N's questions at Yahoo Answers regarding hydrostatic forces

[MarkFL]

Here are the questions:

Hard calculus question involving hydrostatic force?

A swimming pool is 10 ft wide and 20 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and the deep end, 7 ft. Assume the pool is full of water. (Round your answers to the nearest whole number. Recall that the weight density of water is 62.5 lb/ft3.)
(a) Estimate the hydrostatic force on the shallow end.
?? lb

(b) Estimate the hydrostatic force on the deep end.
? lb

(c) Estimate the hydrostatic force on one of the sides.
?? lb

(d) Estimate the hydrostatic force on the bottom of the pool.
?? lb

Please answer all 4 questions. Thanks

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello N,

Let's develop some general formulas which we can use to answer the questions. First, let's find the hydrostatic force on a vertically submerged rectangle, whose width is $w$ and height is $h$. If we orient a vertical $x$-axis along one side of the rectangle, with the origin at the surface and the positive direction down, we may compute the force along one horizontal elemental rectangle having area $A(x)$ making up the entire rectangle as:

\(\displaystyle dF=\rho xA(x)\)

The area of this elemental rectangle is its width at $x$ times its height $dx$, hence:

\(\displaystyle dF=\rho xw(x)\,dx\)

where:

\(\displaystyle w(x)=w\) (the width is constant for a rectangle)

And so we have:

\(\displaystyle dF=\rho w x\,dx\)

Now, if the top edge of the rectangle is at $x=x_1$ and the bottom edge is at $x=x_2$, then by summing all of the elements of the force, we find:

\(\displaystyle F=\rho w\int_{x_1}^{x_2}x\,dx\)

Applying the FTOC, we obtain:

\(\displaystyle F=\frac{\rho w}{2}\left[x^2 \right]_{x_1}^{x_2}=\frac{\rho w}{2}\left(x_2^2-x_1^2 \right)\)

Since the height $h$ of the rectangle is \(\displaystyle h=x_2-x_1\), we may write:

(1) \(\displaystyle F=\frac{\rho wh}{2}\left(2x_1+h \right)\)

Next, let's consider the hydrostatic force on a right triangle whose vertical leg lies along the $x$-axis. Let the horizontal leg, the base of the triangle, be of length $b$, and the vertical leg, the altitude of the triangle be of length $h$.

We may begin as before, computing the elemental force:

\(\displaystyle dF=\rho xw(x)\,dx\)

This time, the width of the triangle varies with the depth, and does so linearly. At a depth of $x=x_1$, we have $w(x)=0$ and at a depth of $x=x_1+h$, we have $w(x)=b$. Hence:

\(\displaystyle w(x)=\frac{b}{h}\left(x-x_1 \right)\)

And this allows us to write:

\(\displaystyle dF=\frac{\rho b}{h} x\left(x-x_1 \right)\,dx\)

\(\displaystyle dF=\frac{\rho b}{h}\left(x^2-x_1x \right)\,dx\)

Summing all of these elements of force, we find:

\(\displaystyle F=\frac{\rho b}{h}\int_{x_1}^{x_1+h} x^2-x_1x\,dx\)

Applying the FTOC, we find:

\(\displaystyle F=\frac{\rho b}{6h}\left[x^2\left(2x-3x_1 \right) \right]_{x_1}^{x_1+h}=\frac{\rho b}{6h}\left(\left(x_1+h \right)^2\left(2\left(x_1+h \right)-3x_1 \right)-\left(x_1 \right)^2\left(2\left(x_1 \right)-3x_1 \right) \right)=\)

\(\displaystyle \frac{\rho b}{6h}\left(\left(x_1+h \right)^2\left(2h-x_1 \right)+x_1^3 \right)=\frac{\rho bh}{6}\left(3x_1+2h \right)\)

And so we have for the right triangle:

(2) \(\displaystyle F=\frac{\rho bh}{6}\left(3x_1+2h \right)\)

Finally, let's consider a rectangle of width $w$ and height $h$ submerged at an angle such that the rectangle is an inclined plane. We may orient an $x$-axis that runs down the surface of the plane, parallel to its height, and whose origin is at the upper edge. Let $d(x)$ be the depth of the fluid, where $d(0)=d_1$ is the depth of the upper edge and $d\left(h \right)=d_2$ is the depth of the lower edge. We may then state:

\(\displaystyle d(x)=\frac{d_2-d_1}{h}x+d_1\)

We may begin as before, computing the elemental force:

