The problem asks to calculate the fluid force on one side of the submerged plate, where the density of the fluid is $62.4 \, \text{lb/ft}^3$. The diagram indicates a triangular plate submerged vertically in a pool, and we are tasked with determining the force due to the pressure of the fluid on this plate.

Let's break down the solution:

Formula for Fluid Force:

The fluid force $F$ on a vertical plate submerged in a liquid is given by:

$F = \rho \cdot g \int_{a}^{b} h(y) \cdot L(y) \, dy$

Where:

• $\rho$ is the density of the fluid (62.4 lb/ft³ in this case),
• $h(y)$ is the depth of the liquid at point $y$,
• $L(y)$ is the width of the plate at depth $y$,
• $a$ and $b$ are the limits corresponding to the range of $y$.

Analyzing the Diagram:

1. The plate is a triangle, with the tip at $y = -3$ and the base at $y = 0$. The horizontal base of the triangle has a length of 12 feet.
2. The width of the plate as a function of depth can be expressed linearly as $L(y)$. Since the width is 12 feet at the bottom and tapers off to zero at the top, we can derive a linear relationship for $L(y)$.

Step 1: Linear Function for Width $L(y)$:

Given the triangle has a base width of 12 feet at $y = -3$ and a width of 0 feet at $y = 0$, we can use the slope-intercept form of a line to express the width $L(y)$:

$L(y) = 4y + 12$

Step 2: Fluid Depth $h(y)$:

The depth of the water at any point $y$ is given by $|y|$, since the plate is submerged vertically. So, $h(y) = -y$ for the range of $y$ between -3 and 0.

Step 3: Set Up the Integral for Force:

The total fluid force is the integral of the pressure times the width over the submerged area. Since the pressure increases with depth, the force is given by:

$F = \rho \int_{-3}^{0} (-y)(4y + 12) \, dy$

Where:

• $\rho = 62.4 \, \text{lb/ft}^3$,
• Limits of integration are $-3$ to $0$ (since the plate goes from 3 feet below the surface to 0).

Step 4: Solve the Integral:

Now we can compute the force by solving the integral. Let me compute that for you.The total fluid force exerted on one side of the triangular plate is approximately 1123.2 lb.

Would you like further details on this solution, or have any questions?

Here are five related questions for further exploration:

1. How would the force change if the plate were deeper in the water?
2. What happens to the force if the fluid density changes?
3. How does the shape of the plate (e.g., rectangular) affect the force?
4. How would you calculate the center of pressure for this triangular plate?
5. What is the effect of fluid velocity on fluid pressure?

Tip: When dealing with fluid forces, always ensure you account for changes in depth, as pressure increases with depth.