Problem Statement
A closed rectangular tank full of water is 3 m long, 2 m wide and 2 m deep. The pressure at the top of the water is raised to 98.1 kPa. If the tank is accelerated horizontally along its length at 6 m/s2, find the forces on the front and rear ends of the tank. Check your results by Newton’s law.
Solution
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Given:
\( a_x = 6\,\text{m/s}^2 \)
\( P = 98.1\,\text{kPa} = 98\,100\,\text{Pa} \)
Density, \( \rho = 1000\,\text{kg/m}^3 \)
Weight density, \( \gamma = 9810\,\text{N/m}^3 \)
Tank dimensions: Length = 3 m, Width = 2 m, Depth = 2 m.The head due to pressure is:
\( h = \frac{P}{\gamma} = \frac{98100}{9810} = 10\,\text{m} \) -
Determine the Inclination of the Free Surface:
When the tank accelerates, the free surface inclines. The angle \( \theta \) is given by:
\( \tan\theta = \frac{a_x}{g} = \frac{6}{9.81} \)Considering half the tank’s length (i.e. 1.5 m) as the lever arm, the horizontal displacement of the free surface is:
\( \Delta = 1.5\,\tan\theta = 1.5 \times \frac{6}{9.81} \approx 0.917\,\text{m} \) -
Compute the Pressure Forces:
The pressure force on a vertical surface is calculated using the average pressure (which depends on the average depth) times the area.
Rear End (Higher Pressure):
\( F_1 = \gamma \, A \, \bar{x}_1 = 9810 \times (2 \times 2) \times (0.917 + 10 + 1) \)Front End (Lower Pressure):
\( F_2 = \gamma \, A \, \bar{x}_2 = 9810 \times (2 \times 2) \times (10 – 0.917 + 1) \)Evaluating these expressions:
\( F_1 \approx 467623\,\text{N} \quad \text{and} \quad F_2 \approx 395657\,\text{N} \) -
Net Force and Verification by Newton’s Law:
\( F_x = F_1 – F_2 \approx 467623 – 395657 \approx 71966\,\text{N} \)According to Newton’s law:
\( F = m\,a_x = \rho \, (\text{Volume})\,a_x = 1000 \times (3 \times 2 \times 2) \times 6 = 72000\,\text{N} \)The net force calculated from the pressure difference closely matches the force predicted by Newton’s law.
Explanation
Raising the pressure at the top of the water creates an equivalent head of 10 m. When the tank is accelerated, the free surface inclines such that its angle \( \theta \) satisfies \( \tan\theta = \frac{6}{9.81} \). The horizontal displacement over half the tank’s length (1.5 m) is about 0.917 m. This shift changes the effective depth at the rear and front ends of the tank, resulting in different average pressures on these faces. Multiplying these pressures by the respective areas yields the forces on the rear and front ends.
Physical Meaning
This problem illustrates how horizontal acceleration affects the pressure distribution in a fluid-filled tank. The acceleration causes the free surface to tilt, thereby creating a pressure difference between the front and rear ends of the tank. The net force resulting from this pressure difference is consistent with the inertial force calculated using Newton’s law, confirming the analysis.
