Rigid-tank acceleration (non-inertial hydrostatics): A 4 m × 3 m × 2 m closed tank filled with oil (specific gravity = 0.83) is moved vertically with acceleration a = g/2. What is the ratio of bottom pressures when moving upward versus downward?

Introduction / Context:

This problem checks understanding of apparent gravity in accelerating reference frames. A rigid, completely filled tank experiencing vertical acceleration shows an effective hydrostatic gradient based on g_eff = g ± a, which directly scales pressure at a given depth.

Given Data / Assumptions:

• Tank dimensions: 4 m × 3 m × 2 m (depth relevant, but ratio cancels).
• Fluid: oil with specific gravity 0.83 (density scales pressure but cancels in the ratio).
• Vertical acceleration magnitude: a = g/2.
• Bottom pressure considered at the same physical depth during up and down motions.

Concept / Approach:

In a non-inertial frame accelerating vertically, the pressure variation remains hydrostatic but with effective gravity g_eff. When the tank accelerates upward, g_eff = g + a; when accelerating downward, g_eff = g − a. The absolute bottom pressure (for the same free surface reference) scales in proportion to g_eff.

Step-by-Step Solution:

Verification / Alternative check:

Since density ρ and depth h are identical in both cases, they cancel in the ratio, leaving only g_eff terms; hence the numerical result is independent of SG = 0.83 and the tank dimensions.

Why Other Options Are Wrong:

• 2 results from mistakenly using a = g/1 instead of g/2.
• 1/2 and 1/3 invert the intended order (they would be p_down/p_up or use wrong arithmetic).
• 1 would imply no effect of acceleration, which is untrue.

Common Pitfalls:

• Using g instead of g ± a for accelerating containers.
• Trying to include the plan dimensions or density in the final ratio (they cancel).

Final Answer:

3