\(\displaystyle dF=\rho w \left(\frac{d_2-d_1}{h}x+d_1 \right)\,dx\)

Summing the elements, we obtain:

\(\displaystyle F=\rho w\int_0^{h} \frac{d_2-d_1}{h}x+d_1\,dx\)

Application of the FTOC gives us:

\(\displaystyle F=\rho w\left[\frac{d_2-d_1}{2h}x^2+d_1x \right]_0^{h}=\rho w\left(\frac{d_2-d_1}{2h}h^2+d_1h \right)=\rho w\left(\frac{d_2h+d_1h}{2} \right)=\frac{\rho wh\left(d_1+d_2 \right)}{2}\)

If $\ell$ is the horizontal distance through which the plane occupies, then by Pythagoras, we may state:

\(\displaystyle h=\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\)

And so we may state:

(3) \(\displaystyle F=\frac{\rho w\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\left(d_1+d_2 \right)}{2}\)

 

[MarkFL]

We now have sufficient formulas to answer the questions.

(a) Estimate the hydrostatic force on the shallow end.

We may use (1) where:

\(\displaystyle \rho=\frac{125}{2}\frac{\text{lb}}{\text{ft}^3},\,w=10\text{ ft},\,h=4\text{ ft},\,x_1=0\text{ ft}\)

Hence:

\(\displaystyle F=\frac{\left(\frac{125}{2}\frac{\text{lb}}{\text{ft}^3} \right)\left(10\text{ ft} \right)\left(4\text{ ft} \right)}{2}\left(2\left(0\text{ ft} \right)+4\text{ ft} \right)=5000\text{ lb}\)

(b) Estimate the hydrostatic force on the deep end.

We may use (1) where:

\(\displaystyle \rho=\frac{125}{2}\frac{\text{lb}}{\text{ft}^3},\,w=10\text{ ft},\,h=7\text{ ft},\,x_1=0\text{ ft}\)

Hence:

\(\displaystyle F=\frac{\left(\frac{125}{2}\frac{\text{lb}}{\text{ft}^3} \right)\left(10\text{ ft} \right)\left(7\text{ ft} \right)}{2}\left(2\left(0\text{ ft} \right)+7\text{ ft} \right)=15312.5\text{ lb}\approx15313\text{ lb}\)

(c) Estimate the hydrostatic force on one of the sides.

We may first consider the rectangular portion of the side, and using (1) where:

\(\displaystyle \rho=\frac{125}{2}\frac{\text{lb}}{\text{ft}^3},\,w=20\text{ ft},\,h=4\text{ ft},\,x_1=0\text{ ft}\)

Hence:

\(\displaystyle F_1=\frac{\left(\frac{125}{2}\frac{\text{lb}}{ \text{ft}^3} \right)\left(20\text{ ft} \right)\left(4\text{ ft} \right)}{2}\left(2\left(0\text{ ft} \right)+4\text{ ft} \right)=10000\text{ lb}\)

Now for the triangular portion we may use (2) where:

\(\displaystyle \rho=\frac{125}{2}\frac{\text{lb}}{\text{ft}^3},\,b=20\text{ ft},\,h=3\text{ ft},\,x_1=4\text{ ft}\)

\(\displaystyle F_2=\frac{\left(\frac{125}{2}\frac{\text{lb}}{ \text{ft}^3} \right)\left(20\text{ ft} \right)\left(3\text{ ft} \right)}{6}\left(3\left(4\text{ ft} \right)+2\left(3\text{ ft} \right) \right)=11250\text{ lb}\)

Thus the total force exerted on one of the sides is:

\(\displaystyle F=F_1+F_2=21250\text{ lb}\)

(d) Estimate the hydrostatic force on the bottom of the pool.

Here we may use (3) where:

\(\displaystyle \rho=\frac{125}{2}\frac{\text{lb}}{\text{ft}^3},\,w=10\text{ ft},\,\ell=20\text{ ft},\,d_1=4\text{ ft},\,d_2=7\text{ ft}\)

\(\displaystyle F=\frac{\left(\frac{125}{2}\frac{\text{lb}}{ \text{ft}^3} \right)\left(10\text{ ft} \right)\sqrt{\left(20\text{ ft} \right)^2+\left(7\text{ ft}-4\text{ ft} \right)^2}\left(4\text{ ft}+7\text{ ft} \right)}{2}=\frac{6875}{2}\sqrt{409}\text{ lb}\approx69519\text{ lb}\)