A vertical plane with a width of 2.5 ft and a height of 3.5 ft is submerged at a depth of 4 ft in a tank of water and is subjected to a gage air pressure of 11 psi, as illustrated in Figure EP 2.28. (a) Determine the magnitude of the resultant hydrostatic force acting on the vertical plane. (b)

Question

A vertical plane with a width of 2.5 ft and a height of 3.5 ft is submerged at a depth of 4 ft in a tank of water and is subjected to a gage air pressure of 11 psi, as illustrated in Figure EP 2.28. (a) Determine the magnitude of the resultant hydrostatic force acting on the vertical plane. (b) Determine the location of the resultant hydrostatic force acting on the vertical plane.

Accepted Answer

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

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

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

Alternatively, the magnitude of the hydrostatic force acting on the vertical plane may be determined by applying Equation 2.164 [latex]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) [/latex] as follows:

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

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

[latex]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 [/latex] [latex]h(F):= y_{o}+ y_{F}=5.783 ft [/latex]

Alternatively, the location of the resultant force acting on the vertical plane may be determined by applying Equation 2.169 [latex]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} [/latex] as follows:

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

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

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