so that 2 ft (60.96 cm) of the member will be above the surface? Use these specific
weights: timber = 38 lb/ft^3 (5969 N/m^3 ); saltwater = 64 lb/ft^3 (10,053 N/m^3 ); concrete =
145 lb/ft^3 (22,777 N/m^3 ).
Calculation Procedure:
- Express the weight of the body and the volume of the displaced
liquid in terms of the volume of concrete required
Archimedes' principle states that a body immersed in a liquid is subjected to a vertical
buoyant force equal to the weight of the displaced liquid. In accordance with the equa-
tions of equilibrium, the buoyant force on a floating body equals the weight of the body.
Therefore,
W= Fw (1)
Let x denote the volume of concrete. Then W= (90/144)(12)(38) + 145* .= 285 + 145*;
V= (90/144)(12 - 2) + jc = 6.25 +jc.
- Substitute in Eq. 1 and solve for x
Thus, 285 + 145 = (6.25 + )64; x = 1.42 ft^3 (0.0402 m^3 ).
HYDROSTATIC FORCE
ON A PLANE SURFACE
In Fig. I, AB is the side of a vessel containing water, and CDE is a gate located in this
plane. Find the magnitude and location of the resultant thrust of the water on the gate
when the liquid surface is 2 ft (60.96 cm) above the apex.
Calculation Procedure:
- State the equations for the resultant magnitude and position
In Fig. 1, FH denotes the centroidal axis of area CDE that is parallel to the liquid surface,
and G denotes the point of application of the resultant force. Point G is termed the pres-
sure center.
Let A = area of given surface, ft^2 (cm^2 ); P ~ hydrostatic force on given surface, Ib (N);
pm = mean pressure on surface, lb/ft^2 (kPa); yCA and ypc = vertical distance from cen-
troidal axis and pressure center, respectively, to liquid surface, ft (m); ZCA and zpc = dis-
tance along plane of given surface from the centroidal axis and pressure center, respec-
tively, to line of intersection of this plane and the liquid surface, ft (m); ICA = moment of
inertia of area with respect to its centroidal axis, ft^4 (m^4 ).
Consider an elemental surface of area dA at a vertical distance y below the liquid sur-
face. The hydrostatic force dP on this element is normal to the surface and has the magni-
tude
dP = wydA (2)
By applying Eq. 2 develop the following equations for the magnitude and position of
the resultant force on the entire surface:
