To find the hydrostatic force on one end of the tank, we can use the concept of hydrostatic pressure, which depends on the depth of the fluid, the density of the fluid, and the area of the region in question. The hydrostatic force $F$ on a surface submerged in a fluid is given by the integral:

$F = \int_A \rho y \, dA$

where:

• $F$ is the hydrostatic force,
• $\rho$ is the density of the fluid (gasoline in this case),
• $y$ is the depth of the point on the surface,
• $dA$ is an infinitesimal area element of the submerged surface.

Step 1: Define the curves and the area

We are given the following:

• The curves are $y = \frac{x^2}{2}$ and $y = 12$.
• The depth of the fluid is 8 ft, and we are to find the hydrostatic force on the submerged area between these curves.

The first curve is a parabola, and the second curve is a horizontal line at $y = 12$. The width of the region at a particular depth depends on the equation of the parabola.

Step 2: Set up the integral

We need to integrate with respect to $y$, so it’s useful to express $x$ in terms of $y$ from the equation $y = \frac{x^2}{2}$. Solving for $x$, we get:

$x = \sqrt{2y}$

The total width of the region at a given height $y$ is the distance from $-x$ to $x$, which is $2x = 2\sqrt{2y}$.

Next, we will calculate the force. The hydrostatic pressure at depth $y$ is $\rho y$, and the infinitesimal area element is $dA = 2\sqrt{2y} \, dy$.

Thus, the hydrostatic force is given by:

$F = \int_0^8 \rho y (2\sqrt{2y}) \, dy$

Substitute $\rho = 42$ lb/ft³:

$F = \int_0^8 42 y (2\sqrt{2y}) \, dy$

Simplify the integrand:

$F = 84 \sqrt{2} \int_0^8 y^{3/2} \, dy$

Step 3: Solve the integral

The integral of $y^{3/2}$ is:

$\int y^{3/2} \, dy = \frac{2}{5} y^{5/2}$

Now, evaluate the integral from 0 to 8:

$F = 84 \sqrt{2} \left[ \frac{2}{5} y^{5/2} \right]_0^8$

At $y = 8$, we have:

y^{5/2} = 8^{5/2} = (2^3)^{5/2} = 2^{15/2} = 128 \quad \text{(since 8 = 2^3)}

At $y = 0$, $y^{5/2} = 0$.

So,

$F = 84 \sqrt{2} \times \frac{2}{5} \times 128 = 84 \sqrt{2} \times \frac{256}{5}$

$F = \frac{84 \times 256 \times \sqrt{2}}{5} = \frac{21504 \sqrt{2}}{5}$

Finally, approximate $\sqrt{2} \approx 1.414$:

$F \approx \frac{21504 \times 1.414}{5} \approx 6060.2 \, \text{lb}$

Final Answer:

The hydrostatic force on one end of the tank is approximately 6060.2 lb.

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Would you like to see more details on any step, or do you have any questions?

Additional questions to explore:

1. How does the hydrostatic force change if the fluid’s density were different?
2. How can we compute the hydrostatic force for a tank with a different cross-sectional shape?
3. What would happen to the hydrostatic force if the depth of the fluid increased beyond the tank’s height?
4. How would the force calculation change if we used gravity in the force equation?
5. How can the area between the curves $y = \frac{x^2}{2}$ and $y = 12$ be computed?

Tip: The hydrostatic force is more significant when the density of the fluid and the depth increase, as both contribute to the pressure on the surface.