Thursday, November 28, 2013

While watching the Macy’s Thanksgiving Day Parade, I started thinking about the forces acting on the balloon.
I decided to take a closer look at the Snoopy Balloon.

The forces acting on this balloon would be the force of gravity, the force of buoyancy, and the force of tension in the cables keeping the balloon from floating away.

The balloon is in equilibrium when it is being held, so the summation of forces in the y direction should be zero (define down as negative).

ΣFy = 0 = Fg + FT + FB

I decided to calculate the force of tension in the cables that are used to keep the balloon in equilibrium.

First, I calculated the buoyant force on the balloon. The density of air is 1.293 kg/m3. To find the volume, I assumed the dimensions of the balloon. To simplify things I imagined the balloon as a cylinder that was 12.2 m long, with a diameter of 4.6 m.

V=π(4.6 m/2)2(12.2 m)
V= 202.75 m3

FB= ρgV
FB=(1.293 kg/m3)(9.8 m/s2)(202.75 m3)
FB= 2569.15 N

Next, I calculated the force of gravity on the balloon. I assumed that the mass of the balloon before it was filled was 150 kg.
I then calculated the mass of the hydrogen gas it is filled with. Hydrogen has a density of 0.164 kg/m3

m=(0.164 kg/m3)(202.75 m3)
m=33.25 kg

Fg=(150 kg + 33.25 kg)(9.8 m/s2)
Fg= 1795.86 N

Using these forces I calculated the force of tension. (Gravity is a negative force and Buoyancy is a positive force because they are acting in opposite directions).

0 = FB - Fg + FT
FT = -FB + Fg
FT = -2569.15 N + 1795.86 N
FT = -773.29 N

I assumed that there are about 40 people holding cables attached to the balloon, so the average force of tension in each cable would be about -19.33 N.

-773.29 N / 40 = -19.33 N

How thick do the cables need to be to keep from breaking?
I assumed that the cables were made of steel.
The ultimate strength of steel is 500 x 106 N/m2

F/A = 500 x 106 N/m2
19.33 N/(πr2) = 500 x 106 N/m2
r = 0.0001 m2

The cables must be at least 0.2 mm in diameter to keep from breaking.

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