Calc07_5day2

# Calc07_5day2 - 7.5 Fluid Pressure and Forces Greg Kelly...

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Unformatted text preview: 7.5 Fluid Pressure and Forces Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2006 Georgia Aquarium, Atlanta What is the force on the bottom of the aquarium? 3 ft 2 ft 1 ft Force weight of water = density volume = ⋅ 3 lb 62.5 2 ft 3 ft 1 ft ft = ⋅ ⋅ ⋅ 375 lb = → If we had a 1 ft x 3 ft plate on the bottom of a 2 ft deep wading pool, the force on the plate is equal to the weight of the water above the plate . 3 lb 62.5 ft density 2 ft ⋅ depth pressure 3 ft 1 ft ⋅ ⋅ area 375 lb = All the other water in the pool doesn’t affect the answer! → y dy What is the force on the front face of the aquarium? 3 ft 2 ft 1 ft Depth (and pressure) are not constant. If we consider a very thin horizontal strip, the depth doesn’t change much, and neither does the pressure. 3 ft 2 ft 2 62.5 3 y F y dy = ⋅ ⋅ density depth area 2 62.5 3 F y dy = ⋅ ⋅ ∫ 2 2 187.5 2 F y = 375 lb = It is just a coincidence that this matches the first answer!...
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## This note was uploaded on 02/12/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.

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Calc07_5day2 - 7.5 Fluid Pressure and Forces Greg Kelly...

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