Problem 88 - Fundamentals of Physics

Question

A \(8.60 \mathrm{~kg}\) sphere of radius \(6.22 \mathrm{~cm}\) is at a depth of \(2.22 \mathrm{~km}\) in seawater that has an average density of \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). What are the (a) gauge pressure, (b) total pressure, and (c) corresponding total force compressing the sphere's surface? What are (d) the magnitude of the buoyant force on the sphere and (e) the magnitude of the sphere's acceleration if it is free to move? Take atmospheric pressure to be \(1.01 \times 10^{5} \mathrm{~Pa}\).

Step-by-Step Solution

Answer

a) \(22236015 \; \text{Pa}\), b) \(22337015 \; \text{Pa}\), c) \(1083510 \; \text{N}\), d) \(10.056 \; \text{N}\), e) \(-8.64 \; \text{m/s}^2\).

Step 1: Convert Depth to Meters

The depth is given as \(2.22 \; \text{km}\). First, convert this depth into meters. Since \(1\; \text{km} = 1000 \; \text{m}\), we have:\[ 2.22 \; \text{km} = 2.22 \times 1000 \; \text{m} = 2220 \; \text{m} \]

Step 2: Calculate Gauge Pressure

The gauge pressure is the pressure due to the weight of the water above the sphere. It can be calculated using the formula:\[ P_{ ext{gauge}} = \rho \cdot g \cdot h \]where \(\rho = 1025 \; \text{kg/m}^3\) is the density of seawater, \(g = 9.81 \; \text{m/s}^2\) is the acceleration due to gravity, and \(h = 2220 \; \text{m}\) is the depth.\[ P_{ ext{gauge}} = 1025 \; \text{kg/m}^3 \cdot 9.81 \; \text{m/s}^2 \cdot 2220 \; \text{m} = 22236015 \; \text{Pa} \]

Step 3: Calculate Total Pressure

Total pressure is the sum of atmospheric pressure and gauge pressure:\[ P_{ ext{total}} = P_{ ext{atmospheric}} + P_{ ext{gauge}} \]Using \(P_{ ext{atmospheric}} = 1.01 \times 10^5 \; \text{Pa}\):\[ P_{ ext{total}} = 1.01 \times 10^5 \; \text{Pa} + 22236015 \; \text{Pa} = 22337015 \; \text{Pa} \]

Step 4: Calculate Total Force on Sphere

The force on the sphere is given by the total pressure times the surface area of the sphere. The surface area, \(A\), for a sphere of radius \(r\) is:\[ A = 4\pi r^2 \]Convert radius from centimeters to meters (\(6.22\; \text{cm} = 0.0622 \; \text{m}\)):\[ A = 4 \pi (0.0622)^2 = 0.0485 \; \text{m}^2 \] (approximately)Then the force is:\[ F_{ ext{total}} = P_{ ext{total}} \cdot A = 22337015 \; \text{Pa} \cdot 0.0485 \; \text{m}^2 \approx 1083510 \; \text{N} \]

Step 5: Calculate Buoyant Force

The buoyant force is equal to the weight of the displaced fluid. This can be calculated using Archimedes' principle:\[ F_{\text{buoyant}} = \rho \cdot V \cdot g \]Here, the volume \(V\) of the sphere is:\[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.0622)^3 \approx 0.001 \; \text{m}^3 \]So:\[ F_{\text{buoyant}} = 1025 \cdot 0.001 \cdot 9.81 \approx 10.056 \; \text{N} \]

Step 6: Calculate Sphere's Acceleration

The net force acting on the sphere when it is free to move is the difference between the buoyant force and the gravitational force (its weight). The weight of the sphere is calculated as:\[ W = m \cdot g = 8.60 \; \text{kg} \cdot 9.81 \; \text{m/s}^2 = 84.366 \; \text{N} \]Then, the net force is:\[ F_{ ext{net}} = F_{\text{buoyant}} - W = 10.056 \; \text{N} - 84.366 \; \text{N} \approx -74.31 \; \text{N} \]The acceleration, \(a\), can be found by using Newton's second law \(F = ma\):\[ a = \frac{F_{ ext{net}}}{m} = \frac{-74.31}{8.60} \approx -8.64 \; \text{m/s}^2 \]

Key Concepts

Gauge Pressure

Gauge pressure is the pressure exerted by a fluid on an object, excluding atmospheric pressure. Unlike absolute pressure, gauge pressure zeroes out the atmospheric effect and focuses on how water or any other fluid compresses an object.
Consider gauge pressure at an ocean's depth: water's weight increases pressure on objects immersed. The deeper you dive, the higher the gauge pressure. This pressure depends primarily on three factors:

• Density of the fluid (\(\rho\))
• Gravitational acceleration (\(g\))
• Height of the fluid column above the object (\(h\))

In formula form: \( P_{\text{gauge}} = \rho \cdot g \cdot h \).
For instance, at a depth of 2220 meters in seawater with a density of 1025 kg/m³, the gauge pressure equates to a substantial 22,236,015 Pascals, highlighting the strength of water pressure forces at such depths.

Buoyant Force

A buoyant force lifts any object submerged in a fluid, caused by pressure differences at varying depths. This force can be puzzling until we break it down with an example. Imagine a bathtub filled with water.
When you submerge a beach ball, it tends to float – that's buoyancy! Key factors influencing buoyant force are:

• The fluid's density
• The submerged volume of the object

The magnitude of this force is described by the formula: \( F_{\text{buoyant}} = \rho \cdot V \cdot g \), where\( V \) is the volume of fluid displaced by the sphere.
For our sphere, with its small volume, the buoyant force is roughly 10.056 Newtons, considerably less than the weight of the sphere, showcasing why it may sink if left untethered.

Archimedes' Principle

Archimedes' principle is a fundamental law in fluid mechanics. It states that any submerged object experiences an upward buoyant force equivalent to the fluid's weight displaced by the object.
Think of it like this: when you push a beach ball underwater, it fights to float back up because the water displaced creates a force pushing upward. This principle explains why ships float and how submarines adjust buoyancy.
For our sphere at depth, this principle helps calculate the buoyant force by finding the water volume displaced, using the sphere's volume. The principle simplifies the understanding of buoyant forces and helps design objects that float effectively.

Newton's Second Law

Newton's second law of motion is a cornerstone of physics. It links forces acting on an object with its mass and the resulting acceleration. It's succinctly captured in the equation \( F = m \cdot a \). For fluids, understanding how forces influence motion is crucial.
In fluid dynamics, the interplay between the buoyant force and the weight of an object determines acceleration. When applied to our spherical object, the net force (buoyant force minus gravitational force) dictates the object's movement as per Newton's law.
The net force calculated here was -74.31 Newtons, implying the sphere accelerates downward. By dividing the net force by the sphere's mass, we find its acceleration \( a \approx -8.64 \text{ m/s}^2 \), a downward acceleration resulting from greater weight than buoyant force.