The force on one of the ends due to hydrostatic pressure is 332.8lb
Data;
- Density = 62.4 lb/ft^3
- length = 4ft
- radius = 2ft
Force Due to Pressure
The force due to hydrostatic pressure can be calculated as
From the attached diagram;
[tex]F = pressure * area\\density = 62.4 lb/ft^3\\depth of water = 2 - y\\pressure = (2 - y)(62.4)\\pressure = 124.8 - 62.4y\\[/tex]
We can proceed as
[tex]x^2 + (y - 2)^2 = 2^2\\x^2 = 4 - (y - 2)^2\\x = +- \sqrt{4y - y^2}\\[/tex]
this implies that
[tex]2x = 2\sqrt{4y - y^2}[/tex]
The area is given as
[tex]\delta A = (2x)*\delta y\\\delta A = 2\sqrt{4y - y^2 \delta y}[/tex]
The force would be given by
[tex]\delta F = (2-y)(62.4)(2\sqrt{4y - y^2})\delta y[/tex]
The total force is given by
[tex]F = \int\limits^2_0 {(2-y)(62.4)(2\sqrt{4y - y^2}) } \, dy\\F = 124.8\int\limits^2_0 {(2-y)(\sqrt{4y - y^2}) } \, dy\\F = 124.8[-\frac{1}{3}y(y -4)(\sqrt{4y -y^2}]_0^2\\F = 124.8[-\frac{1}{3}(2)(2-4)\sqrt{4(2)-2^2}\\ F = 332.8lb[/tex]
The force on one of the ends due to hydrostatic pressure is 332.8lb
Learn more on hydrostatic pressure here;
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