Question 2.7: A tank of oil has a right-triangular panel near the bottom, ...

Part (a): The triangle has properties given in Fig. 2.13c. The centroid is one-third up (4 m) and onethird over (2 m) from the lower left corner, as shown. The area is
\frac{1}{2} (6m)(12m)=36m^4
The moments of inertia are
I_{xx} =\frac{bL^3}{36} =\frac{(6 m)(12 m)^3}{36} =288 m^4
and I_{xy}=\frac{b(b-2s)L^2}{72}=\frac{(6 m)[6 m-2(6 m)](12 m)^2}{72} = -72 m^4
The depth to the centroid is h_{CG} =5 + 4 = 9 m; thus the hydrostatic force from Eq. (2.26) is
F= \rho gh_{CG} A=(800 kg /m^3)(9.807 m /s^2)(9 m)(36 m^2)=2.54 \times 10^6 (kg .m)/s^2=2.54 \times 10^6 N = 2.54 MN
Part (b) : The CP position is given by Eqs. (2.29):
y_{CP}=-\frac{I_{xx} \sin \theta }{h_{CG}A}=-\frac{(288 m^4)(sin 30°)}{(9 m)(36 m^2)} = -0.444m

x_{CP}=-\frac{I_{xy}\sin \theta }{h_{CG}A} =-\frac{(-72 m^4)(sin 30°)}{(9 m)(36 m^2)}=+0.111m
The resultant force F= 2.54 MN acts through this point, which is down and to the right of the centroid, as shown in Fig. E2.7.