathematician and engineer. You'd think he'd be reasonable. So I couldn't imagine him reacting that absentmindedly to such a relatively basic discovery. But if u consider that he spent an immense amount of effort trying to work out the areas and volumes of weird shapes, having discovered the volume of only a few (The sphere for example), then it becomes much more believable that he'd become euphoric after finding a general way to measure the volume of ANY object using some previously unknown function of nature. Which in that era suggested there may be some relationship between nature and mathematics.
Hehe
Not a basic discovery at all.
I love each one of your lecture, each one. Thank you so much!
Because of atmosphere pressure, the juice will not fall out when the glass turned over. I knew and tried this when I was 8 years old.
For the iceburg in salt water, you have not accounted for the lower concentration of salt in the ice vs concentration of salt in the water. The iceburg should float higher than the 92% formula indicates.
I think the level of water will go down because when the stone was in the boat , the boat had to replace an amount of water which was equal to the weight of the man , the boat and the stone itself but later when the stone was thrown , the stone didn't replace anymore water than its equal volume . I presumed that the density of the stone is greater than the density of water. So the water level goes down . I am not sure if this is an appropriate answer or not .
correct
Thank you :) .
@@mdsakhawathossen7088 yes you are right. The nice point in this problem is that everything depends on the density difference between the object you throw and the liquid of the swimming pool. If you empty a bottle of liquid which has the same density of the liquid in the swimming pool, the level remains the same. If you throw a stone the level goes down. Nice problem prof. Lewin! Inspiring as usual
@Jeremy Nathan Well, I'm not so sure about that.. I mean, if you fill the boat with water, the water in the pool still has to displace the same volume that the filled water weitghts, making the boat to sink a lil bit more and therefore, making the water level to rise... When you remove the water inside the boat, the only volume to be displaced is the one from the weight of the boat, making the water level to go down. So I don't think that the density of the rock (or any object inside the boat) plays an important role (as long as it is higher than the density of the air).
Since the density of the rock is so high. it’s volume is quite small, for a given mass, therefore, the water displaced by the rock is much less. The water level will go down
Should the Navier-Stokes equation be covered for this chapter?
no , i think this video is only for +1 or +2 only , but what are you asking is a undergraduate concept :-) !!!
7:00
"Consider a spherical cow"
For the upside down glass trick: Upward Force caused by atmospheric pressure exerted on the cardboard is higher than the downward force produced by weight of water (plus the small force caused by the trapped air on top of the fluid).
the open air pressure and the air pressure in the glass cancel out. they are both 1 atmosphere.
The waterline must go down as the buoyant force isn't changing as the boat displaces the volume, however, the weight of the system(man and boat and rock) is reducing so the upthrust is greater now and with respect to the boat, the man shall see the water going down.
correct
47:55 Sir please crt me if I'm wrong! In this case..when there is no cardboard covered upon the glass the pressure at any point on the surface of the juice must be 1 atm. After the cardboard has been kept on it and closed ,still it is 1 atm. When I think about what happens when we flip the glass which is closed with a cardboard the air molecules inside the glass goes up and juice comes down where the pressure of air inside the glass is not 1atm anymore and it would have become very low such that you can neglect the pressure of the air . Now if you look at the cardboard there are two pressures acting on it from both the sides,one is 1atm due to the atmosphere and the other one is, whatever the hpg of the juice is. And the hpg is way more less than 1 atm because it is not 10 metres ,if so, it would have pushed the cardboard and as result the cardboard is pushed towards the glass by 1atm pressure outside ! Correct me if I'm wrong sir!🤓
you have not explained why I was able to suck up liquid over a distance of about 5 meter.
@@lecturesbywalterlewin.they9259 sir that was because you were just sucking out the air inside the tube and when you do that the atmospheric pressure pushes the fluid up you can do it until you are at 10 meters of height...the reason that you were not able to suck it above 1 metre in the previous demonstration was ... you sucked it at a single breath. you could have done that if you do that by taking a breath in between .otherwise you will not be sucking out all the air inside the tube which would actually resist the water coming up....the main reason behind that is... Your lung capacity..you may not be able to suck it more than 1metre at a time but you can do it more than 1 metre by breathing through your nose in between... correct me if I'm wrong sir
@@lecturesbywalterlewin.they9259 sir please reply sir😅🤓
How did the cranberry juice not fall when it was tilted over in 48:00 min?
