A new pool is to be built and filled so that the water is 0.4m below the rim of the pool. If the town has 3000 residents, what dimensions would the pool have to fit them all in so they can swim comfortably and not overflow the pool?

Guest Aug 17, 2017

#1**+1 **

Assuming each resident have a total body mass of \(M\space(kg)\), they all have a density of \(D\) (\(D\le 1g/cm^3\)) , and the swimming pool being a square prism (i.e. A square with height) with side \(L (meters)\).

The law of physics tell us that any object with volume displaces water in water, and the displaced water causes the container's water level to rise, in this case, a swimming pool.

The remaining non-overflowing volume of the pool will be \(0.4L^2\space(meter\space cubed)\)

Since \(Volume=Mass/Density\)

Each resident's volume is equal to \(\frac{M}{D}\).

Remember that \(D\le 1g/cm^3\), that means part of their body won't sink under the water when they are in the pool, therefore:

Each residents' displaced water volume: \(\frac{M}{D}\cdot D=M\space(cm^3)\) (Because the density of water = \(1g/cm^3\))

For a total of 3,000 residents, their total volume would be \(3000M\)

In order for the pool to not overflow, the water displaced must be smaller than the remaining volume of the pool:

\(3000M\space(cm^3)\le 0.4L^2\space(m^3)\)

\(0.003M\space(m^3)\le0.4L^2\space(m^3)\)

\(L^2\ge 0.0075 M\)

\(L\ge \frac{1}{20}\sqrt{3} M\)

So the side of the pool have to be longer than \(\frac{1}{20}\sqrt{3}\) of each resident's mass in kilograms.

(Somehow I feel the units have some fatal mistakes, correct me if anything is wrong in my process :P)

Jeffes02
Aug 17, 2017