conical tank has height 10 ft and radius 8 ft. The tank is buried vertex-down so that the top is 6 ftunderground. The tank is filled with kerosene (of density 51.2 lb/ft3).(a) Write down an integral for the total fluid force on the side of the tank.(b) Suppose all the kerosene is pumped to ground level. Write down an integral for the work requiredto pump the oil.

a. F = 26214.4π lb/ft-s²∫₋₁₀⁰(y + 10)dy

b. W = 104857.6π lb-ft/s²∫₋₁₀⁰(y + 10).dy

Step-by-step 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

the answer to this question is in the pic