A thin square plate of side $4 m$ is placed horizontally $2 m$ below the surface of water. Calculate the thrust on one of the surface of the plate.
(Given: Atmospheric pressure, $P_{a t m} = 1.013 \times 10^{5} N / m^{2}$
Density of water, $ρ_{w}$ = $1000 k g / m^{3}$ ,
Acceleration due to gravity, $(g) = 9.8 m s^{- 1}$ )

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Solution

Given :
Depth of plate in the water from the surface, $h = 2 m$
Density of water, $ρ_{w} = 1000 k g / m^{3}$ ,
Acceleration due to gravity, $g = 9.8 m s^{- 1}$
Atmospheric pressure, $P_{a t m} = 1.013 \times 10^{5} N / m^{2}$

Area of plate,
$A = 4 m \times 4 m$
$A = 16 m^{2}$
Total pressure at a depth $h$ below the surface of water is given by:
$P = P_{a t m} + h ρ_{w} g$
$P = (1.013 \times 10^{5}) + (2 \times 10^{3} \times 9.8)$
$P = 1.209 \times 10^{5} P a$

Thrust on surface of plate due to liquid above it is:
$F = P A$
$F = 1.209 \times 10^{5} P a \times 16 m^{2}$
$F = 19.344 \times 10^{5} N$

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