The problem asks to find the fluid force (in newtons) on a vertical plate submerged in water. The plate is a square with dimensions of 5 meters by 5 meters, and the weight-density of water is given as 9,800 newtons per cubic meter. The square is positioned vertically, submerged in water.

Solution Steps:

The fluid force $F$ on a submerged vertical surface can be calculated using the formula:

$F = \gamma \int_{a}^{b} h(y) \cdot w(y) \, dy$

Where:

• $\gamma = 9,800 \, \text{N/m}^3$ is the weight-density of water.
• $h(y)$ is the depth of the point $y$ below the water surface.
• $w(y)$ is the width of the plate at the point $y$.
• $a$ and $b$ are the limits of the submerged plate.

For a vertical square plate submerged in water, with its top edge 5 meters below the surface, the force distribution depends on the depth at any point $y$ along the height of the plate.

Given information:

• The plate is a square, so the width is constant, $w(y) = 5$ meters.
• The depth $h(y)$ varies linearly with $y$. At the top of the plate, the depth is $5$ meters, and at the bottom, it's $10$ meters.

Step-by-step:

1. Set up the integral: $F = \gamma \cdot w \int_{5}^{10} h(y) \, dy$ Since $w(y) = 5$ meters and $h(y) = y$, we get: $F = 9,800 \cdot 5 \int_{5}^{10} y \, dy$

2. Solve the integral: $\int_{5}^{10} y \, dy = \left[ \frac{y^2}{2} \right]_{5}^{10} = \frac{10^2}{2} - \frac{5^2}{2} = \frac{100}{2} - \frac{25}{2} = 50 - 12.5 = 37.5$

3. Compute the fluid force: $F = 9,800 \cdot 5 \cdot 37.5 = 1,837,500 \, \text{newtons}$

Thus, the fluid force on the plate is 1,837,500 newtons.

Would you like more details or have any questions about the solution?

Here are some related questions for further understanding:

1. How would the fluid force change if the weight-density of the liquid were different?
2. What happens to the force if the square is submerged at a different depth?
3. How is the force distributed over the surface of the plate?
4. Can the same method be applied for irregularly shaped plates?
5. How does the concept of center of pressure relate to this problem?

Tip: When working with fluid force problems, remember that pressure increases linearly with depth, which is why integrals are used to calculate the total force on submerged surfaces.