Force of Fluid by Integration: Total hydrostatic force acting on the slant side of trapezoidal trough
Submitted by Roydon Jude S. ... on March 16, 2016 - 5:06pm
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.
- Log in to post comments
- 869 reads
Re: FORCE FLUID by Integration
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
$dF = \gamma h \, dA = \gamma (y \sin 45^\circ)(6 \, dy)$
$F = 374.4 \sin 45^\circ {\displaystyle \int_0^{2\sqrt{2}}} y \, dy$
$F = 1058.96 ~ \text{lb}$