Question

A tank with a square base of area 1.0 m${}^2$ is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm${}^2$. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.

Answer 1

Base area of the given tank, $A=1.0m^2$

Area of the hinged door, $a=20c{m}^2=20\times {10}^{-4}{m}^2$

Density of water, ${\rho }_1={10}^{3}kg/m^3$

Density of acid, ${\rho }_2=1.7\times {10}^{3}kg/m^3$

Height of the water column, $h_1=4m$
Height of the acid column, $h_2=4m$
Acceleration due to gravity, $g=9.8$
Pressure due to water is given as:
$P_1=h_1{\rho }_{1}g$
$=4\times {10}^3\times 9.8=3.92\times {10}^{4}Pa$
Pressure due to acid is given as:
$P_2=h_2{\rho }_{2}g$
$=4\times 1.7\times {10}^3\times 9.8=6.664\times {10}^{4}Pa$
Pressure difference between the water and acid columns:
$△P=P_2-P_1$
$=6.664\times {10}^4-3.92\times {10}^4$
$=2.744\times {10}^4$
Hence, the force exerted on the door $=△P\times a$
$=2.744\times {10}^4\times 20\times {10}^{-4}$
$=54.88N$
Therefore, the force necessary to keep the door closed is $54.88N$.