Problem Statement
What distance must the sides of a tank be carried above the surface of the water contained in it if the tank is to undergo a uniform horizontal acceleration of 3 m/s² without spilling any water?
(Express your answer as a fraction of the tank’s length, L; expected answer ≈ 0.153L.)
Solution
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With a horizontal acceleration of:
aₓ = 3 m/s² -
The free surface of the water tilts such that its slope is determined by:
tanθ = aₓ / g = 3 / 9.81 ≈ 0.306Thus, the tilt angle is:
θ ≈ arctan(0.306) ≈ 17° -
In a tank of length L, the vertical rise at the end of the tank (which is at half its length) is:
d = 0.5 L × tanθSubstituting tanθ ≈ 0.306:
d ≈ 0.5 L × 0.306 ≈ 0.153L
Therefore, the sides of the tank must be carried approximately 0.153L above the water surface to prevent spillage.
Explanation
When a tank accelerates horizontally, the water inside does not immediately follow the container’s motion due to inertia. As a result, the free surface of the water tilts and aligns perpendicular to the effective gravitational field—the vector sum of the true gravitational acceleration and the horizontal acceleration.
Here, the tilt is determined by the ratio of the horizontal acceleration to the gravitational acceleration. Over half the tank’s length, this tilt results in a vertical rise given by 0.5L times the tangent of the tilt angle. To avoid any spillage during acceleration, the tank’s sides must be at least this high above the water surface.
Physical Meaning
This problem demonstrates the impact of inertial forces on a fluid’s free surface within a container. Under acceleration, the free surface tilts, causing a difference in the water level across the container. The derived clearance (0.153L) is the minimum height the container’s sides must extend above the undisturbed water level to prevent spillage.
Understanding this effect is crucial in the design of tanks and containers used in moving vehicles or vessels, ensuring that dynamic conditions do not lead to loss of fluid.
