Question 3.SP.6: The cubic tank shown in Fig. S3.6 is half full of water. Fin......

(a)                p_{\text {bott }}=p_{\text {air }}+\gamma_{\text {water }} h_{\text {water }}=8 \mathrm{kN} / \mathrm{m}^{2}+\left(9.81 \mathrm{kN} / \mathrm{m}^{3}\right)(1 \mathrm{~m})

=17.81 \mathrm{kN} / \mathrm{m}^{2}=17.81 \mathrm{kPa}

(b) The force acting on the tank end is divided into two components, labeled A and B on the pressure distribution sketch. Component A has a uniform pressure distribution, due to the pressure of the confined air, which acts throughout the water:

F_{A}=p_{\text {air }} A_{\text {air }}=\left(8 \mathrm{kN} / \mathrm{m}^{2}\right)\left(4 \mathrm{~m}^{2}\right)=32.0 \mathrm{kN}

For component  B , i.e., the varying water pressure distribution on the lower half of the tank wall, the centroid C of the area of application is at

h_{c}=y_{c}=0.5(1 \mathrm{~m})=0.5 \mathrm{~m} \text { below the water top surface, }

so, from Eq. (3.16)

F_{B}=\gamma_{\text {water }} h_{c} A_{\text {water }}=9.81(0.5) 2=9.81 \mathrm{kN}

So the total force on the tank wall is

F=F_{A}+F_{B}=32.0+9.81=41.8 \mathrm{kN} \quad A N S

(c) The locations of the centers of pressure of the component forces, as distances y_{p} below the water top surface, are

\left(y_{p}\right)_{A}=0 \mathrm{~m}

below the water top surface, to the centroid of the 2 -m-square area for the uniform air pressure.

\left(y_{p}\right)_{B}=\frac{2}{3} h_{\text {water }}=\frac{2}{3}(1 \mathrm{~m})=0.667 \mathrm{~m}

below the water top surface for the varying pressure on the rectangular wetted wall area. We could also find this using Eq. (3.18) with y_{c}=0.5 \mathrm{~m}, I_{c}=b h^{3} / 12

Taking moments:          F\left(y_{p}\right)=F_{A}\left(y_{p}\right)_{A}+F_{B}\left(y_{p}\right)_{B}

from which y_{p}=0.1565 \mathrm{~m} below the water top surface .

   F=\gamma h_c A    (3.16)

y_p=y_c+\frac{I_c}{y_c A}      (3.18)