2006-12-26 05:27:49 · 10 answers · asked by david j k 2 in Science & Mathematics ➔ Physics
No clue...but here're the equations to work the problem yourself.
f = Ma = B - W; where f = net forces acting on the person, M = mass of the person plus the rubber and gas in all the party balloons, a = the acceleration (rise rate) of the person and balloons, B = the buoyancy force of lift, and W = the weight of the person, gas, and balloon rubber.
When you have sufficient ballons so that B > W, the net force will be positive; so that f = Ma > 0 and a > 0 meaning that the total mass (person, balloons, and gas) will lift skyward with an acceleration of a. Note that Mp = Wp/g; where Mp is the mass of the person and Wp is that person's weight; g = 32.2 ft/sec^2 or 9.81 m/sec^2. M = Mp + Mb = Wp/g + Wb/g = 1/g(Wp + Wb); where Wb = the total weight of all the balloons.
You can measure the bouyancy of each balloon (b) and the weight (w) of each one. Assuming these are uniform across all the balloons (N of them), we would have B = Nb and Wb = Nw; where N = the number of balloons you would need. Wb = the total weight of N balloons. Thus, W = Wp + Wb; where Wp is the weight of the person, and M = W/g = 1/g(Wp + Wb).
You can measure the bouyancy of a balloon by attaching a fish scale or similar to the balloon and marking how many ounces upward force you get (that's b, the per balloon bouyancy). You can discount the weight of the He in each balloon; so just weigh an empty ballon to find w. fup = b - w = net upward force per balloon.
Now you can solve f = Ma = B - W = Nb - (Wp + Nw) = N(b - w) - Wp. Set a = 0 (no acceleration) so that the balloons are up in the air, but the person has not yet been lifted. Then B = W or N = Wp/(b - w); where N is just enough balloons that one more would start to lift the person of weight Wp. (b - w) is the net upward force per balloon.
So, as one might imagine and as the derived equation shows, if the weight of the person (Wp) is greater, we need more balloons. Or, if the net force upward is less per balloon, we need more balloons.
Have fun working the problem once you measure the b and w for your balloons. I think you'll find N to be a very large number for a person because the net upward force per balloon will be very very small compared to the person's weight.
PS: I'm adding an EXAMPLE to give you (and me) a sense of the scale of the issue. None of the numbers used is real, but I think the assumptions are reasonable.
Assume (b - w) = 1 ounce; where b = buoyancy and w = weight of a party balloon filled with He. The 1 ounce is the net upward force per balloon. Thus, each balloon by itself would rise up. Also assume you weigh Wp = 150 pounds after Holiday dinner.
Then, from the earlier derived equation, N = Wp/(b - w) = 150 X 16/1 = 2,400 = number of balloons needed to just barely keep you on the ground. (The 16 in the numerator converts your weight to ounces.) Thus, to begin liftoff, you'd need one more balloon, or 2,401 balloons given the assumptiions made.
2006-12-26 06:27:56 · answer #1 · answered by oldprof 7 · 1⤊ 0⤋
the version in density greater or much less explains why the helium balloon rises. yet it would not answer the subject of the upward push in PE. All meaning is that the balloon has aquired greater capability by using fact it has moved that lots extra from the floor. the two are no longer the comparable ingredient, and that they do no longer seem to be suitable interior the assumption test which you have conceived, which does no longer recommend that that's a detrimental concept. in case you have been to place a hollow interior the balloon so as that all of the helium could get away into the ambience, what ever replaced into left could rush at a speedier and ever greater desirable velocity by using fact the PE replaced into switched over to KE.
2016-10-19 00:16:34 · answer #2 · answered by Anonymous · 0⤊ 0⤋
A little more than 150 lb of helium-filled balloons.
2006-12-26 09:31:12 · answer #3 · answered by Wufei 3 · 0⤊ 1⤋
I have no idea, but I did see a fellow in Arab Alabama take two or three dozen condoms, fill them with helium and stuff then into a fish net he had tied to a aluminum chaise lounge, and float teathered above the ground.
Made the local news in '96 or'97
2006-12-26 05:32:38 · answer #4 · answered by Anonymous · 0⤊ 1⤋
I am not sure, but I know that someone has done that before I think in Cal., where a guy attached many of them to a lawn chair and it lifted him up a bit too far, he took a BB gun along to shoot out enough of them to gently lower himself down but got into some trouble, I think he is alright but the Police had some issues with him.
2006-12-26 05:31:20 · answer #5 · answered by Anonymous · 0⤊ 0⤋
99 red balloons :-)
-Take 3 ballons with helium.Tighed them togeder
- atached a small plastic bag to tem.
- take a spoon of salt
- put some salt in the plastic bag
- see what is the minimum amount of salt for wich the balloons hover motionless.
- measure the wheight of the plastic bag and the salt inside.
- if 3 ballons could lift X grams of salt, (150 pound = 68 038.8555 grams), than how many ballons needs to lift 68038.85 grams ?
I let you to find out....!
2006-12-26 06:49:08 · answer #6 · answered by eagle 2 · 0⤊ 0⤋
Well, Mythbusters tackled this issue. It took several thousand to lift a 40 pound child. So probably 100000 or so?
2006-12-26 05:30:27 · answer #7 · answered by Mr. Goodkat 7 · 1⤊ 0⤋
20 balloons
2006-12-26 05:35:41 · answer #8 · answered by todd s 4 · 0⤊ 1⤋
Depends on the volume of the balloons...
2006-12-26 05:35:50 · answer #9 · answered by Smoothie 5 · 0⤊ 1⤋
A lot, my friend, a lot, so many they won't fit a regurlar room, so quit the project.
2006-12-26 11:29:07 · answer #10 · answered by Arc 2 · 0⤊ 1⤋