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is it possible to make a free body diagram of this equation? *liquid pressure experienced by the bottom-most part of the house using the equation:

is it possible to make a free body diagram of this equation? *liquid pressure experienced by the bottom-most part of the house using the equation:

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is it possible to make a free body diagram of this equation?

*liquid pressure experienced by the bottom-most part of the house using the equation:

Given- Density of the fluid (p) = 62.4 g/ mL Height (h) = 7. 10 ft Maximum Pressure cement can withstand (Pmax) = 28728. 2 Pa Requirement- The liquid pressure experienced(P). Step 2 Answer- It is known that – P = pgh 62.4x10-3 kg = 62. 4 g/mL 62.4x10-3 kg 10-3x10-3 m3 = 62400 kg /m3 %3D 10-3 L 1 kg 1000 g 1L = 1000 mL 1 m3 = 1000 L g = 9. 81 m/s² h = 7. 10 ft = (7. 10 × 0. 3048) m 1 ft = 0. 3048 m Now, P = (62400 x 9. 81 × (7. 10 × 0. 3048)) P: 1324728. 588 Pa
Transcribed Image Text:Given- Density of the fluid (p) = 62.4 g/ mL Height (h) = 7. 10 ft Maximum Pressure cement can withstand (Pmax) = 28728. 2 Pa Requirement- The liquid pressure experienced(P). Step 2 Answer- It is known that – P = pgh 62.4x10-3 kg = 62. 4 g/mL 62.4x10-3 kg 10-3x10-3 m3 = 62400 kg /m3 %3D 10-3 L 1 kg 1000 g 1L = 1000 mL 1 m3 = 1000 L g = 9. 81 m/s² h = 7. 10 ft = (7. 10 × 0. 3048) m 1 ft = 0. 3048 m Now, P = (62400 x 9. 81 × (7. 10 × 0. 3048)) P: 1324728. 588 Pa
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