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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