Question

A right triangular plate of base 2.0 m and height 1.0 m is submerged vertically in water, with top vertex 3.0 m below the surface. Find the force on one side of the plate.

Answer 1

Given,

Base =$2.0 \space m$

Height = $1.0 \space m$

We know Density of water = $w=9800 N/m^3$

Force = $?$

Here, x = length =

y = depth = $1$

So, ordered pair =$(0,1)$

One more ordered pair = $(2, 4)$

Equation of a line is $y = mx+ c$

Plug $(0, 1)$

$1 = 0 +c \\ c= 1 \\$

The equation becomes $y=mx+ 1$

Plug $(2, 4)$ in the above equation

$4 = 2m + 1 \\ 2m = 3 \\ m = \frac {3}{2}$

so the equation is $y =\frac {3}{2} x + 1$

So, $x = \frac {2y-2}{3}$

Force =$w \int_a ^b x \space y \space dy$

$= 9800 \int_1 ^4 \frac {2y-2}{3} y \space dy$

$$= \frac {9800 \times 2}{3} \int _1 ^4 ( y^2 - y) dy$$

$$= \frac {9800 \times 2}{3} ( \frac {y^3} {3} - \frac {y^2}{2})_1 ^4$$

$$= \frac {19600} {3} (\frac {64}{3} - \frac {16} {2} - \frac {1}{3} + \frac {1}{2} )$$

$$= \frac {19600} {3} (\frac {128}{6} - \frac {48} {6} - \frac {2}{6} + \frac {3}{6} )$$

$$= \frac {19600} {3} (\frac {128-48-2+3}{6} ) =\frac {19600} {3} (\frac {81}{6})$$

$$Force = 88200 \space N$$