Find Work: Empty 6ft Hemisphere Tank of Oil (50lb/ft^3)

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SUMMARY

The work done in emptying a full hemisphere tank of oil with a diameter of 6 feet and a density of 50 lb/ft³ is calculated using the work-energy theorem. The oil must be raised to a height of 4 feet above the top of the tank, resulting in a total height of 7 feet from the base of the tank. To find the work, the tank is divided into horizontal slices, and the mass of each slice is determined to calculate the energy required to raise it to the specified height, followed by integration over the variable z.

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Find the work done in emptying a tank containing oil if the tank is hemisphere with a diameter of 6 feet. The tank is full and the density of the oil is 50lb/ft^3. The oil is raised to a point 4 ft about the top of the tank. Thank you for helping. I have no idea how to start.
 
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Hi idle_09! :smile:
idle_09 said:
Find the work done in emptying a tank containing oil if the tank is hemisphere with a diameter of 6 feet. The tank is full and the density of the oil is 50lb/ft^3. The oil is raised to a point 4 ft about the top of the tank. Thank you for helping. I have no idea how to start.

(i assume you mean "4 ft above …"?)

work done = change in energy (that's the work-energy theorem) …

so divide the tank into horizontal slices of thickness dz, calculate the mass of each slice, how much energy is needed to raise it to z = 7, and integrate over z. :wink:
 
you'll have to find the size of a slice of the hemisphere as a function of z... just figured I'd throw that hint out there.
 

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