A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. 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.
Base area of the given tank, A=1.0m2 Area of the hinged door, a=20cm2=20×10−4m2 Density of water, ρ1=103kg/m3 Density of acid, ρ2=1.7×103kg/m3 Height of the water column, h1=4m Height of the acid column, h2=4m Acceleration due to gravity, g=9.8 Pressure due to water is given as: P1=h1ρ1g =4×103×9.8=3.92×104Pa Pressure due to acid is given as: P2=h2ρ2g =4×1.7×103×9.8=6.664×104Pa Pressure difference between the water and acid columns: △P=P2−P1 =6.664×104−3.92×104 =2.744×104 Hence, the force exerted on the door =△P×a =2.744×104×20×10−4 =54.88N Therefore, the force necessary to keep the door closed is 54.88N.
