# An open tank has a vertical partition and on one side contains gasolin

An open tank has a vertical partition and on one side contains gasoline with a density $p=700\frac{kg}{{m}^{3}}$ at a depth of 4m. Arectangular gate that is 4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly added to the empty side of the tank. At what depth, h, will the gate start to open?

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esfloravaou
The depth of water possible that will cause for the gate to open can be solved by summing up the forces acting on the wall to the hinge.
The forces acting on the wall are the weight and the hydrostatic forces cause by the water and the gasoline.
Use $\gamma$ for water which is equal to $9.80\frac{kN}{{m}^{3}}$ trom Table 1.5.
The specific weight of the gasoline can be solved by multiplying its density by the acceleration due to gravity which is equal to $9.81\frac{m}{{s}^{2}}$
${\gamma }_{gas}=p\left\{gas\right\}×g$
$=700\frac{kg}{{m}^{3}}×9.81\frac{m}{{s}^{2}}$
$=6867\frac{N}{{m}^{3}}$
$=6.87\frac{kN}{{m}^{3}}$
The hydrostatic force on the gate can be solved using the formula ${F}_{R}=\gamma {h}_{c}A$ where $\gamma$ is the specific weight of liquid, ${h}_{c}$ is the centroid of the gate with liquid and A is the area of the gate with liquid.
Solve for the hydrostatic force exerted by the gasoline.
${F}_{R1}={\gamma }_{gas}{h}_{c}A$
$=6.87\frac{kN}{{m}^{3}}\left(\frac{4m}{2}\right)\left(4m×2m\right)$
$=109.92kN$
Solve for the hydrostatic force exerted by the water.
${F}_{R2}={\gamma }_{water}{h}_{c}A$
$=9.80\frac{kN}{{m}^{3}}\left(\frac{h}{2}\right)\left(h×2m\right)$
Sum up the moment of the two hydrostatic forces.
Note that the location of the hydrostatic force for rectangular plane surfaces can be found $\frac{1}{3}$ of the height of the liquid from the base.
${F}_{R1}×\frac{1}{3}\cdot 4m={F}_{R2}×\frac{1}{3}\cdot h$
$=109.92kN×\frac{1}{3}\cdot 4m=9.80{h}^{2}\frac{kN}{{m}^{2}}×\frac{1}{3}\cdot h$
$146.56kN-m=3.27{h}^{3}\frac{kN}{{m}^{2}}$
$\frac{146.56kN-m}{3.27\frac{kN}{{m}^{2}}}$
$44.87{m}^{3}={h}^{3}$
$\sqrt{3}\left\{44.87{m}^{3}\right\}=\sqrt{3}\left\{{h}^{3}\right\}$
$3.55m=h$
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intacte87
${F}_{Rg}={\gamma }_{gh}{C}_{g}{A}_{g}$
where g refers to gasoline..
${F}_{Rg}=\left(700\frac{kg}{{m}^{3}}\right)\left(9.81\frac{m}{{s}^{2}}\right)\left(2m\right)\left(4m\star 2m\right)$
$=110\star {10}^{3}N$
$=110kN$
${F}_{Rw}={\gamma }_{w}{h}_{cw}{A}_{w}$
where w refers to water...
${F}_{Rw}=\left(9.80\star 103\frac{N}{{m}^{3}}\right)\left(\frac{h}{2}\right)\left(2m\star h\right)$
where h is depth of the water...
${F}_{Rw}=\left(9.80\star 103\right){h}^{2}$
For Equilibrium,
$\sum MH=0$
so that,
${F}_{Rw}\star {I}_{w}={F}_{Rg}\star {I}_{g}$
with ${l}_{w}=\frac{h}{3}$ and ${I}_{p}=\frac{4}{3}m$
Thus,
$\left(9.80\star {10}^{3}\right)\left({h}^{2}\right)\left(\frac{h}{3}\right)=\left(110\star {10}^{3}N\right)\left(\frac{4}{3}m\right)$
$=h=3.55m$