# Fluid mechanics

A rectangular surface is placed in a vertical position in fresh water with its top and bottom edges horizontal, as indicated in Examination Figure 1. If the dimensions b and d of the immersed surface are 8.00ft and 12.00ft and the vertical distance h from the surface of the water to the top edge of the immersed surface is 10.00ft, what is the magnitude of the resultant force on one side of the immersed surface?

The total force on a face of the immersed surface is $F= P*S$ where P is the pressure on the surface and S is the surface area.

The pressure at a height h below the surface submerged is $P(h) =\rho*g*h$ where $\rho$ is the density of the water.

Let y be the height below the submerged surface and $x = 8ft =2.4 m$ the horizontal dimension of the surface. $y(max) =12+10 =22 ft =6.6 m$

$y(min) =10 ft = 3 m$

$S =x*dy$

$F =\int_3^{6.6} (\rho*g*y*(x*dy)) =\rho*g*x*(y^2/2)|_3^{6.6} =$

$= 1000*9.81*2.4*(6.6^2/2 -3*3/2) = 406840 N =406840/9.81 = 41472 kgforce$

$=41427*2.2 =91238 lb$