A rectangular swimming pool 60 ft long, 20 ft wide, and 14 ft deep is filled with water to a depth of 11 ft. Use an integral to find the work required to pump all the water out over the top. (Take as the density of water δ=62.4lb/ft3.)
Work=??????
the weight of the water is 60*20*14*62.4 = ____ lbs
the center of mass is at a depth of 7 ft
work = weight * distance
or, you can use an integral, as indicated by your text.
google can provide many examples.
Post your work if you get stuck.
To find the work required to pump all the water out over the top of the pool, we can calculate the force required to lift each small volume of water and then integrate it over the entire volume of water in the pool.
The force required to lift a small volume of water can be calculated using Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. In this case, the weight of the fluid is equal to its density times its volume times the acceleration due to gravity.
First, we need to find the volume of water in the pool. The pool is rectangular, so its volume can be calculated as the product of its length, width, and depth:
Volume of water in the pool = length * width * depth
= 60 ft * 20 ft * 11 ft
= 13,200 ft^3
Now, we can calculate the weight of the water in the pool:
Weight of water = density * volume * acceleration due to gravity
= 62.4 lb/ft^3 * 13,200 ft^3 * 32.2 ft/s^2
= 26,145,600 lb*ft^2/s^2
To find the work required to pump all the water out, we need to calculate the force required to lift each small volume of water and then integrate it over the entire volume of water in the pool. The force required to lift a small volume of water is equal to the weight of that water, so we need to integrate the weight of the water over the volume of the pool.
The work required to pump all the water out can be calculated using the integral:
Work = ∫[weight of water] dV
Where dV is the differential volume element. Since the pool is rectangular, we can split the integral into three separate integrals representing the three dimensions (length, width, and depth) of the pool.
Work = ∫∫∫[62.4 lb/ft^3 * acceleration due to gravity * dx * dy * dz]
Here, the variables x, y, and z represent the coordinates in each dimension, and the limits of integration are as follows:
For x: 0 to 60 ft
For y: 0 to 20 ft
For z: 0 to 11 ft
Evaluating this triple integral will give us the total work required to pump all the water out over the top of the pool.
To find the work required to pump all the water out of the swimming pool, we can calculate the weight of the water and then use the equation for work: Work = Force x Distance.
First, let's determine the weight of the water. The weight of an object is given by the equation: Force = mass x acceleration due to gravity. In this case, the mass is the volume of water times its density, and the volume of water is the difference in depth between the initial and final levels.
Given:
Length of the swimming pool (l) = 60 ft
Width of the swimming pool (w) = 20 ft
Initial depth of water (h1) = 14 ft
Final depth of water (h2) = 11 ft
Density of water (δ) = 62.4 lb/ft^3
Acceleration due to gravity (g) = 32.2 ft/s^2 (approximately)
Volume of water = (l * w * (h1 - h2))
= (60 ft * 20 ft * (14 ft - 11 ft))
= 60 ft * 20 ft * 3 ft
= 3600 ft^3
Mass of water = (Volume of water * Density)
= (3600 ft^3 * 62.4 lb/ft^3)
= 224,640 lb
Force = (Mass of water * Acceleration due to gravity)
= (224,640 lb * 32.2 ft/s^2)
= 7,237,248 lb*ft/s^2
Next, we need to find the distance over which the force is applied. This is the height difference between the initial and final levels of water, which is (h1 - h2) = (14 ft - 11 ft) = 3 ft.
Now we can calculate the work using the equation: Work = Force * Distance.
Work = (7,237,248 lb*ft/s^2 * 3 ft)
= 21,711,744 lb*ft
Thus, the work required to pump all the water out of the swimming pool over the top is approximately 21,711,744 lb*ft.