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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 1 — The Floating Timber

Distinguish between the center of gravity and the center of buoyancy to analyze a floating object's vertical, spring-like restoring forces and its rotational stability against capsizing.

A uniform rectangular wooden block of mass \(m = 60\text{ kg}\), height \(H = 1\text{ m}\), and horizontal cross-sectional area \(A = 0.1\text{ m}^2\) is floating vertically in a calm pool of fresh water.

(Use \(g = 10\text{ m/s}^2\) and the density of water \(\rho_w = 1000\text{ kg/m}^3\).)

Part A: Depth of Floating

  1. Calculate the submerged depth \(h\) (the draft) of the floating block.

  2. Determine the vertical coordinates (measured from the flat bottom of the block, \(y = 0\)) of:

    • The block's Center of Gravity (\(CG\)).
    • The block's Center of Buoyancy (\(CB\)).

Part B: Vertical Restoring Forces

Suppose a swimmer gently pushes straight down on the top of the block, displacing it downward by an additional distance of \(y = 0.05\text{ m}\) (5 cm) from its equilibrium position, and holds it there.

  1. Calculate the new buoyant force acting on the block.

  2. Calculate the net vertical restoring force acting on the block while it is held at this displaced position.

  3. Show that this restoring force follows Hooke's Law (\(F = -ky\)), and find the effective "water spring constant" \(k\) for this system.

Part C: Cargo and Rotational Stability

To transport supplies, a heavy, flat steel plate of mass \(m_{\text{cargo}} = 20\text{ kg}\) is bolted directly to the very top of the wooden block (\(y = 1.0\text{ m}\)).

  1. Calculate the new submerged depth \(h'\) of the combined system.

  2. Calculate the new vertical positions of both the combined \(CG'\) and the new \(CB'\) (measured from the bottom of the block).

  3. If this loaded block is tilted slightly by a wave, describe how the lines of action of the gravitational force (acting at \(CG'\)) and the buoyant force (acting at \(CB'\)) create a torque, and explain why placing the heavy cargo on top of the block instead of the bottom affects the system's rotational stability.

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

- Topic: Floating objects and stability.
- Difficulty Level: 4/5 (challenging with common pitfalls).
- Numbers: Optimized for no-calculator use.
- Previously Challenging Parts I Want to Study: Center of gravity vs. center of buoyancy, depth of floating, and restoring forces by buoyancy.

Prompt generated by Jens Mahlmann for Gemini 3.5 Flash.

Part A: Depth of Floating

  1. Submerged depth \(h\)

    For a floating object, the buoyant force equals the weight of the block.

    \[\rho_w A h g = mg\]
    \[h = \frac{m}{\rho_w A} = \frac{60}{(1000)(0.1)} = 0.6\text{ m}\]

  2. Vertical coordinates of \(CG\) and \(CB\)

    The Center of Gravity is at the geometric center of the block:

    \[y_{CG} = \frac{H}{2} = \frac{1.0}{2} = 0.5\text{ m}\]

    The Center of Buoyancy is at the geometric center of the submerged portion:

    \[y_{CB} = \frac{h}{2} = \frac{0.6}{2} = 0.3\text{ m}\]

Part B: Vertical Restoring Forces

  1. New buoyant force

    Pushing the block down by \(y = 0.05\text{ m}\) increases the submerged depth to \(h + y = 0.65\text{ m}\).

    \[F_B' = \rho_w A (h+y) g = (1000)(0.1)(0.65)(10) = 650\text{ N}\]

  2. Net restoring force

    The weight of the block is:

    \[mg = (60)(10) = 600\text{ N}\]

    The net force is the difference between the new buoyant force and the weight:

    \[F_{net} = F_B' - mg = 650 - 600 = 50\text{ N}\]

    This force acts upward, opposing the downward displacement — hence it is a restoring force.

  3. Hooke's Law form

    Writing the buoyant force in general as \(F_B = \rho_w A g (h + y)\), the net force is:

    \[F_{net} = \rho_w A g (h+y) - mg = \rho_w A g\, y\]

    since \(\rho_w A g h = mg\) at equilibrium. Taking downward displacement \(y\) as positive, the restoring force acts in the negative (upward) direction:

    \[F_{net} = -(\rho_w A g)\, y = -k y\]
    \[k = \rho_w A g = (1000)(0.1)(10) = 1000\text{ N/m}\]

Part C: Cargo and Rotational Stability

  1. New submerged depth \(h'\)

    The total mass is now \(M = m + m_{\text{cargo}} = 60 + 20 = 80\text{ kg}\).

    \[h' = \frac{M}{\rho_w A} = \frac{80}{(1000)(0.1)} = 0.8\text{ m}\]

  2. New positions of \(CG'\) and \(CB'\)

    \(CG'\) is found by taking the weighted average of the block's CG (60 kg at 0.5 m) and the cargo's CG (20 kg at 1.0 m):

    \[y_{CG'} = \frac{(60)(0.5) + (20)(1.0)}{80} = \frac{30 + 20}{80} = 0.625\text{ m}\]

    \(CB'\) is at the center of the new submerged depth:

    \[y_{CB'} = \frac{h'}{2} = \frac{0.8}{2} = 0.4\text{ m}\]

  3. Rotational stability

    When the loaded block tilts slightly, the buoyant force (acting upward through \(CB'\)) and the gravitational force (acting downward through \(CG'\)) are no longer vertically aligned. This creates a couple (torque pair) between the two forces.

    Because \(CG'\) (0.625 m) is now higher above \(CB'\) (0.4 m) than in the unloaded case, the horizontal separation between the lines of action of these two forces grows quickly as the block tilts. This torque acts to increase the tilt rather than correct it, since the center of gravity is high relative to the center of buoyancy. If the cargo were instead placed at the bottom of the block, \(CG'\) would be lower and closer to \(CB'\), producing a torque that pushes the block back toward the upright position instead of away from it. This is why placing heavy cargo on top significantly reduces stability and makes the system prone to capsizing.

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!