Can YOU solve these Tricky Hydrostatic Physics Puzzles? | Hydrostatic Paradox + more
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- Опубліковано 14 лип 2024
- Three tricky hydrostatics physics puzzles / problems are introduced, the solution is going to be covered in the second part. The first is the famous hydrostatic paradox, the hydrostatic paradox involves some initially strange forces due to equal pressures even with different volumes. This is a great physics problem
The second is the famous ping pong ball and golf ball balance. It is counterintuitive to those not experienced with fluid statics. Hydrostatics are really quite tricky (it even has the hydrostatic paradox named after it). I believe this is in 200 Puzzling Physics Problems which is quite a coincidence (since the names are similar). Also this was in the Moscow Physics Olympiad a long time ago. International Physics Olympiad (IPhO) really have some great problems.
The third is a puzzling perpetual motion machine. Nothing seems wrong with it, except that it produces infinite energy which doesn't make sense. Another great, tricky physics problem.
These hydrostatic physics puzzles are very fun. There are many more! I encourage you to find more on your own :D.
This video was sort of inspired by veritasium's puzzling physics problems that he uploaded a while back.
Outro:
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Finally by Loxbeats / loxbeats
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Solutions are now out! ua-cam.com/video/Zo5_8KURECg/v-deo.html
For the first question we should realise that the sidewalls contribute to the vertical force equilibrium. The extra weight of water due to the extra volume is compensated by the vertical component of the pressure force which is always perpendicular to the surface. The different glasses will however excert different pressures on the ground.
The left side of the scale will tip upwards. Since both sides carry an equal amount of weight of water. But a buoyant foce is applied to the left side as well. If it was somehow possible to drill a hole through the bottom of the left container and tie the rope to the floor. I'd think the scale would remain in equilibrium.
The last question is tricky. One way to see if our setup will rotate is to check if a net torque is excerted onto it. We can neglect the weight of our ping pong balls since the torques due to their weight would cancel out for the left and right side. Therefore we'll only have to look at the force applied due to the water.
We'll start our perpetuum mobile machine with a couple of balls already submerged and another approching our magical tank bottom. That allows for ping pong balls to flow through while keeping the water inside. All is well and the submerged balls contribute a positive torque to our machine and it starts rotating. Until a new ball approaches our barrier. The force on this new ping pong ball will be downwards and thus apply a torque in the opposite sense. But is it enough to cancel the positive contributions by all the other submerged ping pong balls? Unfortunately yes...
As the new ping pong ball flows through the barrier it will experience an increasing downwards force equal to the weight of the fluid column on it's cross section that has already passed. The maximum downward foce it will experience is at a time when it's halfway through. At that point the buoyant force of the other ping pong balls is always less. Since the volume displaced by the submerged ping pong balls is less than the water column. Because they are a part of the column. So an equilibrium will have been reached before this moment and our beautiful perpetuum mobile machine will have come to a standstill.
excitedly awaiting the solutions
:D
You have quite improved the voice from the first video..nice it's understandable now.... And problems are nice
Thanks! I definitely tried to speak more clearly this time :D
@@SlippingHexagons where are you from? US or Canada?
@@akiiiphysics3345 I'm from the US :)
@@akiiiphysics3345 Orz gawd
@@SlippingHexagons This video is 3years ago, but seen a 1yr ago video, that one also more hard to detect the words. Speak slowly & clearly when you record the video then make it faster by editing. Ok, the contents are good, keep going on.
In the second puzzle, the scale deviates to the right. Because the thread is connected to the bottom of the container on the left side, and the tension force of the thread is equal to the difference between the buoyancy force and the weight of the object, and this force enters the container upwards. On the other hand, the reaction of the buoyancy force on the object is added to the water and the whole set downwards. So, the added forces in the set on the right are added to the size of the weight of the light object. On the right side, however, the tension force of the thread does not enter the container because it is not connected to the container. The only force added to the set is the buoyant force that enters the object. Now, since the volume of the two objects is the same, then the buoyancy force vector is the same for both objects, and the weight of the light object on the left side is less than the buoyancy force, so by comparing, we can see that the force added to the container on the right side is more.
