Mechanical Properties of Fluids

Question : 21

Total: 31

A tank with a square base of area 1.0m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20cm2. 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 closed.

Solution:

For compartment containing water,
h1=4 m, ρ1=103kg m−3
The pressure exerted by water at the door provided at bottom,
P1=h1 ρ1g
=4×103×9.8
=3.92×104Pa
For compartment containing acid,
h2=4 m, ρ2=1.7×103kg∕ m3
The pressure exerted by acid at the door provided at bottom,
P2=h2 ρ2g
=4×1.7×103×9.8
=6.664×104 Pa
∴ Difference of pressure =P2–P1
=6.664×104−3.92×104
=2.744×104Pa
Given, area of door, A=20 cm2=20×10−4 m2
Force on the door = difference in pressure × area
=(P2−P1)×A
=(2.744×104)×(20×10−4)
=54.88 N=55 N
To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.

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