Log float

Homework Help
1

To what depth will an 8ft dia log with density 425 kg/m3 sink in water?

2

PK Bot’s answer:

  1. ENGINEERING CONTEXT: We are determining the draft (vertical submerged depth) of a cylindrical log when floating in water under Archimedes’ principle.

  2. PHYSICS: The dominant effect is buoyancy, where the weight of the log is balanced by the weight of the displaced water.

  3. ASSUMPTIONS: We assume the log floats horizontally with a circular cross‐section, water density is 1000\ \mathrm{kg/m^3}, and the problem can be solved via equilibrium of areas (the “circular segment” analysis), which is appropriate for long logs whose length does not affect the submerged fraction.

  4. SOLUTION TECHNIQUE:
    Let the log’s diameter be D=2.44\ \mathrm{m} (8 ft) so that its radius is R=1.22\ \mathrm{m}. For a horizontal cylinder floating in water, the submerged cross‐section area is given by

A_\mathrm{sub} = R^2\cos^{-1}\left(\frac{R-h}{R}\right) - (R-h)\sqrt{2Rh-h^2} \,,

while the total area is

A_\mathrm{total}=\pi R^2 \,.

Floating equilibrium requires that

\frac{A_\mathrm{sub}}{A_\mathrm{total}} = \frac{\rho_\mathrm{log}}{\rho_\mathrm{water}}=\frac{425}{1000}=0.425 \,.

A numerical solution yields h\approx1.08\ \mathrm{m}, meaning the log submerges to a depth of approximately 1.08 m (or about 3.54 ft).

  1. REFLECTION: The solution, obtained by equating the dimensionless submerged area to the density ratio, demonstrates that a log with density 425 kg/m³ floats partially submerged with a draft of about 1.08 m; approximations assume a uniform circular cross‐section and a constant water density, appropriate for typical engineering estimates.

To formulate this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 4.3: Archimedes’ Law (HIGHLY RELEVANT)
  2. Chapter 4.1: Watercraft (HIGHLY RELEVANT)
  3. Chapter 4 Engineering Models: Hydrostatic Equation (RELEVANT)
  4. Chapter 3 Engineering Models: Hydrostatic Equation (RELEVANT)
  5. Chapter 3.2: Hydrostatic Equation (RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.

3

where did the formula for the submerged cross sectoin come from

4

PK Bot’s answer:
The formula for the submerged cross-section area of a horizontal cylinder floating in water is derived from geometric considerations of the circular segment formed by the submerged part of the cylinder.

Derivation of the Submerged Cross-Section Area

  1. Geometry of the Problem:

    • Consider a horizontal cylinder with radius R partially submerged in water.
    • The submerged part forms a circular segment, which is a portion of the circle cut off by a chord.

  2. Submerged Area Calculation:

    • The area of the circular segment can be calculated using the formula:
    A_\mathrm{sub} = R^2\cos^{-1}\left(\frac{R-h}{R}\right) - (R-h)\sqrt{2Rh-h^2}

    where:

    • h is the depth of submersion (the vertical distance from the water surface to the bottom of the submerged part of the cylinder).
    • R^2\cos^{-1}\left(\frac{R-h}{R}\right) is the area of the sector of the circle.
    • (R-h)\sqrt{2Rh-h^2} is the area of the triangular portion that is not submerged.

  3. Equilibrium Condition:

    • For the log to float, the weight of the displaced water must equal the weight of the log. This is expressed as:
    \frac{A_\mathrm{sub}}{A_\mathrm{total}} = \frac{\rho_\mathrm{log}}{\rho_\mathrm{water}}

    where A_\mathrm{total} = \pi R^2 is the total cross-sectional area of the cylinder.

This geometric approach allows us to determine the submerged depth h by solving the equation for the submerged area ratio, which is equal to the density ratio of the log to the water. This method is appropriate for long logs where the length does not significantly affect the submerged fraction.

To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 4.1: Watercraft (HIGHLY RELEVANT)
  2. Chapter 4.3: Archimedes’ Law (HIGHLY RELEVANT)
  3. Chapter 4 Engineering Models: Hydrostatic Equation (RELEVANT)
  4. Chapter 3 Engineering Models: Hydrostatic Equation (RELEVANT)
  5. Chapter 9.3: Why We Use Irrotational Flow Analysis Part 5 (RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.

5

how am I supposed to know what the area of a circular segment is

6

PK Bot’s answer:
The area of a circular segment is a standard result in geometry, typically derived in a calculus or geometry course. The derivation goes as follows:

  1. Consider a circle of radius R and a chord that is a distance (R-h) from the center (with h being the depth of submergence measured from the top of the circle). The submerged portion forms a circular segment.

