SteemSTEM physics challenge #1 - Up!
The steemSTEM crew is happy today to initiate a series of physics challenges. Yep, physics is included in STEM too (even if there is no ‘P’ in STEM)!
The rules are very simple:
- A complicated problem will be asked.
- The problem will be well defined, but not many details will be given.
- A solution will have to be found by the community before the payout of the post.
- The solution itself is worthless. The proof and the logic used to tackle the problem are what matters.
- The challenge lasts 7 days.
So far, nothing crazy. However we insist: the solution will have to be found by the community. This means that the community will share some rewards.
In this way, anyone making a comment allowing the group to get closer to the solution will be rewarded. Yes, you read that correctly: anyone trying to help will be rewarded (and by rewarded we mean with money $$$).
For this first challenge, the rewards consist of 100 STEEM as well as all SBD this post earns.
The way this will be shared depends on the relevance of the comments that will be made, and this will be assessed by the steemSTEM crew: @lemouth, @justtryme90, @anarchyhasnogods, @jaki01, @hagbardceline. No member of the steemSTEM crew is eligible for these rewards, only any other participating member of the steemit community.
Note that ALL rewards will always be paid to participants in the form of Steem Power. That is our usual way to contribute to the community. :)
If after three days, the community is still in the dark, a first hint will be given (and the pot will be accordingly reduced). If after 5 days, the community is still lost, a second clue will be given (and the pot will be reduced a second time).
Very importantly, there is no guarantee that the entire pot will be split among the participants. For instance, we can imagine a situation where the solution is not found, or where the comments are not helpful enough. In this case, a fraction of the money will be shared, and the rest will be used for the subsequent challenge.
NOW THE CHALLENGE
[image credits: Pixabay]
Check the image above. How many balloons do we need to make this happen?
From the discussion in the comments, please find below some further specifications:
- We just want to lift the house out of the ground by few centimeters. We do not assume it is fixed.
- We consider a real house and real balloons, so that the picture is only illustrative.
- We make the crude and wrong assumption that all balloons can be inflated simultaneously in no time.
Have fun! And remember: we don’t care about the exact solution. Only the deduction and logic that you all use matters!
Okay.... will try and answer in parts... however the solution entirely depends on the followings:
The weight of the balloon and ropes can be considered negligible if we are considering a real house
Mathematically,
We need to calculate the net upward Force (Buoyancy) generated by balloon of volume V, by replacing equal amount of air (or water) around it.
F (net) = Upward force of balloon - Force of medium displaced
In order for home to get lifted
F(net) has to balance F (home)
F(net) is directly proportional to Volume of air replaced
F (net) ~ V (volume)
V = nv (n = number of balloons, v is volume of balloons)
n = V/v
Let's start with the weight of the house, most house movers I know say the max weight of a moveable house is 70,000kg. Let's say ours is a little smaller, 40,000kg.
A standard 1 liter balloon can lift 1 gram. This is found by using the ideal gas law, first explained by Émile Clapeyron in 1834, PV = nRT. or:
pressure multiplied by volume equals the amount of gas multiplied times the temperature. In other words, the temperature multiplied by the amount equals the pressure in an equal volume
You see, gas is just particles flying around, so if the same amount of particles get flying around in a smaller space or volume, they bounce around more, causing more force, and they get hotter(like they're in the gym turning up the treadmill). Helium has smaller particles bouncing around than air.
Therefore, helium is less dense than air, and it is contained in the body of the balloon. As explained in archimedes principle, a less dense body will float, or experience a buoyant force in a more dense one. And, 1 liter of helium at the density of helium, .16 grams per liter, is less than the density of air, 1.18 grams per liter. So, at the same temperature, the force of gravity(mass times acceleration of gravity) is 1.18-.16 = roughly 1.02grams lifted against gravity, or floating.
So, for 40,000kg, lifted by 2 liter helium balloons each lifting 2.4 grams, you'd need at least 17,000,000 balloons.