Abjo Das I’m unsure if this is correct but here’s my reasoning. When he inverts the cup, the empty portion above the juice is essentially a weak vacuum with very low pressure. By contrast, the piece of cardboard still feels the full atmospheric pressure pressing it against the cup. Interestingly, however, even though I haven’t done the maths for this claim - I don’t think this will work if the volume or density of liquid is too high. Essentially, you need a fine balance where Patmosphere > (Pcup + weight/area). Is this correct Dr. Lewin?
When the cup is turned upside down, the water wants to fall out. The air-filled cavity is therefore stretched a bit as the gravity pulls down the water. This reduces the air pressure inside the cup, since increasing volume reduces pressure.
Eventually this lower pressure pulls upwards with the same force as the weight of the water pulls downwards. The water is now kept in place and the pressure inside is lower than atmospheric pressure outside.
We usually did it with the glass which was full [of water].
Surprising that it works also with a glass which is 25% empty.
5:26 how did he know about gravitational acceleration?
Good evening professor. I've tried different experiments about the upended cup. First I assumed it's because of the shape of the cup because the base is narrow and the opening is wide and when upended the Pressure*Area inside is lower than the P*A outside. However, the opposite shape also worked... I tried different quantity of water inside. But it worked no matter it's fully filled or with just little water. It has nothing to do with the covering material whether it's water-absorbable such as cardboard, a piece of paper or a piece of plastic sheet. But when I made a hole at the bottom of the cup and upended it, it didn't work. So It's absolutely due to the air pressure difference between the inside and outside. Since 1/100 difference of 1 atm produces 10 gw/cm2, it's possible. I noticed when upended if some liquid was intentionally let out (just some drops on the paper) it had high success rate. So my conclusion is that the volume of air inside increases due to the loss of some liquid (leaked out or absorbed in the paper) which decreases the pressure of air so the pressure difference supports the paper. Could you verify this ? Thanks a lot !
IN ADDITION adhesion is CRUCIAL!
Thank you professor! I forgot the existance of adhesion force. :p
Professor, your lectures are thought-provoking. I want to share with you an application of this phenomenon in my field. I'm a dentist and when we fabricate a upper complete denture we have to make sure the borders are in close contact with the soft tissue so the saliva can adhere in between. We call this "border sealing" and when the patient put on the denture they will squeeze out some liquid or air inside and good border seal produces negative pressure therefore the denture will be sucked tight on the hard palate (just like the cardboard). There's a syndrome called xerostomia which means lack of saliva. Edentulous patients with this syndrome suffer from denture dislodgment when chewing because there'e no saliva to provide good adhesion.
Hello. If you have a leak out of fluid it could not work any more I think. And if you have a Piece of plastic, water can not be absorbed. In my opinion the Explanation is another one: if cardboard Surface is 20cm^2 than from atmospheric pressure you get 20Kg (200N) pushing on the paper. Instead, inside the galss, the weight of the water can be for example 0.5Kg. So you have 20Kg pushing against 0.5 Kg for example. The air inside the glass is pushing on the water with atmospheric pressure but the total weight that you get pushing down on the paper is always the weight of the water!
What do you think of this solution?
43:00 what a prank!!
Regarding the Problem of the cranberry Juice!!
If cardboard/paper Surface is 20cm^2 than from atmospheric pressure (100000 Pa) we get 20Kg (200N) pushing on the paper. Instead, inside the galss, the weight of the water can be for example 0.5Kg. So we have 20Kg pushing against 0.5 Kg for example. The air inside the glass is pushing on the water with atmospheric pressure but the total weight (air + juice) that is pushing down on the paper is the weight of the water: so much smaller that force produced by air pressure!
Professor Lewin is this correct?
Thank you so much, you will be remembered sir
Thank you, Professor Lewin!
45:04 I lost my mind when the ping pong ball wasn't falling
In your questions about the boat and the rock, does the water drop to level ?! I concluded that, is it correct?
sir I think that the water level will go down because the weight of the boat decreases. and the density of the water becomes higher than that of the density of the boat
U r mind blowing professor lewin
sir in your earlier fluid mechanics video in which 5 meter hose magic was shown
could we have even generated 0 atm with continuous block and inblow method
Equation of continuity states that A1V1=A2V2 . If A2V1 . How does the velocity increase(how does the fluid gain energy)?
