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Chapter 3: Problem 23

Suppose a spherical floating buoy has radius \(1 \mathrm{m}\) and density \(\frac{1}{4}\) that of sea water. Given that the formula for the volume of a spherical cap is \(V_{\mathrm{cap}}=\frac{\pi x}{6}\left(3 a^{2}+x^{2}\right),\) to what depth does the buoy sink in sea water?

Short Answer

Expert verified
The buoy sinks to a depth of approximately 0.8 meters.

Step by step solution

01

- Determine Volume and Density

First, note that the radius of the sphere is \( r = 1 \text{ meter} \). The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \). Plug in \( r = 1 \) to get \( V = \frac{4}{3} \pi \times 1^3 = \frac{4}{3} \pi \text{ cubic meters} \). The density of the buoy relative to sea water is \( \frac{1}{4} \), leading to the buoy having \( \rho_b = \frac{1}{4} \rho_{sw} \), where \( \rho_{sw} \) is the density of sea water.
02

- Calculate Buoyant Force

The buoyant force is equal to the weight of the displaced water. The weight of the displaced water is calculated by the formula \( F_b = \text{Volume} \times \text{Density} \times g \), where g is the acceleration due to gravity. Let the volume of displaced water be \( V_{displaced}\) and its density be \( \rho_{sw} \), then the buoyant force \( F_b = V_{displaced} \rho_{sw} g \).
03

- Find the Weight of the Buoy

The weight \( W_b \) of the buoy can be computed using density and volume: \( W_b = V \rho_b g \). Substituting the values we get, \( W_b = \frac{4}{3} \pi \times \frac{1}{4} \rho_{sw} \times g = \frac{1}{3} \pi \rho_{sw} g \).
04

- Equate Buoyant Force to Weight of Buoy

For the buoy to float, the buoyant force must equal the weight of the buoy: \( F_b = W_b \). So, \( V_{displaced} \rho_{sw} g = \frac{1}{3} \pi \rho_{sw} g \). This simplifies to \( V_{displaced} = \frac{1}{3} \pi \).
05

- Use the Volume of Spherical Cap Formula

We use the given formula for the volume of the spherical cap: \( V_{\text{cap}} = \frac{\pi x}{6}(3a^{2}+x^{2}) \), where \(a = 1\) (radius), to equate \( V_{\text{cap}} = \frac{1}{3} \pi\). Substituting \(a = 1\), we get: \( \frac{\pi x}{6}(3(1)^2 + x^2) = \frac{1}{3} \pi \). Simplifying, we obtain \( \frac{\pi x}{6}(3 + x^2) = \frac{1}{3} \pi \).
06

- Solve for x (Depth)

Solve the equation: \( \frac{\pi x}{6}(3 + x^2) = \frac{1}{3} \pi \). Cancel \( \pi \) and multiply both sides by 6 to get \( x(3 + x^2) = 2 \). This simplifies to: \( x^3 + 3x - 2 = 0 \). Use numerical methods or algebraic factorization to solve the cubic equation for \(x\).
07

- Approximate Solution

By solving the cubic equation \( x^3 + 3x - 2 = 0 \), we find that \( x \approx 0.8 \) meters after using a numerical solver or graphing method.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of a Sphere
The volume of a sphere is a crucial concept in many physics and engineering problems. It's calculated using the formula: \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere.
To find the volume, just plug in the value of the radius and follow the formula.
For instance, with a radius of 1 meter, the volume calculation would look like this:
  • Plug in \( r = 1 \)
  • Calculate \( (1)^3 = 1 \)
  • Multiply by \( \pi \)
  • Then multiply by \frac{4}{3}\text{ yielding } V = \frac{4}{3}\pi \text{ cubic meters}
This concept is fundamental as it applies to many real-world objects, such as balloons, planets, and buoys.
Density
Density is another core concept, defined as mass per unit volume: \( \rho = \frac{m}{V} \). It's an essential property of materials that tells us how much mass is packed into a given volume.
In our problem, the buoy has a density that is \( \frac{1}{4} \) of sea water's density. This means if sea water has a density \( \rho_{sw} \), then the buoy's density will be \( \rho_b = \frac{1}{4} \rho_{sw} \).
Understanding density helps in determining why some objects float while others sink. And it forms the basis of calculating buoyant forces.
Spherical Cap Volume Calculation
The volume of a spherical cap is a bit more involved but equally important. A spherical cap is essentially a 'cap' or 'hat' on top of a sphere, sliced off by a plane.
The formula for the volume of a spherical cap is: \( V_{\text{cap}} = \frac{ \pi x}{6}(3a^2 + x^2) \), where \( x \) is the height of the cap and \( a \) is the radius of the sphere.
  • Plug in the height \( x \) and radius \( a = 1 \)
  • Compute \( 3(1)^2 + x^2 = 3 + x^2 \)
  • Multiply and simplify to find the volume of the displaced water
This helps us determine how much of the buoy is submerged in water, which is key to solving the exercise.

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Most popular questions from this chapter

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