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Midterm III Practice

This collection features solved problems covering work and energy, linear momentum and its conservation, and rotational dynamics—including rotational motion, moments of inertia, and angular momentum.

Problem 5 — The Two Water Tanks

By comparing two water tanks of different shapes and analyzing a submerged sensor, this problem demonstrates that the fluid force on a container's base depends strictly on depth and area rather than total water weight, while illustrating that pressure acts equally in all directions.

Two different water tanks, Tank 1 and Tank 2, are sitting on a flat table. Both tanks have the exact same flat, circular bottom base area of \(A = 0.2\text{ m}^2\). Both tanks are filled with fresh water (\(\rho = 1000\text{ kg/m}^3\)) to the exact same vertical height of \(h = 1.0\text{ m}\).

  • Tank 1 has perfectly vertical side walls (a straight cylinder). It contains exactly \(200\text{ kg}\) of water.
  • Tank 2 has outward-sloping walls (it is wider at the top than at the bottom). Because of its wider top, it holds more water, containing exactly \(300\text{ kg}\) of water.

(Use \(g = 10\text{ m/s}^2\). Use gauge pressure for all calculations.)

Part A: The Hydrostatic Paradox

  1. Calculate the total weight (\(W_1\) and \(W_2\)) of the water inside each tank.

  2. Calculate the hydrostatic gauge pressure (\(P_1\) and \(P_2\)) at the bottom surface of each tank.

  3. Calculate the downward force (\(F_1\) and \(F_2\)) exerted by the water on the flat bottom surface of each tank.

  4. For Tank 2, the downward force of the water on the bottom base is only \(2000\text{ N}\), but the total weight of the water is \(3000\text{ N}\). Explain where the "missing" \(1000\text{ N}\) of force is being supported.

Part B: Isotropic Pressure (Direction Independence)

A small, flat square pressure sensor of area \(a = 0.01\text{ m}^2\) is lowered into Tank 1 to a depth of \(d = 0.5\text{ m}\) below the water's surface.

  1. Calculate the magnitude of the force exerted by the water on the face of this sensor if it is oriented:

    • Facing straight UP (water pushes down on it).
    • Facing straight DOWN (water pushes up on it).
    • Facing straight SIDEWAYS (water pushes horizontally on it).

  2. State the physical principle that explains why the force magnitude is the same in all three directions, even though gravity only pulls vertically.

I need an exercise for exam practice for a General Physics I lecture at the college level (algebra-based).

- Topic: Pressure inside of a fluid, force in a stationary fluid is the same in all directions
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Same force for same area at same height regardless of shape.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Part A: The Hydrostatic Paradox

  1. Weight of the water

    • Tank 1: \(W_1 = m_1 g = (200\text{ kg})(10\text{ m/s}^2) = 2000\text{ N}\)

    • Tank 2: \(W_2 = m_2 g = (300\text{ kg})(10\text{ m/s}^2) = 3000\text{ N}\)

  2. Gauge pressure at the base

    The pressure at a depth \(h\) in a stationary fluid is determined solely by \(P = \rho g h\):

    For both tanks, \(h = 1.0\text{ m}\).

    \[P_1 = P_2 = (1000\text{ kg/m}^3)(10\text{ m/s}^2)(1.0\text{ m}) = 10{,}000\text{ Pa}\]

  3. Downward force on the base

    The force on the flat base is \(F = P \cdot A\):

    • Tank 1: \(F_1 = (10{,}000\text{ Pa})(0.2\text{ m}^2) = 2000\text{ N}\)

    • Tank 2: \(F_2 = (10{,}000\text{ Pa})(0.2\text{ m}^2) = 2000\text{ N}\)

    Both bases experience the exact same downward force of \(2000\text{ N}\), despite Tank 2 holding \(1000\text{ N}\) more water.

  4. Resolving the paradox

    In Tank 1 (vertical walls), the walls do not support any of the water's weight. Therefore, the force on the bottom (\(2000\text{ N}\)) is exactly equal to the weight of the water (\(2000\text{ N}\)).

    In Tank 2 (slanted walls), the walls slope outward. Because the water pushes perpendicular to these slanted walls, the walls must push back on the water with a normal force that has an upward vertical component. These slanted walls physically support the "extra" \(1000\text{ N}\) of water, preventing that extra weight from pressing down on the bottom base.

Part B: Isotropic Pressure

  1. Force on the submerged sensor

    First, find the pressure at the depth of the sensor (\(d = 0.5\text{ m}\)):

    \[P = \rho g d = (1000\text{ kg/m}^3)(10\text{ m/s}^2)(0.5\text{ m}) = 5000\text{ Pa}\]

    Now, calculate the force magnitude on the sensor area (\(a = 0.01\text{ m}^2\)):

    \[F = P \cdot a = (5000\text{ Pa})(0.01\text{ m}^2) = 50\text{ N}\]
    • Facing UP: The water pushes down with a force of \(50\text{ N}\).
    • Facing DOWN: The water pushes up with a force of \(50\text{ N}\).
    • Facing SIDEWAYS: The water pushes horizontally with a force of \(50\text{ N}\).

  2. Physical principle

    This occurs because fluid pressure is a scalar quantity, not a vector, and acts isotropically (equally in all directions). At a microscopic level, pressure is caused by the random, multi-directional thermal collisions of water molecules. Therefore, at any specific depth, the fluid exerts the exact same pressure, resulting in the same force magnitude perpendicular to any surface it contacts, regardless of the surface's orientation.

Feedback & Contributions

Have feedback, want to share your experience, or found a fun variation of this exercise? Email jens.f.mahlmann@dartmouth.edu — we'd love to hear from you!