It doesn't gain energy, the velocity of the fluid increases, but the mass traveling through it in a set time will decrease.
bernaolli eq at 29:00
In bernoullis,while solving at the same Y...(33min probably)..u are just saying like P1
Always learn something new.
Excellent lecture Sir. Thanks and Regards 🙏🙏🙏🙏🙏🙏🙏🙏
Love the way he says Boyant (Buoyant)
@22.22 what happens when i suck out all the air inside (vacuumed) that room? because the balloon has some pressure inside, will it explode? just curious. Btw I always enjoy your lectures.
Most definitely, for any real balloon.
Isn't coanda effect better explanation for the stability of the ping pong ball than bernoulli principle???
At, 35:08 P1
Cranberry juice ::
Because of the difference in pressure of inside air, in glass and outside the water will hang on .Also adhesion is critical as I tried many times with different cardboard... Plz correct me professor If i am wrong
incorrect
As the water will slip down a little bit the volume available for air will be increased and according to BOYLE'S LAW the pressure will decrease and the pressure difference will balance weight of water..Is this right professor ?
no
Lectures by Walter Lewin. They will make you ♥ Physics. Sir plzz tell me and I did a rigorous search on Google ,,just tell me why the pressure inside the glass decreases ..plzzzzzzz
question unclear. How many minutes into the lecture?
In which lecture surface tension and viscosity are included?
*the water lavel will go down.*
*At first the mass of the rock is displacing the water not the volume. As the rock is highly densed, when put into water it will displace the water same as it's volume which is lesser than when the water is displaced by it's mass.
Dear Prof. Walter Lewin,
The Ping ball- Funnel experiment was awesome. In case of inverted position (blowing down), what happens when the velocity is kept on increasing ? Will the ball fall down, stick to the top or stabilizes at level below the initial level ?
It will probably depend on the kind of funnel that is used but given enough air pressure (from above) it will probably fall out.
Thank you.
Dear professor for me it was astonished, it's plead to u please let me know as we r always taking direction of acceleration due to gravity upwards or downwards only irrespective of what is the direction of external acceleration whether horizontal, vertical but g l never studied about horizontal direction of acceleration due to gravity as u mentioned in case of Apple and balloon.
Please reply
Thank u so much sir
Most welcome
The level of water will go down cause in the begging to counter total weight(man+stone+boat) greater amount of buoyant force will act and thus greater amount of water will be displaced and later weight(man+boat) decreases thus amt. Of water displaced also decreases and thus level decreases🤟
Can I check if this is correct for swimming pool question thats why less water is displaced when the rock is in the pool?
Volume(rock) x Density(rock) x G > Volume(water displaced) x Density(water) x G
use google
sorry but wouldn't it be the inertia that keeps the apple fall backwards or stay in position while the comparment moves
loved the lecture btw
NOOOOOOOOOOO
@@lecturesbywalterlewin.they9259 There is no force to accelerate the apple only the string so inertia must play a role.
I think you are correct!
Sir you like our super sir
Is there some intuitive reason for why the pressure is lower at a point in a fluid where the velocity is higher? Thanks very much for the videos.
The demonstrations with angular momentum and torques were less of a shock to me than the helium balloon going against gravity. Never really gave any thought to why it goes to the perceived "up".By the way, if you took the balloon out of the case would it still go against the perceived gravity? If you accelerated it? Thank you for the lecture series.
>>>a shock to me than the helium balloon going against gravity.>>>
does it also shock you that if you release a piece of wood at the bottom of a swimming pool that it will go up against gravity? It's the same physics.
True, but when I see that balloon going to the front when accelerated will always seem strange, even with me understanding the physics behind it. Also I think that if you took the balloon outside the case it would behave just like the apple, because there wouldn't be a pressure difference. Am I correct?
when the closed container is accelerated forward the air pressure will be larger in the back than in front. That's the key. when you release the He filled balloon in your room it goes up becoz the atmospheric pressure below the balloon is large than above.