Problem 3.59

Determine the hydrostatic pressure force Fp due to water acting on a 4 m by 5 m rectangular area AB. Locate its line-of-action LOA.

Work:

Fp = -(integral) pdA, where p = pressure and A = area
p = -(sw)*h, where sw = specific weight and h = height
A = AB = (9.84 ft)*(16.4 ft) = 161.38 ft^2 Fp = -(sw)hA, where h = 9.84 ft
Fp = (sw)*[(9.84 ft)*(161.38 ft^2), where ft = feet
Fp = (sw)*(2,117.25 ft^3), where ^ = raise to the power of
(assuming fresh water at sea level at 60 degrees Fahrenheit, sw = 62.4 lbf/ft^3)
Fp = (64.4 lbf/ft^3)*(2,117.25 ft^3)
Fp = 132,116 lbf

= Coordinates of the line-of-action are Xc.p. and Yc.p., where c.p. = center of pressure
Yc.p.*Fp = (integral) YpdA
Yc.p.*(-sw*h*A) = (integral) YpdA
Yc.p. = (1/A)*(integral) YdA = Ybar, where Ybar = distance in the y-direction to the centroid of the planar surface.
Xc.p. = (1/A)*(integral) XdA = Xbar, where Xbar = distance in the x-direction to the centroid of the planar surface.

Solution

Fp = 132,116 lbf
Yc.p. = (1/A)*(integral) YdA = Ybar
Xc.p. = (1/A)*(integral) XdA = Xbar