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Here is some background theory and a very similar example that should help you. \ \ \ \ The increase in pressure at depth h due to any static fluid of density ρ \ \ Consider rectangular container volume V = base area A * depth h \ \ P = (Force due to weight)/Area = mg/A =mg* h/V \ \ but m/V = density ρ \ \ Pressure = ρgh \ \ If ρ is in kg/m^3, g ~ 9.81 m/s^2 and h is in metres then P will be in Pascals \ \ The Pascal is the unit of pressure. 1Pa = 1N/m^2 \ \ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \ \ Calculation increase of pressure at the bottom of swimming pool: \ \ a) What is the absolute pressure in N/m2 on the bottom of a swimming pool 1.0 m by 1.6 m whose uniform depth is 2.0 m? Take atmospheric pressure to be 1.013×10^5 N/m^2. \ \ b) What is the total downward force in N. \ \ (c) What will be the pressure against the vertical side of the pool near the bottom? \ \ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \ \ a) Density of water ρ = 1gm/cm^3 = 1000 kg/m^3 \ \ Increase in pressure P at depth h = 9.81*10^3*h Pa = 9.81h kPa \ \ Or in short, for water only increase in P (in kPa)= hg \ \ Absolute pressure includes the atmospheric pressure 101.3 kPa \ \ Absolute pressure at bottom of pool P (in kPa)= (hg + 101.3) kPa \ \ In this example h = 2 metres, so absolute pressure = (19.62 + 101.3) = 120.92 kPa \ \ \ \ b) Calculate downward force as Area* pressure = 1.0 m * 1.6 m * 120920 N/m^2 = 19347.2 N \ \ c) Fluid pressure in all directions, so same answer as a).

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Dimensions of a swimming pool are 28.0 m by 9.5 m and its uniform depth is 2.9 m . The atmospheric pressure is 1.013×105N/m2.

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Part A

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Determine the absolute pressure on the bottom of the swimming pool.

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Part B

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Calculate the total force on the bottom of the swimming pool.

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