Part (a): Expression for Hydrostatic Pressure $p(y)$

The hydrostatic pressure $p(y)$ at a height $y$ from the base of the dam is given by the formula:

$p(y) = \rho g h(y)$

Where:

• $\rho$ is the density of water ($\rho = 1000 \, \text{kg/m}^3$),
• $g$ is the acceleration due to gravity ($g = 9.8 \, \text{m/s}^2$),
• $h(y)$ is the depth of the water at height $y$ from the base.

The height of the water level is 10 m, so the depth of the water is $h(y) = 10 - y$ for $0 \leq y \leq 10$.

Thus, the formula for the hydrostatic pressure becomes:

$p(y) = 1000 \cdot 9.8 \cdot (10 - y)$ $p(y) = 9800 \cdot (10 - y) \, \text{Pa} \quad (\text{Pascal, which is N/m}^2)$

This is the hydrostatic pressure at a given height $y$ in the water, where $y$ is measured from the base of the dam. The pressure decreases linearly as we move upward from the bottom.

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Part (b): Total Hydrostatic Force on the Dam

To calculate the total hydrostatic force $F$ exerted on the dam, we integrate the pressure $p(y)$ along the height of the water. The force exerted by the water on a thin strip of the dam at a height $y$ is given by:

$\Delta F = p(y) \cdot A(y) \cdot \Delta y$

Where:

• $p(y)$ is the hydrostatic pressure at height $y$,
• $A(y)$ is the area of the strip of the dam at height $y$,
• $\Delta y$ is the small vertical height increment.

To compute the area $A(y)$, we note that the dam has the shape of an isosceles trapezoid, and the width of the water at height $y$ varies linearly as we go up. The width $w(y)$ at height $y$ is:

$w(y) = 20 + \left( \frac{40 - 20}{10} \right) y = 20 + 2y$

So, the area of the strip at height $y$ is:

$A(y) = w(y) \cdot \Delta y = (20 + 2y) \Delta y$

Now, we can express the total force by integrating over the height of the water (from $y = 0$ to $y = 10$):

$F = \int_0^{10} p(y) \cdot w(y) \, dy$

Substituting the expressions for $p(y)$ and $w(y)$:

$F = \int_0^{10} 9800 \cdot (10 - y) \cdot (20 + 2y) \, dy$

Let's expand the integrand:

$F = 9800 \int_0^{10} (10 - y)(20 + 2y) \, dy$ $F = 9800 \int_0^{10} \left( 200 + 20y - 20y - 2y^2 \right) \, dy$ $F = 9800 \int_0^{10} \left( 200 - 2y^2 \right) \, dy$

Now, we integrate term by term:

$F = 9800 \left[ 200y - \frac{2y^3}{3} \right]_0^{10}$ $F = 9800 \left[ (200 \times 10) - \frac{2 \times 10^3}{3} \right]$ $F = 9800 \left[ 2000 - \frac{2000}{3} \right]$ $F = 9800 \left[ \frac{6000}{3} - \frac{2000}{3} \right]$ $F = 9800 \times \frac{4000}{3}$ $F = \frac{39200000}{3}$ $F \approx 13066666.67 \, \text{N}$

Thus, the total hydrostatic force exerted on the dam is approximately 13,066,667 N.

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Would you like further details or clarifications on any part? Here are some related questions to explore:

1. How does the shape of the dam affect the hydrostatic force?
2. What other factors could influence the hydrostatic pressure on a dam besides water depth?
3. How can the formula for the total force be adapted if the dam's shape changes?
4. Why is the pressure at a given height $y$ dependent on the depth of the water, not just the height of the dam?
5. How would the total hydrostatic force change if the water height was increased to 12 m?

Tip: When solving integrals of physical quantities, always ensure that your limits correspond to the actual physical domain (e.g., $y$ from 0 to the water height) and verify the units for consistency throughout.