Question 2.28: A vertical plane with a width of 2.5 ft and a height of 3.5 ...

(a) The magnitude of the hydrostatic resultant force acting on the vertical plane is determined by applying Equation 2.163 F= \int{pdA} =\int_{0}^{L}{(\gamma h(y)+ p_{g} )(wdy)} =\int_{0}^{L}{(\gamma (y_{o}+y )+ p_{g} )(wdy)} , integrating over the length of the vertical plane using the top of the plane as a reference point for integration as follows:

W:= 2.5ft              L: 3.5 ft              A:=w.L = 8.75 ft^{2}              y_{o} := 4 ft              \gamma :=62.417 \frac{Ib}{ft^{3} }

p_{g} :=11 \frac{Ib}{in^{2} } \frac{(12in)^{2} }{(1in)^{2}} =1.584 \times 10^{3} \frac{Ib}{ft^{2} }                               h(y):= y_{o} +y
F:= \int_{0}^{L}{(\gamma .h(y)+ p_{g}).wdy} =1.7 \times 10^{4} Ib

Alternatively, the magnitude of the hydrostatic force acting on the vertical plane may be determined by applying Equation 2.164 F= (p_{ca} +p_{g})A=(\gamma h_{ca}+ p_{g})A=(\gamma (y_{o}+ y_{ca} )+p_{g} )A= (\gamma (y_{o}+L/2 )+ p_{g} )(wL) as follows:

y_{ca}:= \frac{L}{2} =1.75 ft                             h(ca):= y_{o}+ y_{ca}=5.75 ft
F:= (\gamma .h(ca)+ p_{g} ).A=1.7 \times 10^{4} Ib

(b) The location of the resultant hydrostatic force acting on the vertical plane is determined by applying Equation 2.166 y_{F} =\frac{M}{F} as follows:

y_{F}:= \frac{\int_{0}^{L}{y.(\gamma .h(y)+ p_{g}).wdy} }{\int_{0}^{L}{(\gamma .h(y) + p_{g} ).wdy} } =1.783 ft                    h(F):= y_{o}+ y_{F}=5.783 ft

Alternatively, the location of the resultant force acting on the vertical plane may be determined by applying Equation 2.169 h_{F} =\frac{\gamma (I_{x-ca}+Ah^{2}_{ca} )+p_{g} h_{ca}A}{(\gamma h_{ca}+ p_{g})A} = \frac{\gamma (wL^{3}/12 + (wL)(y_{o}+L/2 )^{2})+ p_{g}(y_{o}+L/2) (wL)}{(\gamma (y_{o}+L/2)+p_{g})wL} as follows:

I_{xca} := \frac{w. L^{3} }{12} =8.932 ft^{4}                           h(F):= \frac{\gamma .(I_{xca} +A. h^{2}_{ca} )+ p_{g}. h(ca).A}{(\gamma .h(ca)+ p_{g}).A} = 5.783 ft