Now, you'd have to take into account that some balloons would be popped, and wind may create downward force you would need to overcome.
first let's look at the wind pressure. The Ensweiler equation shows the pressure of wind. Let's say the bunch of balloons is hit by 50mph wind. 50mph^2 times the constant .00256 = 6.4 pounds for square foot. converted to 31 kg/square meter. let's say that our balloon mass is100 balloons long, 100 balloons wide, 1700 balloons tall, that's roughly 17million. If a 2 liter balloon is 1 ft in diameter, or .3meters, then our balloon mass is roughly 1000 meters squared area. then that would be 31,000 kg of force, almost doubling the weight of the house!
I think it's safe to say that we'd need double the balloons, using the old engineering 2X safety factor. That's 32million balloons. Imagine that a flock of birds could take out 30%, and let's go ahead and jack that up to 50 million regular balloons.
Great comment! That is the first time I see someone discussing the wind pressure issue! I was about to comment on extra stuff like the ropes, the weight of the ballons, etc... but this is probably well covered by your safety factor.
PS: why using so small balloons?
thanks, I thought it would be more fun to calculate it using a common sized balloon. Very cool idea for a physics challenge series, enjoyed it, thanks!
you are welcome! Fun is what matters :)
Jamhuery that is impressive with the exact numbers. I was just going to walk through how to solve without numbers because I surely wouldn't have known that without extensive research... Anyway here is my take on it:
First you would need to figure out how heavy the house is. This isn't an easy task to find a scale large enough to weigh an entire house. Rather I would take the individual components and sum the weight of the materials used to construct it.
Next, The Balloons. How much weight are they capable of lifting and at what speed? The simple F = ma can come into play ( Force equals Mass Times acceleration). From this photo alone, we do not know how quickly the house is rising. More balloons = a faster lift into the sky because there is more force opposing gravity's pull on the mass of the house ( IE Weight). Similarly to what Jamhuery mentioned you figure out how much helium is contained in one balloon. Then figure out the air mass compared to oxygen. The tricky part to this is that it will change the higher you go because of the density of atmospheric pressure and additionally temperature. Heat causes air molecules to expand, while cooling causes them to condense. Think of a car tire. When it is hot outside, your air pressure in your tires is high, when it first starts to cool every fall/winter, your tires appear slightly less full and the pressure is decreased.
Once you have figured out how much 1 balloon can lift under the circumstances given and how much the house weighs, and at what rate you would like the house to lift, you can determine how many balloons you need.
To Actually make this happen, it is going to take a whole (excuse the french) Shit ton of steemians to get this done. Huge numbers of balloons will need to be filled with helium within a very short period of time (as we are all familiar with it only takes a couple days for helium balloons to fall to the ground and prior to that they are constantly loosing force, which would then necessitate the use of more balloons to make up for the loss due to inflation time. Using Jamhuery's number of 69 balloons for 1 kg and assuming the average house weighs roughly 50,000 Kg ( This I googled for a very rough estimate, as all houses vary) That is literally over 3 million balloons. But that number only holds if these are inflated simultaneously...
How long does it take to fill a balloon with helium and connect it to the house? Well once we figure that out, then we can figure out exactly how many Steemians we would need. Lets say we are incredibly efficient and can complete a balloon and attach it in 40 Seconds. We probably want to have this project done in less than 5 hours so the balloons don't start sagging (as those in the front of the picture are doing). One steemian can do 750 balloons in that timeframe.. With needing 3 million We would need 4,000 steemians who are all equally efficient at inflating and attaching balloons.
So In sum (and after making several assumptions)
We need lots of mathematical numbers to plug and chug through equations, but more importantly we need the Man power to do this efficiently otherwise the number of balloons changes with variation in effort and time. In fact countless variables throughout this can change the number needed to make that happen!
thanks @lovewild, i asked it on my first comment "Great question, the first we have to know mass of the house so we know how much energy to up this house" , So i just estimate it , amount of ballon in 1 Kg
Thanks a lot for this very well detailed answered. You demonstrate exactly why I love asking this kind of problems. There is no unique answer and there are gazillions of ways to go through the problem. Everything you said is correct, and the problem was made intentionally short (so that the challenge has a chance to last for a week). I was planning to give more information with the comments.
Here are more specifications (I will edit the post accordingly):
As I said, this is not realistic but this is for the fun of the challenge!
