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## A rectangular tank that is 4 feet long, 2 feet wide and 15 feet deep is filled with a heavy liquid that weighs 60 pounds per cubic foot. In

Question

A rectangular tank that is 4 feet long, 2 feet wide and 15 feet deep is filled with a heavy liquid that weighs 60 pounds per cubic foot. In each part below, assume that the tank is initially full. Your answers must include the correct units.

(a) How much work is done pumping all of the liquid out over

the top of the tank?

(b) How much work is done pumping all of the liquid out of a

spout 3 feet above the top of the tank?

(c) How much work is done pumping two-thirds of the liquid

out over the top of the tank?

(d) How much work is done pumping two-thirds of the liquid

out of a spout 3 feet above the top of the tank

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Physics
6 months
2021-07-30T04:49:57+00:00
2021-07-30T04:49:57+00:00 1 Answers
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## Answers ( )

Answer:a) W₁ = 54000 Lb-ftb) W₂ = 77760 Lb-ftc) W₃ = 24000 Lb-ftW₄ = 40560 Lb-ftStep by stepW= ∫₁² ydF 1 and 2 are the levels of liquidWhere dF is the differential of weight of a thin layer

y is the height of the differential layer and

ρ*V = F

Then

dF = ρ* A*dy*g

ρ*g = 60 lb/ft³

A= Area of the base then

Area of the base is:A(b) = 4*2 = 8 ft²Now we have the liquid weighs 60 lb/ft³Then the work is:a)W₁ = ∫₀¹⁵ 8*60*y*dy ⇒ W₁ =480* ∫₀¹⁵ y*dyW₁ =480* y² /2 |₀¹⁵ ⇒ 480/2 [ (15)² – 0 ]W₁ = 240*225W₁ = 54000 Lb-ftb) The same expression, but in this case we have to pump 3 feet higher, then:W₂ = ∫₀¹⁸ 480*y*dy ⇒ 480*∫₀¹⁸ydy ⇒ 480* y²/2 |₀¹⁸W₂ = 480/2 * (18)²W₂ = 240*324W₂ = 77760 Lb-ftc) To pump two-thirds f the liquid we have2/3* 15 = 10W₃ = 480*∫₀¹⁰ y*dy ⇒ W₃ = 480* y²/2 |₀¹⁰W₃ = 240*(10)²W₃ = 24000 Lb-ftd)W₄ =480*∫₀¹³ y*dyW₄ =480* y²/2 |₀¹³W₄ = 240*(13)²W₄ = 240*169W₄ = 40560 Lb-ft