Answer to Question #108764 in Calculus for Edward

Question

A right triangular plate of base 3.0 m and height 1.5 m is submerged vertically in water, with top
vertex 3.5 m below the surface. Find the force on one side of the plate.

Answer 1

The total force on the plate is given by

$$F=w\displaystyle\int_{a}^bxydy$$

where

$x$  is the length (in m) of the element of area (expressed in terms of y)

$y$  is the depth (in m) of the element of area

$w$  is the density of the liquid (in N m^-3)

(for water, this is w = 9800 N m^-3)

$a$  is the depth at the top of the area in question (in m)

$b$  is the depth at the bottom of the area in question (in m)

Before we can proceed, we need to find $x$ in terms of $y.$

Now when $x=0,y=3.5,$ and when $x=3,y=5.$

So we have: $y=mx+b$

$x=0,y=3.5: b=3.5$

$x=3,y=5:5=3m+3.5=>m=0.5$

$$y=0.5x+3.5$$

This gives us 

$$x=2y-7$$

Hence

$$F=9800\displaystyle\int_{3.5}^{5}(2y-7)ydy=9800\bigg[{2y^3\over 3}-{7y^2\over 2}\bigg]\begin{matrix} 5 \\ 3.5 \end{matrix}=$$ $$=99225\ N$$

$F=99225\ N$