The diameter of the toy balloon = 30 cm -> r = 15 cm (we assume perfectly round)
Balloon volume = 4/3 x pi x r ^ 3 = 14137 cm3 = 0.014137 m3
The diameter of the toy balloon
30 cm -> r = 15 cm (we assume perfect round)
Balloon volume = 4/3 x pi x r ^ 3 = 14137 cm3 = 0.014137 m3
This balloon will be filled with helium
Helium mass = 0.1785 kg / m3
Air type mass = 1.204 kg / m3 (at a temperature of 20 degrees Celsius)
Difference of mass type = 1.0255 kg / m3 (mass of balloon skin while we ignore)
Power lift per balloon = 0.014137 x 1.0255 = 0.014498 kg
So to lift 1 kg of goods need = 1 / 0.014498 = 69 balloons!
If the weight of House is 200 kg, The Balloon we need is 13800 Balloons
That's a good first try, but that's not the entire story. First, could you please add enough text so that anyone, even without any physics background, could follow. Thanks in advance (the idea is to share with others the solving of the problem).
Next, and more important: you are making approximations. Please justify them.
this is just some estimate, the question above without weight mass
What I meant is that estimates are OK, as long as they are justified (quantitatively). Can you please justify them? Thanks in advance.
The picture above shows a toy balloon that lifts the house, so the question is how many balloons are needed to lift the house. We just assume energy or lift (number of balloons) per 1kg, thanks @lemounth
Sorry for being unclear, as said below. Let us assume real balloons and a real house :)
If we assume the original house and ballon, what thread or rope will we use? Balloon made of soft rubber and gell, if only we use a strong rope may be a balloon that will tear. i think this is impossible
You have different types of balloons and ropes. IT is up to you to chose the right ones :)
That would really depend on the size of the balloon.
One really big one.
I am happy with the answer, but... you must then:
You know I have never done anything like this. I did not even get the opportunity to finish high school. I'll try my best but I feel as though I'm way behind the curve.
With very few basic physics laws, the problem can be started. I will not really help now (too early), but feel free to ask other Steemians taking part to the challenge to help you :)
It is kind of easy to get somewhere, and I hope you will enjoy learning how :)
OK, you beat me to this.
Great question, the first we have to know mass of the house so we know how much energy to up this house
True, but you can also estimate it :)
estimated, for example
Lol, this is going to be fun for me!.... As the solution in this case, is just 1 balloon!, a big one of course, as the image only shows a fake house (not a real one), i will not solve the problem for a real house as it is not relevant...
Haha!
Please consider a real house and real balloons. Why not relevant? We may show that this is possible to achieve= and possibly do it, just for fun.
Scientist work on observation, if you show me a fake house, i will solve the problem for a fake house LOL... And of course it is possible, i think i saw a TV show where someone did it...
I suppose that if I show you a picture of a real house, you may tell me it is a fake picture of a real house? At the end of the day, this will not help to solve anything.
This being said, the idea behind this challenge is not to make real observations and achieve calculations to be used in state-of-the-art research. Instead, we aim to use basic physics principles and show they can be put together to manage solving something funny. And solving this problem will not change our lives (although with 100 STEEM, this is arguable). Just for fun!
I would not say is a fake picture because i have seen this happen for real!...
The thing is, the way our mind works (physics mind), if you ask me something like "how many", i will try to give you the most certain answer with numbers, because that we are used to do so.
Probably i am not understanding exactly what kind of answer are you looking for... But i'm having fun anyway :P
Sorry for being unclear. Let us assume a real situation with a real house and real balloons. Is it better? :D
[Fun is what matters].
The TV show is titled Up! It is such a cute show haha.
The beginning was super sad though.
It is actually done already
According to their experiment it was 300 balloons, but the thing is, the amount of balloons will depend on the balloons size, and the weight of the house, too many variables to give a correct answer without further information... How big is the house you want to lift?, if the house is attached to the ground you probably wont make this happen...
That is right. The answer depends on many parameters that have been intentionally not given. It is up to you to estimate them. This makes the problems funnier (and give more sources for getting rewarded). ^^