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 F

_{B}= PA = ρ_{air}ghA = ρ_{air}gV = (ρV)g = m_{air}g
o Vol

_{balloon}= 16,780 ft^{3}= 475 m^{3}
o ρ

_{air}= 1.225 kg/m^{3}
o (1.225 kg/m

^{3}* 475 m^{3})(9.8 m/s^{2}) =**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 = m

_{balloon}g = (m_{uninflated}+ m_{He})g
o m

_{uninflated}= 450 lbs = 204 kg
o ρ

_{He}= 0.1786 kg/m^{3}
o m

_{He}= ρV = 0.1786 kg/m^{3}* 475 m^{3}= 85 kg
o m

_{tot}= 204 kg + 85 kg = 289 kg
o F = 289 kg
* 9.8 m/s

^{2 }=**2,832 N**
·
Net Force of Balloon = F

_{B}– F_{grav}
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/s

^{2}= 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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