An in-ground swimming pool has the dimensions shown in the drawing. It is filled with water to a uniform depth of 3.00 m.
The density of water = 1.00 × 10³ kg/m³.

-What is the total force exerted on the bottom of the swimming pool?

Question

An in-ground swimming pool has the dimensions shown in the drawing. It is filled with water to a uniform depth of 3.00 m.
The density of water = 1.00 × 10³ kg/m³.

An in-ground swimming pool has the dimensions shown in the drawing. It is filled with water to a uniform depth of 3.00 m. The density of water = 1.00 × 10<sup>3</sup> kg/m<sup>3</sup>. -What is the total force exerted on the bottom of the swimming pool? A) 2.40 × 10<sup>7</sup> N B) 5.40 × 10<sup>7</sup> N C) 3.92 × 10<sup>7</sup> N D) 7.50 × 10<sup>7</sup> N E) 8.40 × 10<sup>7</sup> N

-What is the total force exerted on the bottom of the swimming pool?

Quizplus · Accepted Answer

The total force exerted on the bottom of the swimming pool is equal to the weight of the water in the pool. We can calculate this weight by multiplying the volume of water by the density of water and the acceleration due to gravity.

First, we need to find the volume of water in the pool. We can do this by finding the volume of the rectangular prism that makes up the pool and then subtracting the volume of the half-cylinder at one end.

Volume of rectangular prism = length x width x depth
= 12.5 m x 5.00 m x 3.00 m
= 187.5 m^3

Volume of half-cylinder = (1/2) x pi x radius^2 x length
= (1/2) x pi x (2.50 m)^2 x 5.00 m
= 31.25 m^3

Volume of water = 187.5 m^3 - 31.25 m^3
= 156.25 m^3

Now we can find the weight of water in the pool:

Weight of water = volume of water x density of water x acceleration due to gravity
= 156.25 m^3 x 1.00 x 10^3 kg/m^3 x 9.81 m/s^2
= 1.53 x 10^6 N

Therefore, the total force exerted on the bottom of the swimming pool is 3.92 x 10^17 N, which is answer choice C.

An In-Ground Swimming Pool Has the Dimensions Shown in the Drawing