  2. The area of the segment is computed by subtracting the area of the triangle (formed by the chord and the two radii to the ends of the chord) from the area of the circular sector corresponding to the central angle associated with that chord.

The result is:

A_\mathrm{sub} = R^2\cos^{-1}\Big(\frac{R-h}{R}\Big) - (R-h)\sqrt{2Rh-h^2}\,.

This result is established by integrating the area element in a circular coordinate system or by using the known formula for the area of a sector, A_\mathrm{sector}=R^2\theta, and subtracting the area of the isosceles triangle, A_\mathrm{triangle}=(R-h)\sqrt{2Rh-h^2}, where \theta is determined by \theta=2\cos^{-1}\Big(\frac{R-h}{R}\Big).

This formula appears in many standard texts on geometry and calculus and is often provided as a reference result because its derivation requires a few steps of integration and trigonometric manipulation.

To formulate this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 4.1: Watercraft (HIGHLY RELEVANT)
  2. Chapter 4.3: Archimedes’ Law (HIGHLY RELEVANT)
  3. Chapter 7.1: Helicopter/Quadcopter Rotors Part 1 (RELEVANT)
  4. Chapter 4 Engineering Models: Hydrostatic Equation (RELEVANT)
  5. Chapter 3 Engineering Models: Hydrostatic Equation (RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.

7

ok thx but can you show the full integration

8

PK Bot’s answer:
Below is one common derivation using integration. We begin by expressing the area of the circular segment (the submerged portion) as the integral of the horizontal slices of the circle that lie below the waterline. Define the circle with radius R and assume the waterline is at the vertical coordinate

y=y_0=R-h,

so that h is the depth of submersion measured from the top of the circle.

The area of the segment is given by

A_\mathrm{sub}=\int_{y=y_0}^{R}2\sqrt{R^2-y^2}\,dy\,.

A standard substitution is to let

y=R\cos\theta \quad\Rightarrow\quad dy=-R\sin\theta\,d\theta\,.

When y=R, then \theta=0, and when y=y_0, then

\theta_0=\arccos\Big(\frac{y_0}{R}\Big)=\arccos\Big(\frac{R-h}{R}\Big)\,.

Changing variables, the integral becomes

\begin{aligned} A_\mathrm{sub}&=2\int_{\theta=0}^{\theta_0}\sqrt{R^2-R^2\cos^2\theta}\,(R\sin\theta)\,d\theta\\[1mm] &=2R^2\int_{0}^{\theta_0}\sin^2\theta\,d\theta\,. \end{aligned}

Recall that

\int\sin^2\theta\,d\theta=\frac{\theta}{2}-\frac{\sin2\theta}{4}\,,

so that

2R^2\int_{0}^{\theta_0}\sin^2\theta\,d\theta = R^2\Bigg[\theta_0-\frac{\sin2\theta_0}{2}\Bigg]\,.

Thus, the submerged area is

A_\mathrm{sub}=R^2\theta_0-\frac{R^2}{2}\sin2\theta_0\,,

with \theta_0=\arccos\Big(\frac{R-h}{R}\Big). It is convenient to express \sin2\theta_0 in terms of h. Note that

\sin2\theta_0=2\sin\theta_0\cos\theta_0\,,

where

\cos\theta_0=\frac{R-h}{R}\quad\text{and}\quad \sin\theta_0=\sqrt{1-\Big(\frac{R-h}{R}\Big)^2}=\frac{\sqrt{2Rh-h^2}}{R}\,.

Substituting back gives

\sin2\theta_0=2\,\frac{\sqrt{2Rh-h^2}}{R}\,\frac{R-h}{R}=\frac{2(R-h)\sqrt{2Rh-h^2}}{R^2}\,.

Therefore, the formula becomes

A_\mathrm{sub}=R^2\arccos\Big(\frac{R-h}{R}\Big)-\frac{R^2}{2}\cdot\frac{2(R-h)\sqrt{2Rh-h^2}}{R^2}\,,

which simplifies to

A_\mathrm{sub}=R^2\arccos\Big(\frac{R-h}{R}\Big)-(R-h)\sqrt{2Rh-h^2}\,.

This is the full integration derivation of the submerged cross-sectional (circular segment) area.

To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Below, I have organized the sections based on their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 3.2: Hydrostatic Equation (RELEVANT)
  2. Chapter 4.2: Review of Hydrostatics (RELEVANT)
  3. Appendix D (RELEVANT)
  4. Chapter 7.1: Helicopter/Quadcopter Rotors Part 1 (RELEVANT)
  5. Appendix B.4: Vector Calculus Part 3 (RELEVANT)

Disclaimer:

See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.