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) corrcsponding total force compressing the sphere's surface? What are (d) the magnitude of the buoyant force on the sphere and (c) 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) 22,301,385 Pa, (b) 22,402,385 Pa, (c) 1,088,958 N, (d) 10.1 N, (e) -8.63 m/s².

Step 1: Calculate Gauge Pressure

The gauge pressure at a certain depth in a fluid is calculated by the equation: \( P = \rho gh \), where \( \rho \) is the density of the fluid, \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h \) is the depth. Here, \( \rho = 1025 \, \text{kg/m}^3 \) and \( h = 2220 \, \text{m} \). So, the gauge pressure is: \[ P = 1025 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 2220 \, \text{m} \approx 22301385 \, \text{Pa}. \]

Step 2: Calculate Total Pressure

To find the total pressure at this depth, add the atmospheric pressure to the gauge pressure. The atmospheric pressure is given as \( 1.01 \times 10^5 \, \text{Pa} \). Thus, the total pressure \( P_{\text{total}} \) is: \[ P_{\text{total}} = P + P_{\text{atm}} \approx 22301385 \, \text{Pa} + 1.01 \times 10^5 \, \text{Pa} = 22402385 \, \text{Pa}. \]

Step 3: Calculate Total Force on the Sphere's Surface

The force exerted on the surface of the sphere is given by \( F = PA \), where \( A \) is the surface area of the sphere. The surface area \( A \) is \( 4\pi r^2 \) where \( r = 0.0622 \, \text{m} \). Thus, \[ A = 4\pi (0.0622)^2 \, \text{m}^2 \approx 0.04861 \, \text{m}^2. \] The total force is: \[ F = 22402385 \, \text{Pa} \times 0.04861 \, \text{m}^2 \approx 1088958 \, \text{N}. \]

Step 4: Calculate the Buoyant Force on the Sphere

The buoyant force is given by Archimedes' principle: \( F_{\text{buoyant}} = \rho V g \), where \( V \) is the volume of the sphere. The volume \( V \) is \( \frac{4}{3}\pi r^3 \). So, \[ V = \frac{4}{3}\pi (0.0622)^3 \, \text{m}^3 \approx 0.001007 \, \text{m}^3. \] Then, the buoyant force is: \[ F_{\text{buoyant}} = 1025 \, \text{kg/m}^3 \times 0.001007 \, \text{m}^3 \times 9.81 \, \text{m/s}^2 \approx 10.1 \, \text{N}. \]

Step 5: Calculate Sphere's Acceleration when Free to Move

The net force acting on the sphere, \( F_{\text{net}} \), when it is free to move, is the buoyant force minus the gravitational force: \( F_{\text{net}} = F_{\text{buoyant}} - mg \), where \( m = 8.60 \, \text{kg} \). Thus: \[ mg = 8.60 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 84.366 \, \text{N}. \] Therefore, \[ F_{\text{net}} = 10.1 \, \text{N} - 84.366 \, \text{N} = -74.266 \, \text{N}. \] The sphere's acceleration \( a \) is given by Newton's second law: \( a = \frac{F_{\text{net}}}{m} \), so: \[ a = \frac{-74.266 \, \text{N}}{8.60 \, \text{kg}} \approx -8.63 \, \text{m/s}^2. \] The negative sign indicates acceleration is directed upwards.

Key Concepts

Buoyant Force

The buoyant force is a phenomenon encountered in fluid mechanics, which makes objects immersed in a fluid seem to weigh less. This force is the result of the pressure difference between the top and bottom of the object. It acts in the upward direction, opposing gravity.

The buoyant force can be calculated using the formula:

• \( F_{\text{buoyant}} = \rho V g \)

where:

• \( \rho \) is the density of the fluid,
• \( V \) is the volume of fluid displaced by the object,
• \( g \) is the acceleration due to gravity, typically \( 9.81 \, \text{m/s}^2 \).

This means that the buoyant force is directly proportional to the volume of the fluid displaced. As the volume increases, so does the buoyant force.

In our exercise, a sphere submerged in seawater experiences a buoyant force of approximately 10.1 newtons. The buoyancy slightly attempts to lift the sphere, counteracting its weight.

Archimedes' Principle

Archimedes' Principle is a fundamental principle in fluid mechanics that states any object submerged in a fluid is subject to an upward, or buoyant, force equal to the weight of the fluid displaced by the object.

Mathematically, Archimedes' Principle can be expressed as:

• \( F_{\text{buoyant}} = \rho V g \)

This principle applies to objects diversely, such as whether they float or sink. For an object to float, the buoyant force must equal the object's weight. Meanwhile, to sink, the object's weight must be greater than the buoyant force.

By using Archimedes' Principle, we calculated the sphere's buoyant force when submerged in seawater. Recognizing this helps us understand how objects move or stabilize in fluids.

Gauge Pressure

Gauge pressure is a term used in fluid mechanics to describe the pressure relative to atmospheric pressure. When you dive into water, you feel an increase in pressure due to the weight of the water above, known as gauge pressure. It’s crucial because it helps measure how much pressure is exerted solely by the fluid.

It can be found using the equation:

• \( P_{\text{gauge}} = \rho gh \)

where:

• \( \rho \) is the density of the fluid,
• \( g \) is the acceleration due to gravity,
• \( h \) is the depth of the fluid.

This is a measure of how much more pressure exists at a certain depth than at the surface.

In the exercise, we calculated the gauge pressure on the sphere to be approximately 22,301,385 Pascal at a depth of 2220 meters underwater.

Total Pressure

Total pressure is the sum of the gauge pressure and the atmospheric pressure. It accounts for all pressures exerting on a submerged object. This is important because it represents the absolute pressure experienced by the object, not just the added pressure from the fluid.

The calculation is simplified as:

• \( P_{\text{total}} = P_{\text{gauge}} + P_{\text{atm}} \)

where \( P_{\text{atm}} \) is the atmospheric pressure, often given as \( 1.01 \times 10^{5} \, \text{Pa} \).

For the sphere submerged in seawater, the total pressure calculates to about 22,402,385 Pascal, considering both the gauge and atmospheric pressures. This comprehensive evaluation allows us to grasp the effective pressure acting on the sphere, influencing its mechanical state.

Net Force

In physics, the net force is the total, or resultant, force acting on an object. It determines whether and how an object will move. A net force can cause an object to accelerate, stay still, or maintain constant velocity.

The net force equation can be captured by

• \( F_{\text{net}} = F_{\text{buoyant}} - mg \)

where \( mg \) is the gravitational force acting downward, typically expressed as \( mg \). In this scenario:

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity.

The net force considers the upward buoyant force and the downward gravitational force, providing insight into the object's movement.

For our sphere, the net force amounted to \(-74.266 \, \text{N}\), indicating the sphere would accelerate upward if unrestrained, contrary to the downward pull of gravity.

Newton's Second Law

Newton's Second Law is a foundational principle in physics that describes how the velocity of an object changes when it is subjected to an external force. This law is particularly significant in calculating the acceleration of an object when multiple forces are acting on it.

The formula is succinctly stated as:

• \( a = \frac{F_{\text{net}}}{m} \)

where:

• \( a \) is the acceleration of the object,
• \( F_{\text{net}} \) is the net force acting on the object,
• \( m \) is the mass of the object.

This law indicates that the net force will accelerate an object proportionally to the force exerted and inversely proportional to the object's mass.

With regard to the sphere, the calculated acceleration is approximately \(-8.63 \, \text{m/s}^2\), directed upwards. The negative sign implies the acceleration opposes the force of gravity, consistent with the buoyant force's upward direction.