In the first puzzle. The reason that despite the different amounts of water in the containers, the force on the bottom of the container from the water is the same is the effect of the force of the walls of the container. In the figure on the left, where the amount of water in the container is more than the others, due to the walls facing outwards, part of the weight of the water is neutralized by the walls, and only part of the water weight, which is the same as the weight of the water in the middle container, is applied to the bottom of the container In the middle container, where the force of the wall has no effect and the weight of the water has entered the bottom of the container, and in the third container, because the walls are facing inward, the vertical reaction force of the wall adds to the weight of the water, and the force on the bottom of the container is greater than the weight of the water and It is equal to the weight of the water in the middle container, in this way, all three apply the same force to the bottom of the container
My guesses (not very confident):
1. For the first and third cups, if you looked at the forces from above from each point in the cross-section, you'd get vectors with varying angles, and so while their individual magnitudes may be greater/smaller, the vertical component stays the same between them, even with the increased or decreased mass.
2. The scale tips up on the left and down on the right. The rock is held externally, so its mass won't affect the scale. The mass of water is the same on both sides, but the ping pong ball will apply a buoyancy force lifting the left side up.
3. Well, the diagram makes it look like the water would just flow out. For my actual answer, I think that the ping pong balls would lose energy as they enter the water (though I'm not sure if that just counts as friction/viscosity)
btw nice music at the end :D
Wow, this is a very erudite answer. For the second one, I thought that the sea saw would tip left because the stone wants to settle down and the ping pong ball wants to settle up. But now I see why that is specious. As you said, the stone is held externally, so its mass won't affect the scale, and buoyancy force will lift the left side up. Looks like I forgot all of my high school physics in 3 years of computer science... Anyways, nice video!
@@bigphatballllz Well, I want to leave my guesses as they are, but talking to some other friends, some said that the rock may have an effect.
They said this would be because the water would push up on the rock due to buoyancy, and the rock would push back (newton's third law), which would push the right-side further down. [I'm not sure though, guess we got to wait for the answers video]
@@bigphatballllz Thanks! :D
ill try the second and third, even though its after 3 years and the solutions probably exist
2. i initially thought the see saw would stay even but after reading one of your replies, and considering tension, it will tip right
3. if you look at the string connecting the balls, the mass of the string is 0 so the net force on the string is 0 by F=ma. the forces acting on the string are the same forces acting on each ping pong ball. since the net force on the string is 0, the net force on each ball is also 0 and it stays in equilibrium
For number 3, if the machine were to spin, the ping pong balls on the right would go up and the ones on the left would go down. However, all the balls are pulling upwards, so that will not work. Only if the ping pong balls were swapped with rocks on the left would the machine work, and that one would only work for a small amount of time
Yay! A new slipping hexagons video!
:D
Nice job!Cool video to inspire curiosity
Thanks!
Although this channel has 900 subscribers its videos and animations are top quality. these animations are reminiscent of 3b1b. Which program do you use for animations?
Thank you! I actually use 3b1b's animation library, manim. That's probably why it reminded you of 3b1b ;)
@@SlippingHexagons 👍
last one, when you push in the ping pong balls in through the bottom, it takes work to push them into the tank against the water pressure
MOE
For the first one, the normal force from the slanted walls contribute to the force at the bottom
GIDEON
Please take up some more questions.You really are a life saver..
Thank you! I will try to :)
I didn’t understand the meaning of life till I saw this video
Glad to have helped with that! :)
Very nice video! I'm very bad at physics, but I'm excited to see the answers. Is it possible for you to share the code for this video? I'm also learning Manim, would be interesting to see how you made the conveyer belt (the golf ball machine in Problem 3)
Sure, I'll reply to your comment again with a link to the repo soon
For seesaw: Since the mass of the water is the same in both containers, the only difference is that the mass of the ping pong ball actually affects the mass of the left one while the one on the right does not (I'm ignoring volume displacement since it is the same for both). Thus, the left side will be heavier by the mass of the ping pong ball and the scale tips left side down.
wait but isn’t F net upwards on the ping pong ball which would pull the left up
Cool
:)
new mic kinda nice
Thanks! Omg it's Yuchen Li the IESO gold medalist 😱
2nd one, it’ll tip right because same amount of water but buoyant force from the ping pong ball.
I don't think so, I'd say it's even. The buoyancy is due to displacement but the rock is displacing an equal volume of water but is being supported so the mass doesn't add to the right side.
Still, could be wrong.
@@AshleyBaker75 There's one more force to consider on the left side (hint: string)
This looks exactly like the design of 3Blue1Brown's animations. Using the same program?
yep, I'm using manim :)
Nice 4K resolution!
Thanks!
Physics Wallah best ❤️❤️❤️❤️❤️❤️
Do you have mental health problems