Friday, November 29, 2013


Over Thanksgiving break, I was swimming a practice at the aquatic center at Eisenhower Park that was built for the 1998 Goodwill Games. The pool is huge, and it made me wonder how much force and pressure was acting on the pool bottom. First, the formula to find the pressure would be P=Patm + pgh. I assumed that the pool depth is uniformly 10 feet, although it varies throughout. I also assumed that the water in the pool is not chlorinated. Thus,

P=Patm +pgh
P=(1.013x10^5 N/m^2) + (1.00x10^3 kg/m^3)(9.81 m/s^2)(10ft)(1m/3.28084ft)
P=131,201 N/m^2

To find the force acting on the bottom of the pool, I used the equation P=F/A. To find the area of the bottom of the pool, I found that the length of the pool is 68 meters and the width of the pool is 25 meters (http://www.nassaucountyny.gov/agencies/parks/wheretogo/recreation/nc_aqua_ctr.html). Since A=lw,

P=F/A
F=PA
F=(131,201 N/m^2)(25m)(68m)
F=223,041,700 N

I also looked at the difference if this pool was filled with ocean water. I found the density of ocean water at the surface of the water at http://www.windows2universe.org/earth/Water/density.html. Thus, the pressure of the pool filled with ocean water is

P=(1.013x10^5 N/m^2) + (1.027x10^3 kg/m^3)(9.81 m/s^2)(10ft)(1m/3.28084ft)
P=132,008 N/m^2

The force acting on the bottom of the pool filled with ocean water is

F=(132,008 N/m^2)(25m)(68m)
F=224,413,600 N

Thus, the salinity of the ocean water causes a large force increase in a pool of this size.

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