Pressure Forces on Cylinder

Consider a solid, closed, right circular cylinder of length 2 m 2 \, \text{m} and radius 1 m 1 \, \text{m} . The cylinder is placed below the surface of a body of water.

The water surface is the plane z = 0 z = 0 . The center of the cylinder is at ( C x , C y , C z ) = ( 0 , 0 , 2 ) (C_x,C_y,C_z) = (0,0,-2) . The central axis of the cylinder (along its length) points in the direction of the vector ( v x , v y , v z ) = ( 1 , 1 , 1 ) (v_x,v_y,v_z) = (1,1,1) .

The density of the water is 1000 kg/m 3 1000 \, \text{kg/m}^3 , and the ambient gravitational acceleration is 10 m/s 2 10 \, \text{m/s}^2 in the negative z z direction. Neglect atmospheric pressure.

The total water pressure force acting on the cylinder can be represented as a vector sum of the pressure forces on the top disk face, bottom disk face, and side:

F w a t e r = F T + F B + F S \large{\vec{F}_{water} = \vec{F}_T + \vec{F}_B + \vec{F}_S }

Determine the following ratio of force magnitudes:

Ratio = F T + F B + F S F w a t e r = F T + F B + F S F T + F B + F S \large{\text{Ratio} = \frac{\Big|\vec{F}_T\Big| + \Big|\vec{F}_B\Big| + \Big|\vec{F}_S\Big|}{\Big|\vec{F}_{water}\Big|} = \frac{\Big|\vec{F}_T\Big| + \Big|\vec{F}_B\Big| + \Big|\vec{F}_S\Big|}{\Big|\vec{F}_T + \vec{F}_B + \vec{F}_S\Big|} }


The answer is 2.817.

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