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.