--- title: "Net hydrostatic force on conical surface" subject: "Fluid Mechanics" chapter: 3 key: "semester_3/fluid_mechanics:net-force-conical-surface" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/net-force-conical-surface site: BCT Engineering --- # Net hydrostatic force on conical surface Fluid Mechanics · Chapter 3: Hydrostatics - Asked in: 2079 Baisakh, 8 marks - Video: [Fluid Mechanics: Topic 4.3 - Hydrostatic force on a curved surface](https://www.youtube.com/watch?v=zV-JO-l7Mx4) — CPPMechEngTutorials - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/net-force-conical-surface - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-3 ## Question Pressurized water fills the tank as shown in figure below. Compute the net hydrostatic force on the conical surface ABC. [Figure: water tank with a gauge reading 100 kPa; conical surface ABC with A and C at the top, 3 m horizontal width AC, apex B 6 m below AC; tank water height 7 m] ## Short answer Net hydrostatic force on the conical surface is vertical downward, equal to the pressure force on horizontal plane AC (776.2 kN) plus the weight of water in the cone (138.7 kN), totalling 915 kN. ## Answer **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. ```text 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.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the net force on the conical surface vertical? A. By symmetry the horizontal components cancel B. Water pressure acts only downward C. The surface is horizontal D. The air pressure acts vertically only Answer: A. A symmetric cone has equal opposing horizontal forces, leaving a vertical resultant. (Key point: Direction of net force) 2. The net force on the cone ABC is found as: A. Pressure force on AC minus the weight of the cone water B. Pressure force on the plane AC plus the weight of water inside the cone C. Pressure at B times area of AC only D. Weight of water inside the cone only Answer: B. The vertical force equals the weight of fluid above the surface up to the free surface: the plane AC force plus the cone volume of water. (Key point: Method) 3. The water surface is 1 m above AC and the gauge pressure on it is 100 kPa. What is the pressure at AC? A. 100 kPa B. 158.86 kPa C. 109.81 kPa D. 9.81 kPa Answer: C. p_AC = 100 + 9.81 x 1 = 109.81 kPa. (Key point: Pressure at AC) 4. With F1 = 776.2 kN on plane AC and cone water volume 14.137 m^3, what is the net force? A. About 915 kN downward B. About 637 kN C. About 776 kN D. About 138.7 kN Answer: A. F2 = 9.81 x 14.137 = 138.7 kN; F = 776.2 + 138.7 = 914.9 kN. 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