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1 year ago

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an open tank has the shape of a right circular cone (see figure). the tank is 8 feet across the top and 6 feet high. how much wo

rk is done in emptying the tank by pumping the water over the top edge? (the weight-density of water is 62.4 pounds per cubic foot.) ft-lb

1 answer:

Liono4ka [1.6K]1 year ago

7 0

The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.

Given that, the tank is 8 feet across the top and 6 feet high

By the property of similar triangles, 4/6 = r/y

6r = 4y

r = 4/6*y = 2/3*y

Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²

The weight of each disc is m = ρw* A

m = 62.4* 4π/9*y² = 87.08*y²

The distance pumped is 6-y.

The work done in pumping the tank by pumping the water over the top edge is

W = 87.08 ∫(6-y)y² dy

W = 87.08 ∫(6y³ - y²) dy

W = 87.08 [6y⁴/4 - y³/3]

W = 87.08 [3y⁴/2- y³/3]

The limits are from 0 to 6.

W = 87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs

The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.

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