N's questions at Yahoo Answers regarding hydrostatic forces

In summary, the hydrostatic force on the shallow end is 5000 lb, on the deep end it is approximately 15313 lb, on one of the sides it is 21250 lb, and on the bottom of the pool it is approximately 69520 lb.
  • #1
MarkFL
Gold Member
MHB
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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.
 
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  • #2
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) \(\displaystyle F=\frac{\rho w\sqrt{\ell^2+\left(d_1-d_1 \right)^2}\left(d_1+d_2 \right)}{2}\)
 
  • #3
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:

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

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

Related to N's questions at Yahoo Answers regarding hydrostatic forces

1. What is hydrostatic force?

Hydrostatic force is the force exerted by a fluid, such as water, on an object submerged in it. It is a result of the pressure exerted by the weight of the fluid on the object.

2. How is hydrostatic force calculated?

Hydrostatic force can be calculated using the formula F = ρghA, where ρ is the density of the fluid, g is the acceleration due to gravity, h is the depth of the fluid, and A is the area of the object submerged in the fluid.

3. What factors affect the magnitude of hydrostatic force?

The magnitude of hydrostatic force is affected by the density of the fluid, the acceleration due to gravity, the depth of the fluid, and the area of the object submerged in the fluid. It is also affected by the shape and orientation of the object.

4. What are some real-life applications of hydrostatic force?

Hydrostatic force is used in a variety of real-life applications, such as in the design of dams, submarines, and ships. It is also important in understanding the behavior of fluids in hydraulic systems, such as in pipes and pumps.

5. How does hydrostatic force differ from hydrodynamic force?

Hydrostatic force is a result of the weight of a fluid on an object, while hydrodynamic force is caused by the motion of the fluid. Hydrostatic force is also dependent on the depth of the fluid, while hydrodynamic force is dependent on the velocity of the fluid.

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