Force of Fluid by Integration: Total hydrostatic force acting on the slant side of trapezoidal trough Member for 10 years 1 month Submitted by Roydon Jude S… on Wed, 03/16/2016 - 17:06 A trough of length 6 ft has for its vertical cross section an isosceles trapezoid. The upper and lower base are 6 ft and 2 ft respectively and the altitude is 2 ft. If the trough is full of water, find the force on the slant side of the trough. Tags Fluid Pressure Log in or register to post comments Re: FORCE FLUID by Integration Member for 17 years 6 months Jhun Vert Thu, 03/17/2016 - 20:59 This is a basic problem in Hydraulics $F_H = \gamma \bar{h} A = 62.4(1)(2 \times 6) = 748.8 ~ \text{lb}$ $F_V = \gamma V = 62.4 \times \frac{1}{2}(2)(2)(6) = 748.8 ~ \text{lb}$ $F = \sqrt{{F_H}^2 + {F_V}^2} = 1058.96 ~ \text{lb}$ Solution by Integration $dF = \gamma h \, dA = \gamma (y \sin 45^\circ)(6 \, dy)$ $F = 374.4 \sin 45^\circ {\displaystyle \int_0^{2\sqrt{2}}} y \, dy$ $F = 1058.96 ~ \text{lb}$ Log in or register to post comments
Re: FORCE FLUID by Integration Member for 17 years 6 months Jhun Vert Thu, 03/17/2016 - 20:59 This is a basic problem in Hydraulics $F_H = \gamma \bar{h} A = 62.4(1)(2 \times 6) = 748.8 ~ \text{lb}$ $F_V = \gamma V = 62.4 \times \frac{1}{2}(2)(2)(6) = 748.8 ~ \text{lb}$ $F = \sqrt{{F_H}^2 + {F_V}^2} = 1058.96 ~ \text{lb}$ Solution by Integration $dF = \gamma h \, dA = \gamma (y \sin 45^\circ)(6 \, dy)$ $F = 374.4 \sin 45^\circ {\displaystyle \int_0^{2\sqrt{2}}} y \, dy$ $F = 1058.96 ~ \text{lb}$ Log in or register to post comments
Re: FORCE FLUID by Integration
Member for
17 years 6 monthsThis is a basic problem in Hydraulics
$F_H = \gamma \bar{h} A = 62.4(1)(2 \times 6) = 748.8 ~ \text{lb}$
$F_V = \gamma V = 62.4 \times \frac{1}{2}(2)(2)(6) = 748.8 ~ \text{lb}$
$F = \sqrt{{F_H}^2 + {F_V}^2} = 1058.96 ~ \text{lb}$
Solution by Integration
$dF = \gamma h \, dA = \gamma (y \sin 45^\circ)(6 \, dy)$
$F = 374.4 \sin 45^\circ {\displaystyle \int_0^{2\sqrt{2}}} y \, dy$
$F = 1058.96 ~ \text{lb}$