## Monday, December 9, 2013

### Thanksgiving Day Balloons

This Thanksgiving Day my family went to watch the Thanksgiving Day Parade in New York City.  It was particularly windy and cold at 32 degrees Fahrenheit.  They said that the workers had to hold the parade balloons particularly low.  This made me wonder how much force the workers must apply to the balloons to hold them down.  If the balloon is not moving at all in the Y direction, with up being positive, the sum of the forces equals the buoyant force, the force of the balloon and the force of the people holding it down.

Fy=pair*g*Vairdisplaced + phelium*g*Vballoon - Fpeople=0 N
The buoyant force and the force of the balloon are acting in the positive direction because the buoyant force acts up and the force of the ballon is the force of the helium, whose density is less than that of air so it pulls the balloon upward.  The force of the people is then the only force in the downward direction.

The largest balloon ever was a Shamu balloon with 18000ft^3 or 509.7m^3.  How much force was needed to hold this balloon down?
Fy=(1.269kg/m^3)(509.7m^3)(9.8m/s^2)+(1.786E-4kg/m^3)(509.7m^3)(9.8m/s^2)-Fpeople
Fpeople(total) = 6340 N

If there are approximately 40 people holding down each balloon then each person is applying about 159N of force.  This is only the force in the Y direction.  With the wind there were likely other forces in the X direction which would have put more work on the balloon holders to hold it in place.