Net hydrostatic force on conical surface

Reading of the figure: the conical surface ABC hangs below the plane AC, which is 3 m across (radius 1.5 m); the apex B is 6 m below AC and the water stands 7 m above B, so the free surface is 1 m above AC. The air above the water has a gauge pressure of 100 kPa. The water fills the cone, so it lies on the inside of the surface.

   gauge 100 kPa
  ------------------- free surface
        1 m
  A -------+------- C   (3 m)
    \      |      /
     \     | 6 m /
      \    |    /
       \   |   /
          B

By symmetry the horizontal components cancel; the net force is vertical (downward). It equals the pressure force on the plane AC plus the weight of water in the cone.

Pressure at AC:

$$p_{AC}=100+9.81(1)=109.81\ \text{kPa}$$

Area of AC:

$$A=\pi(1.5)^{2}=7.069\ \text{m}^2$$

Pressure force on AC:

$$F_1=109.81\times7.069=776.2\ \text{kN}$$

Weight of water in the cone ($V=\tfrac13\pi r^{2}h$):

$$V=\tfrac13(7.069)(6)=14.137\ \text{m}^3,\qquad F_2=9.81\times14.137=138.7\ \text{kN}$$ $$F_{net}=F_1+F_2=776.2+138.7=914.9\ \text{kN}$$

Answer: net hydrostatic force on the conical surface $\approx915$ kN, acting vertically downward along the axis. (If the free surface is 7 m above AC instead, use $p_{AC}=100+9.81\times7$ in the same method.)