Question 2.10: A tank 20 ft deep and 7 ft wide is layered with 8 ft of oil,...

Divide the end panel into three parts as sketched in Fig. E2.10, and find the hydrostatic
pressure at the centroid of each part, using the relation (2.26) in steps as in
Fig. E2.10:
p_{CG1} = (55.0 lbf/ft^3)(4 ft) =220 lbf/ft^2

p_{CG3} = (55.0)(8) + 62.4(6) + 846(2) = 2506 lbf/ft^2
These pressures are then multiplied by the respective panel areas to find the force on each
portion:
F_{1}=p_{CG1}A_{1}=(220 lbf/ft^2)(8 ft)(7 ft) =12,300 lbf

F = \sum{F_{i}}=108,800 lbf
Part (b)
Equations (2.32) can be used to locate the CP of each force F_{I} noting that \theta= 90° and
\sin \theta=1 for all parts. The moments of inertia are I_{xx1} =(7 ft)(8 ft)^3/12 = 298.7 ft^4,

I_{xx2} =7(6)^3/12 = 126.0 ft^4 , and I_{xx3}=7(4)^3/12 =37.3 ft^4The centers of pressure are thus at
y_{CP1} =- \frac{ \rho_{1}gI_{xx1} }{F_{1} }= – \frac{(55.0 lbf/ft^3)(298.7 ft^4)}{12,300 lbf}=-1.33 ft

This locates Z_{CP1} =-4 -1.33 = -5.33 ft , Z_{CP2} =-11-0.30 = -11.30 ft, and Z_{CP3} =-16-0.45 = -16.45 ft.Summing moments about the surface then gives
\sum{ F_{i} Z_{CPi}}= F Z_{CPi}
or 12,300(-5.33) + 26,300(-11.30) + 70,200(-16.45) = 108,800Z_{CP}
or Z_{CP}= -\frac{1,518,000}{108,800}= -13.95 ft
The center of pressure of the total resultant force on the right side of the tank lies 13.95 ft below the surface.