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Chapter 15: Problem 38

Solve the given problems. Use a calculator in \(E x\) ercises \(40,41,\) and 44. The specific gravity \(s\) of a sphere of radius \(r\) that sinks to a depth \(h\) in water is given by \(s=\frac{3 r h^{2}-h^{3}}{4 r^{3}} .\) Find the depth to which a spherical buoy of radius \(4.0 \mathrm{cm}\) sinks if \(s=0.50\).

Short Answer

Expert verified
The buoy sinks to a depth of approximately 8.0 cm.

Step by step solution

01

Identify Given Values

We are given that the radius of the spherical buoy is \( r = 4.0 \) cm and the specific gravity \( s = 0.50 \). We need to find the depth \( h \).
02

Write Down the Given Equation

The equation for specific gravity is given by: \[s = \frac{3r h^{2} - h^{3}}{4r^{3}}\]
03

Substitute Known Values

Substitute \( s = 0.50 \) and \( r = 4.0 \) cm into the equation:\[0.50 = \frac{3 \times 4.0 \times h^{2} - h^{3}}{4 \times (4.0)^{3}}\]
04

Simplify the Equation

Calculate \( 4^3 = 64 \) and substitute:\[0.50 = \frac{12h^{2} - h^{3}}{256}\]
05

Cross-Multiply to Eliminate the Fraction

Multiply both sides by 256 to eliminate the fraction:\[0.50 \times 256 = 12h^{2} - h^{3}\]\[128 = 12h^{2} - h^{3}\]
06

Rearrange the Equation

Bring all terms to one side of the equation:\[h^{3} - 12h^{2} + 128 = 0\]
07

Solve the Cubic Equation

Use a calculator to solve the cubic equation for \( h \). The solutions are \( h \approx 8.0 \) cm and two complex roots. Only the real, positive solution is physically meaningful here.

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Key Concepts

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

Sphere
A sphere is a perfectly symmetrical three-dimensional shape. Every point on the surface is equidistant from its center. Spheres are common in geometry and many real-world applications. In this problem, the sphere represents a spherical buoy, which is used to float and indicate measurements or locations.
  • The main characteristic of a sphere is its radius \( r \), which is the distance from the center to any point on its surface.
  • In this exercise, the radius given is \( 4.0 \) cm.
The volume of a sphere is calculated using the formula \( V = \frac{4}{3} \pi r^3 \). This volume plays a central role in understanding the buoyancy of floating objects because it determines how much water the sphere displaces.
Cubic Equation
A cubic equation is a polynomial equation of the form \( ax^3 + bx^2 + cx + d = 0 \). In this exercise, we encounter the cubic equation \( h^3 - 12h^2 + 128 = 0 \). Solving these equations can sometimes be challenging due to the possibility of complex roots.
  • Cubic equations typically have three roots, which can be real or complex.
  • There is always at least one real root for a cubic equation since polynomials with real coefficients have complex roots in conjugate pairs.
To find the depth \( h \) to which the buoy sinks, solving the cubic equation in this context only requires the real, positive root. Using a calculator or computational tools can simplify solving these equations.
Physics Problem Solving
Physics problem solving often involves breaking down complex problems into simpler steps. This ensures you clearly understand what each symbol and formula represents. Here鈥檚 a common approach:
  • **Identify:** Start by identifying what is given and what you need to find. Write down all known values and units. In this case, known values were the radius and specific gravity.
  • **Formulate:** Use relevant equations that relate the known and unknown variables. Write them down clearly.
  • **Substitute:** Replace the variables in your formula with the given numbers.
  • **Calculate:** Solve the equation using algebraic manipulation and computational tools where necessary.
  • **Interpret:** Analyze the solution to evaluate if it makes sense physically. Only real, positive numbers had meaning in this buoyancy problem.
Confidence in physics comes with practicing and honing these problem-solving skills.
Buoyancy
Buoyancy is the upward force that keeps objects afloat in a fluid. In simple terms, it is why boats and balloons float instead of sinking.
The buoyant force on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. Specific gravity (\(s\)) helps determine buoyancy by comparing the object's density to the fluid's density.
  • If an object's specific gravity is less than 1, it will float. This is because it is less dense than the fluid.
  • If an object鈥檚 specific gravity is exactly 1, it will neither sink nor float, but remain suspended.
  • If its specific gravity is more than 1, the object will sink as it is denser than the fluid.
In this problem, the specific gravity of \(0.50\) indicates the buoy will float back up as soon as it has submerged sufficiently to displace a volume of water equal to its own weight.

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Most popular questions from this chapter

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