Answer:
a. F = 26214.4π lb/ft-s²∫₋₁₀⁰(y + 10)dy
b. W = 104857.6π lb-ft/s²∫₋₁₀⁰(y + 10).dy
Explanation:
(a) Write down an integral for the total fluid force on the side of the tank.
We know force, F = ∫PdA where P = pressure and dA = differential area element in tank surface.
Let the position of the bottom of the tank be y = -h where h = height of tank and y = 0 be the top of the tank. So, the height of any point y of the tank H is y - (-h) = y + h = y + 10
The pressure P = ρgH where ρ = density of oil in tank = 51.2 lb/ft³ and g = acceleration due to gravity = 32 ft/s²
P = ρgH
P = ρg(y + 10)
The differential area element of a point on the side of the tank is dA = 2πrdy
So, F = ∫PdA
F = ∫ρg(y + 10)2πrdy
we integrate y from y = -h = -10 to y = 0
So, F = ∫₋₁₀⁰ρg(y + 10)2πrdy
F = ∫₋₁₀⁰2πrρg(y + 10)dy
F = ∫₋₁₀⁰2π(8 ft)(51.2 lb/ft³)(32 ft/s²)(y + 10)dy
F = ∫₋₁₀⁰26214.4π lb/ft-s²(y + 10)dy
F = 26214.4π lb/ft-s²∫₋₁₀⁰(y + 10)dy
(b) Suppose all the kerosene is pumped to ground level. Write down an integral for the work required to pump the oil.
The work done W = ∫F.dy
W = ∫F.dy
F = ρgV where V = volume of tank = πr²H = πr²(y + 10)
F = ρgπr²(y + 10)
So, W = ∫F.dy
W = ∫ρgπr²(y + 10).dy
Since the water is pumped to ground level, we integrate y from y = -h = -10 to y = 0
W = ∫₋₁₀⁰ρgπr²(y + 10).dy
W = ∫₋₁₀⁰π(51.2 lb/ft³)(32 ft/s²)(8 ft)²(y + 10).dy
W = ∫₋₁₀⁰104857.6π lb-ft/s²(y + 10).dy
W = 104857.6π lb-ft/s²∫₋₁₀⁰(y + 10).dy