This **question aims to determine** the **depth of a tank given the density of liquid,** **height,** and **width of the tank. **This article uses the concept of force exerted by the liquid on the **walls of the tank.**

The **magnitude of hydrostatic force** applied to the immersed surface is given by:

\[F = P_{c}A \]

**Expert Answer**

The depth of water that will cause the **gate to open** can be solved by adding the forces acting on wall to the hinge. The **forces acting** on the wall are weight and **hydrostatic** due to **water and gasoline.**

The $\gamma $ for the **water** is given as:

\[\gamma = 9.80 \dfrac { kN }{m ^ {3}} \]

The **specific gravity of gasoline** can be solved by **multiplying its density** by the **acceleration due to gravity,** which equals $9.81 \dfrac{m}{s^{2}}$.

\[\gamma_{gas} = p_{gas} \times g \]

\[ =700 \dfrac{kg}{m^{3}} \times 9.81 \dfrac{m}{s ^ {2}}\]

\[ = 6867 \dfrac{N}{m^{3}} \]

\[ = 6.87 \dfrac{kN}{m^{3}} \]

**Hydrostatic force** on 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 gate with liquid** and $ A $ is area of the gate with liquid.

The** hydrostatic force exerted by the gasoline** is calculated as:

\[ F_{R1} = \gamma _{gas} h_{c} A \]

\[ = 6.87 \dfrac{kN}{m^{3}} (\dfrac {4m}{2}) (4m \times 2m ) \]

\[ = 109.92 kN \]

The hydrostatic force exerted by the water is calculated as:

\[ F_{R1} = \gamma _{water} h_{c} A \]

\[F_{R2} = 9.80 \dfrac { kN }{m^{3}} (\dfrac {h}{2}) (h \times 2m) \]

\[F_{R2} = 9.80 h^{2} \dfrac { kN }{m^{3}} \]

The location of hydrostatic force for rectangular plane surfaces can be found $\dfrac {1}{3} $ height of the liquid from the base.

\[ F_{R1} \times \dfrac{1}{3} .4m = F_{R2} \times \dfrac{1}{3} .h \]

\[ 109.92 kN\times \dfrac{1}{3} .4m = 9.80 h^{2} \dfrac { kN }{m^{3}} \times \dfrac{1}{3} .h \]

\[ 1146.56 kNm = 3.27 h^{3} \dfrac { kN }{m^{2}} \]

\[ h^{3} = 44.87 m^{3} \]

\[ h=3.55m \]

**Numerical Result**

The **depth $ h $ of the tank is** $3.55m$.

**Example**

A tank has a vertical partition and on one side contains gasoline with a density $p = 500 \dfrac {kg}{m^{3}}$ at a depth of $6\:m$. A rectangular gate that is $6\:m$ high and $3\: m$ wide and hinged at one end is located in the partition. Water is added to the empty side of the tank. At what depth, h, will the gate start to open?

**Solution**

The $\gamma $ for the water is given as:

\[\gamma = 9.80 \dfrac { kN }{m ^ {3}} \]

\[\gamma_{gas} = 4.9\dfrac{kN}{m ^ {3}} \]

The** hydrostatic force exerted by the gasoline** is calculated as:

\[F_{R1} = 4.9 \dfrac{kN}{m ^ {3}} (\dfrac {6m}{2}) (6m \times 3m ) \]

\[ = 264.6 kN \]

The **hydrostatic force exerted by the water** is calculated as:

\[F_{R2} = 14.7 h ^ {2} \dfrac { kN }{m ^ {3}} \]

The **height of the tank is calculated** as:

\[ h =4.76m \]