How Do Fluid Forces Affect a Swimming Pool's Structure?

[jenha14]

(a) A swimming pool has dimensions 28.0 m 12.0 m and a flat bottom. When the pool is filled to a depth of 1.90 m with fresh water, what is the force caused by the water on the bottom?
(b) What is the force on each end? (The ends are 12.0 m.)
(c) What is the force on each side? (The sides are 28.0 m.)

 

[olgranpappy]

Attempt at solution? Use your words. Use your equations...

For example, ones that involve pressure and depth and density and pressure and force and area.

Also, you should be able to get part 'a' right off the bat without any fancy fluid equations, eh?

 

[ogb p]

(a) To calculate the force caused by the water on the bottom of the swimming pool, we can use the equation F = ρghA, where ρ is the density of water (1000 kg/m^3), g is the acceleration due to gravity (9.8 m/s^2), h is the depth of the water (1.90 m), and A is the area of the bottom of the pool (28.0 m x 12.0 m = 336 m^2).

Plugging in these values, we get F = (1000 kg/m^3)(9.8 m/s^2)(1.90 m)(336 m^2) = 6.38 x 10^6 N. This is the force caused by the water on the bottom of the pool.

(b) To calculate the force on each end, we can use the equation F = ρghL, where L is the length of the end (12.0 m). Plugging in the same values as before, we get F = (1000 kg/m^3)(9.8 m/s^2)(1.90 m)(12.0 m) = 2.29 x 10^5 N. This is the force on each end of the pool.

(c) Similarly, to calculate the force on each side, we can use the equation F = ρghW, where W is the width of the side (28.0 m). Plugging in the same values, we get F = (1000 kg/m^3)(9.8 m/s^2)(1.90 m)(28.0 m) = 5.15 x 10^5 N. This is the force on each side of the pool.

In summary, the force caused by the water on the bottom of the pool is much greater than the force on each end or each side, due to the larger surface area. This is an important consideration for designing structures that can withstand fluid forces, such as swimming pools.