I'm working the below problem:
Calculate the fluid force on one side of a right-triangular plate with edges 3ft, 4ft and 5ft if the plate sits at the bottom of a pool filled with water to a depth of 6ft on its 3ft edge and tilted at 60 degrees to the bottom of the pool.
Related Info:
The pressure at a point inside a fluid is given by
$$
P=wh
$$
where $w$ is the weight density(wieght per volume) of the fluid and $h$ is the depth from the surface of the fluid at which the point is situated.
The force due to this pressure could be calculated by multiplying the value of pressure by the area on which the pressure is acting.
$$
F=PA=whA
$$
My Solution:
The triangular plate is a right-angled triangle.
Picture 1
The widht $w$ of the triangle at any height $y$ from the origin is given by
$$ w(y)=3-\frac{3y}{4} \tag{1} $$
Picture 2
Since the plate is tilted at an angle of $60^0$ to the bottom of the pool (taken as $x$-axis), to find the width of the plate at some height $y_k$ from the $x$-axis, we'd have to make the following modification to $(1)$: The length of the plate $l$ (looking from the cross-section view) from the origin varies with $y$ as $$ l=\frac{y}{\sin\frac{\pi}{3}}=\frac{2y}{\sqrt{3}} $$ The $y$ in $(1)$ is $l$ in the case of the tilted plate. Hence $(1)$ becomes $$ w(l)=3-\frac{3l}{4}=3-\left(\frac{3}{4}\right)\frac{2y}{\sqrt{3}}=3-\frac{\sqrt{3}}{2}y $$
Hence, the width of the plate, $w_k$ in our tilted set-up at a height of $y_k$ from the bottom of the pool is given by $$ w_k=3-\frac{\sqrt{3}}{2}y_k $$
Next, let us consider a thin strip of this plate of width $w_k$ and height $\Delta d_k$ (along the length of the plate) at a height of $y_k$ from the bottom of the pool.
Picture 3
Hence the area of this thin strip of plate is given by $$ A_k=w_k\Delta d_k=\left(3-\frac{\sqrt{3}}{2}y_k\right)\Delta d_k $$
From Picture 3, it is clear that $$ \Delta d_k=\frac{\Delta y_k}{\sin\frac{\pi}{3}} = \frac{2\Delta y_k}{\sqrt{3}} $$
Therefore, $$ A_k=\left(3-\frac{\sqrt{3}}{2}y_k\right)\frac{2\Delta y_k}{\sqrt{3}} $$
This strip of plate we're talking about is at a depth of $(6-y_k)$ft below the surface of the pool. Hence, the pressure at any point along this strip would be $$ P_k=w(6-y_k) $$ where $w$ is the weight density of water, which is 62.4 lb/cu. ft.
Consequently, the force acting on this strip would be $$ \begin{aligned} F_k &= A_kP_k \\ &=\left(3-\frac{\sqrt{3}}{2}y_k\right)\frac{2\Delta y_k}{\sqrt{3}}\times w(6-y_k) \\ &=\frac{w}{\sqrt{3}}(6-y_k)(6-\sqrt{3}y_k)\Delta y_k \end{aligned} $$
To find the force acting on the entire strip, we'd have to sum the forces acting on such thin strips along the plate, i.e $$ F=\sum_{k=1}^n F_k=\sum_{k=1}^n \frac{w}{\sqrt{3}}(6-y_k)(6-\sqrt{3}y_k)\Delta y_k $$
and making the strips thinner, we'd have $$ F=\lim_{n\rightarrow\infty}\sum_{k=1}^n \frac{w}{\sqrt{3}}(6-y_k)(6-\sqrt{3}y_k)\Delta y_k $$ which is a Riemann sum: $$ F=\int_0^{2\sqrt{3}}\frac{w}{\sqrt{3}}(6-y)(6-\sqrt{3}y)dy\approx 1814\text{ lb} $$ The limits of integration are
- Bottom of the pool, $y=0$.
- Apex of the plate, $y=2\sqrt{3}$, see Picture 2.
Can someone confirm if this solution is correct?