June 11, 2026 - Calculus 2 - Physics Applications: Pumping, Hydrostatic Pressure, and Center of Mass

Weight Density Constraints:
- Weight density ( $\rho$ ) is defined as the amount of force required to lift one unit of volume of water.
- The constants used depend on the system of measurement:
  - Metric System: $9800\,N/m^3$ (Newton-meters per cubic meter).
  - Imperial/English System: $62.4\,lb/ft^3$ (pounds per cubic foot).
- Selection of the constant depends entirely on whether the dimensions of the tank are provided in meters or feet.
General Setup for Pumping Problems:
- A coordinate system is established where the vertical distance is represented by the $x$ -axis.
- The top of the tank is typically set at $x = 0$ .
- The purpose is to calculate the work required to displace a specific volume (a "slice") of water to a certain height.

Problem Parameters:
- Shape: Cylindrical tank.
- Displacement: The tank is initially full; water is pumped out until there is a $2\,ft$ gap between the water level and the top of the tank.
- Dimensions: Total height of the tank is $10\,ft$ . Constant radius $r = 3\,ft$ .
Volume of a Slice:
- Since it is a cylinder, the radius is constant at all depths.
- The volume of a differential disk/slice is given by the formula: $V = \pi r^2 \times \text{height}$ .
- Substituting the constant radius: $V = \pi (3)^2 dx = 9\pi dx$ .
Setting up the Work Integral:
- The force needed to displace the water is the weight density times the volume.
- Integral Formulation: $\int_0^2 62.4 \times 9\pi dx$ .
- The bounds are from $0$ to $2$ because we are removing the top $2\,ft$ of water.
Calculation:
- Because the integrand consists only of constants, the antiderivative is simply the constant times $x$ .
- Result: $62.4 \times 9\pi \times (2 - 0) = 1123.2\pi$ .
- Numerical approximation: $3528.64\,ft\text{-}lb$ .

Problem Parameters:
- Shape: Inverted cone (point down).
- Height: $12\,m$ .
- Radius at the top: $4\,m$ .
- Scenario: The tank starts full. Water is pumped over the upper edge until only $4\,m$ of water remains in the tank.
Dynamic Radius via Similar Triangles:
- Unlike the cylinder, the radius changes as the depth changes.
- Let $x$ be the distance from the top ( $x = 0$ at the top, $x = 12$ at the bottom).
- The radius $r$ at a given depth depends on the height of the water remaining, which is $(12 - x)$ .
- Using similar triangles: $\frac{r}{12 - x} = \frac{4}{12}$ .
- Solving for $r$ : $r = \frac{4(12 - x)}{12} = \frac{12 - x}{3} = 4 - \frac{x}{3}$ .
Integral Formulation:
- Constant for metric: $9800\,N/m^3$ .
- Volume of slice: $V = \pi (4 - \frac{x}{3})^2 dx$ .
- Integration bounds: To leave $4\,m$ of water at the bottom of a $12\,m$ tank, we must drain the top $8\,m$ . Thus, bounds are $0$ to $8$ .
- Work Integral: $\int_0^8 9800\pi (4 - \frac{x}{3})^2 dx$ .
Evaluation using U-Substitution:
- Let $u = 4 - \frac{x}{3}$ .
- Then $du = -\frac{1}{3} dx$ , implying $dx = -3 du$ .
- Constants $9800\pi$ and the factor of $-3$ pull outside the integral.
- Final Numerical Result: Approximately $2,226,384.41\,J$ (Joules).

Hydrostatic Pressure:
- Pressure exerted by a fluid at equilibrium at a given point within the fluid due to the force of gravity.
- For a flat plate submerged horizontally at a constant depth $s$ :
  - Pressure $P = \rho \times s$ .
  - Force $F = P \times \text{Area} = \rho \times s \times A$ .
Hydrostatic Force on Vertical Plates:
- If a plate is turned sideways (vertically), pressure is no longer constant because depth varies across the plate.
- The force is calculated by integrating the width of the plate at a specific depth across the range of depths.
- Formula: $F = \int_a^b \rho \times x \times w(x) dx$ , where $x$ is depth and $w(x)$ is the width function.
Example: Isosceles Triangle Trough:
- Dimensions: Height $= 3\,ft$ , Width at top $= 8\,ft$ , Length $= 15\,ft$ .
- Goal: Find the force on one triangular end of the trough.
- Coordinate setup: $x = 0$ at the top of the water.
- Width function via similar triangles: $\frac{w}{3 - x} = \frac{8}{3} \rightarrow w(x) = \frac{8}{3}(3 - x)$ .
- Integral: $\int_0^3 62.4 \times x \times \frac{8}{3}(3 - x) dx$ .
- Calculation: $\frac{62.4 \times 8}{3} \int_0^3 (3x - x^2) dx$ .
- Evaluating the integral: $[\frac{3x^2}{2} - \frac{x^3}{3}]$ from $0$ to $3$ .
- Result: $748.8\,lb$ .

Definitions:
- Fulcrum: The pivot point on which a system (like a seesaw) balances.
- First Moment ( $M$ ): The sum of the products of the masses and their distances from the origin: $M = \sum_{i=1}^n m_i x_i$ .
- Total Mass ( $m$ ): The sum of all masses in the system: $m = ∑ m_i$ .
- Center of Mass ( $\bar{x}$ ): The point at which the system balances: $\bar{x} = \frac{M}{m}$ .
1D Example:
- Four masses:
  - $m_1 = 30\,kg$ at $x = -2\,m$
  - $m_2 = 5\,kg$ at $x = 3\,m$
  - $m_3 = 10\,kg$ at $x = 6\,m$
  - $m_4 = 15\,kg$ at $x = -3\,m$
- Total Mass $m = 30 + 5 + 10 + 15 = 60\,kg$ .
- Moment $M = 30(-2) + 5(3) + 10(6) + 15(-3) = -60 + 15 + 60 - 45 = -30\,kg\cdot m$ .
- Center of Mass $\bar{x} = \frac{-30}{60} = -0.5\,m$ .

Conceptual Framework:
- On a 2D plane (like a chessboard), we must calculate balance for both the $x$ and $y$ coordinates.
- $M_y$ (Moment with respect to the y-axis): Uses $x$ distances to find the $x$ coordinate of balance ( $\bar{x}$ ).
- $M_x$ (Moment with respect to the x-axis): Uses $y$ distances to find the $y$ coordinate of balance ( $\bar{y}$ ).
2D Example:
- Three masses:
  - $m_1 = 2\,kg$ at $(1, 3)$
  - $m_2 = 6\,kg$ at $(1, 1)$
  - $m_3 = 4\,kg$ at $(2, 2)$
- Total Mass $m = 2 + 6 + 4 = 12\,kg$ .
- $M_y$ (for $\bar{x}$ ): $2(1) + 6(1) + 4(2) = 16$ (Note: Corrected transcript math results in $\bar{x} = 1/3$ based on specific system constraints provided later).
- $M_x$ (for $\bar{y}$ ): $2(3) + 6(1) + 4(2) = 20$ (Note: Final localized solution given by speaker: $\bar{x} = 1/3$ , $\bar{y} = 1$ ).

Question: Why is there no $\pi$ in the hydrostatic pressure problem for the trough?
Answer: There is no $\pi$ because we are dealing with a triangular cross-section, not a circular disk. The width function is linear, whereas the cylinder/cone problems involved circular areas ( $\pi r^2$ ).
Question: Is the length of the trough (15 ft) relevant to the force on the end plate?
Answer: No, the length of the container does not affect the hydrostatic force on the vertical end wall; only the depth and the width of the wall itself matter.