Friday, November 23, 2012

Physics of Giant Balloons



By Evan Angelus
As I was watching the Macy’s Thanksgiving Day Parade and all of the giant balloons go by, I wondered just how hard it actually is to hold those balloons down.






·      Buoyant Force:
·      The floats (filled with helium) are submerged in the air surrounding them.
o   FB = PA = ρairghA = ρairgV = (ρV)g = mairg
o   Volballoon = 16,780 ft3 = 475 m3
o   ρair = 1.225 kg/m3
o   (1.225 kg/m3 * 475 m3)(9.8 m/s2) = 5,704 N

·      Force of Gravity:
·      We will assume that the nylon fabric the balloon is made of does not contribute to the volume of the balloon, but for such a large balloon, its mass is important.
o   F = mballoong = (muninflated + mHe)g
o   muninflated = 450 lbs = 204 kg
o   ρHe = 0.1786 kg/m3
o   mHe = ρV = 0.1786 kg/m3 * 475 m3 = 85 kg
o   mtot = 204 kg + 85 kg = 289 kg
o   F = 289 kg * 9.8 m/s2 = 2,832 N

·      Net Force of Balloon = FB – Fgrav
o   5,704 N – 2,832 N = 2,872 N à 2,800 N

·      Will they be able to hold the Rugrats down?
·      Each float is held down by 2 vehicles weighing 800 pounds, and a number of volunteers. The only requirement to be a balloon handler is to weigh at least 120 pounds. The rugrats float requires a minimum of 30 people to maneuver it.
o   2 vehicles (800 lbs) = 362.874 kg * 2 = 726 kg
o   30 People (at least 120 lbs) = 54.4311 kg * 30 = 1,633 kg
o   Total = 726 kg + 1,633 kg = 2,359 kg
o   2359 kg * 9.8 m/s2 = 23,115 N à ­– 23,000 N
·      This will be more than enough force to keep the balloon from floating away.
o   2,800 N – 23,000 N = – 20,200 N
o   Why is all of this force needed?
o   To maintain the normal force so that the balloon handlers have enough friction to direct the balloon.
o   20,200 N / 23,000 N = 88% à the normal force on each person on average is 88% what it would normally be. This means that they only have 88% as much friction.

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