To determine the total force of the water on the parabolic dam, we need to follow these steps:

1. General Setup:

The equation of the dam is given as: $y = \frac{x^2}{164}$ where $y$ represents the height of the dam at horizontal coordinate $x$. Assume that water is pressing against the dam, and the water level reaches the top of the dam.

2. Hydrostatic Pressure:

Hydrostatic pressure at any depth $y$ below the water's surface is given by: $P(y) = \rho g h(y)$ where:

• $\rho$ is the density of water ($\approx 1000 \, \text{kg/m}^3$),
• $g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$),
• $h(y)$ is the height of the water column above point $y$, which is the difference between the water's surface height and the depth $y$.

Since the water is at the top of the dam, the height of the water column above any point $y$ on the dam is $h(y) = H - y$, where $H$ is the maximum height of the dam.

3. Force on a Strip:

To calculate the total force, we consider a small horizontal strip of the dam at height $y$ with thickness $dy$. The width of the strip can be found from the equation of the dam $y = \frac{x^2}{164}$, which gives: $x = \sqrt{164 y}$ Thus, the width of the strip at height $y$ is $2x = 2\sqrt{164y}$.

The infinitesimal force on the strip is given by: $dF = P(y) \cdot \text{(area of the strip)} = \rho g (H - y) \cdot 2 \sqrt{164y} \, dy$

4. Total Force:

To find the total force, integrate the infinitesimal force over the height of the dam, from $y = 0$ to $y = H$: $F = \int_0^H \rho g (H - y) \cdot 2 \sqrt{164y} \, dy$ Simplifying and substituting values for $\rho$ and $g$: $F = 2 \times 1000 \times 9.8 \times \int_0^H (H - y) \sqrt{164y} \, dy$

5. Solving the Integral:

Now, let's solve the integral: $\int_0^H (H - y) \sqrt{164y} \, dy$ This can be split into two parts: $\int_0^H H \sqrt{164y} \, dy - \int_0^H y \sqrt{164y} \, dy$

Each part can be solved using standard calculus techniques, which involves substitution and power rule integration.

Let me calculate this for you.The integral evaluates to:

$\frac{8 \sqrt{41} H^{5/2}}{15}$

6. Total Force Expression:

Now we can substitute this result back into the equation for the total force:

$F = 2 \times 1000 \times 9.8 \times \frac{8 \sqrt{41} H^{5/2}}{15}$

Simplifying:

$F = \frac{156800 \sqrt{41} H^{5/2}}{15}$

Thus, the total force of the water on the dam is:

$F = 10453.33 \sqrt{41} H^{5/2} \, \text{Newtons}$

Next Steps:

If you know the value of the height $H$, you can plug it in to compute the numerical value of the force. Let me know if you would like to proceed with that or have further questions.

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Here are five related questions for further exploration:

1. What would the total force be if the dam's height $H$ is 10 meters?
2. How does the shape of the dam affect the distribution of force compared to a rectangular dam?
3. What is the relationship between the height of the water and the total force exerted on the dam?
4. How would the force change if the dam was constructed on a different planet with a different gravitational constant?
5. What if the water level only reached half the height of the dam? How would the total force be affected?

Tip: When calculating hydrostatic forces, remember that pressure increases linearly with depth, and the force distribution is not uniform across the surface.