To solve the problem, let's go step by step. We need to calculate the hydrostatic force on one end of the trough under two conditions:

1. When the trough is completely full of oil.
2. When the trough is filled to a depth of 1.8 meters.

The density of the oil is $\rho = 940 \, \text{kg/m}^3$, and the acceleration due to gravity is $g = 9.8 \, \text{m/s}^2$.

1. Hydrostatic Force When the Trough is Full

Step 1: Understanding the Trapezoidal Cross-Section

The area of a trapezoid is calculated by: $A = \frac{1}{2} \times (b_1 + b_2) \times h$ where:

• $b_1$ is the bottom width (1 m),
• $b_2$ is the top width (2 m), and
• $h$ is the height (3 m).

$A = \frac{1}{2} \times (1 + 2) \times 3 = \frac{1}{2} \times 3 \times 3 = 4.5 \, \text{m}^2$

Step 2: Calculating Hydrostatic Force

The hydrostatic pressure at a depth $y$ is given by: $P = \rho \cdot g \cdot y$

The hydrostatic force is: $F = \rho \cdot g \cdot \int_0^h y \cdot L(y) \, dy$ where $L(y)$ is the width of the trapezoid at depth $y$.

From the trapezoid's geometry:

• At $y = 0$ (bottom), the width is 1 m.
• At $y = 3$ m (top), the width is 2 m.

The width increases linearly, so: $L(y) = 1 + \frac{y}{3}$

Now, substitute everything: $F = 940 \times 9.8 \times \int_0^3 y \cdot (1 + \frac{y}{3}) \, dy$

Step 3: Evaluate the Integral

$F = 940 \times 9.8 \int_0^3 (y + \frac{y^2}{3}) dy$

$= 940 \times 9.8 \left[ \frac{y^2}{2} + \frac{y^3}{9} \right]_0^3$

$= 940 \times 9.8 \left( \frac{9}{2} + \frac{27}{9} \right)$

$= 940 \times 9.8 \times (4.5 + 3) = 940 \times 9.8 \times 7.5$

$F = 690900 \, \text{N}$

2. Hydrostatic Force When the Trough is Filled to 1.8 m

We repeat the calculation, but this time the integral limits change to 0 to 1.8 m.

$F = 940 \times 9.8 \times \int_0^{1.8} (y + \frac{y^2}{3}) dy$

Evaluate the integral: $= 940 \times 9.8 \left[ \frac{y^2}{2} + \frac{y^3}{9} \right]_0^{1.8}$

$= 940 \times 9.8 \left( \frac{1.8^2}{2} + \frac{1.8^3}{9} \right)$

Calculate each part: $1.8^2 = 3.24 \quad \Rightarrow \quad \frac{3.24}{2} = 1.62$ $1.8^3 = 5.832 \quad \Rightarrow \quad \frac{5.832}{9} = 0.648$

$F = 940 \times 9.8 \times (1.62 + 0.648)$

$F = 940 \times 9.8 \times 2.268$

$F \approx 20859.7 \, \text{N}$

Final Results

1. Hydrostatic force when the trough is full: $690900 \, \text{N}$
2. Hydrostatic force when the trough is filled to 1.8 m: $20859.7 \, \text{N}$

Would you like further details or have any questions about these calculations?

Related Questions:

1. How would the force change if the oil's density was different?
2. What is the hydrostatic pressure at the bottom of the trough when it is full?
3. How would the force calculation change if the trough's shape was rectangular?
4. Could we use the same method for a tank filled with water instead of oil?
5. How does the depth affect the distribution of pressure across the trough?

Tip: When calculating hydrostatic force, always remember that the pressure increases linearly with depth, so integrals are useful to account for this variation.