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(II) A scuba tank, when fully submerged, displaces \(15.7 \mathrm{~L}\) of seawater. The tank itself has a mass of \(14.0 \mathrm{~kg}\) and, when "full," contains \(3.00 \mathrm{~kg}\) of air. Assuming only a weight and buoyant force act, determine the net force (magnitude and direction) on the fully submerged tank at the beginning of a dive (when it is full of air) and at the end of a dive (when it no longer contains any air).

Short Answer

Expert verified
Net force at the beginning: -8.78 N (downward). Net force at the end: 20.20 N (upward).

Step by step solution

01

Calculate the Weight of the Tank with Air

At the beginning of the dive, the total weight of the tank is the sum of the weight of the tank itself and the weight of the air it contains. The weight can be calculated as the product of mass and gravitational acceleration (\(g = 9.81 \, \text{m/s}^2\)).\[ W = (14.0 + 3.0) \, \text{kg} \times 9.81 \, \text{m/s}^2 = 166.32 \, \text{N} \]
02

Determine the Buoyant Force

The buoyant force acting on the tank is equal to the weight of the water displaced by the tank. The density of seawater is typically given as \(1025 \, \text{kg/m}^3\), and the volume of displaced water is 15.7 liters or 0.0157 cubic meters.\[ F_b = \rho \cdot V \cdot g = 1025 \, \text{kg/m}^3 \times 0.0157 \, \text{m}^3 \times 9.81 \, \text{m/s}^2 = 157.54 \, \text{N} \]
03

Calculate Net Force at the Beginning of the Dive

The net force is the difference between the buoyant force and the weight of the tank with air. Since the buoyant force acts upward and weight acts downward, the net force is:\[ F_{\text{net}, \text{begin}} = F_b - W = 157.54 \, \text{N} - 166.32 \, \text{N} = -8.78 \, \text{N} \]This indicates the net force is downward.
04

Calculate the Weight of the Empty Tank

At the end of the dive, the tank no longer contains any air. Thus, the weight is just the weight of the tank:\[ W_{\text{empty}} = 14.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 137.34 \, \text{N} \]
05

Calculate Net Force at the End of the Dive

With the tank empty, the buoyant force remains the same, but the weight is reduced. Therefore, the net force is:\[ F_{\text{net}, \text{end}} = F_b - W_{\text{empty}} = 157.54 \, \text{N} - 137.34 \, \text{N} = 20.20 \, \text{N} \]This result shows the net force is upward.

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Key Concepts

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

Scuba Diving Physics
When scuba diving, understanding the forces acting on a diver and their equipment is crucial for safety and navigation. One such force is buoyancy, which determines how an object like a scuba tank behaves underwater. Buoyancy refers to the upward force exerted by a fluid that opposes the weight of an object immersed in it. The principle behind this is Archimedes' Principle, which states that the buoyant force is equal to the weight of the fluid that the object displaces.
When a scuba tank is fully submerged, it displaces a certain volume of water. In our example, it displaces 15.7 liters of seawater, which creates an upward buoyant force. This force plays a significant role in scuba diving, as it can make the tank or diver float or sink, depending on the balance with the gravitational force (weight).
Understanding these principles helps divers control their buoyancy and maintain their position in the water column, allowing them to dive safely and enjoyably. Proper buoyancy control enhances maneuverability and conserves energy, which is important for both novice and experienced divers.
Fluid Dynamics
Fluid dynamics is the study of fluids (liquids and gases) in motion. It applies to many fields, including scuba diving, because water is a fluid that exerts forces on submerged objects. When a scuba tank is submerged in water, fluid dynamics helps explain how the water and tank interact.
In our problem, we consider how the moving water around the tank affects and is affected by the tank's shape and surface. The drag force, a concept from fluid dynamics, is a resistance force caused by the motion of the tank through the water. While buoyancy specifically deals with the displaced water creating an upward force, fluid dynamics adds layers, such as drag and flow patterns, that affect how objects move underwater.
  • Water pressure increases with depth, impacting the buoyancy and movement of the tank.
  • Understanding turbulence and flow patterns around the tank can inform design tweaks for efficiency.
Fluid dynamics also ensures that divers know how to minimize unnecessary movements, as moving against water's dense resistance can tire them quickly. This knowledge is applied through streamlining body and equipment to reduce drag.
Net Force Calculation
Calculating net force is a fundamental skill in physics, allowing us to understand how forces interact to cause objects to move. For a scuba tank submerged in water, two main forces interact: the downward gravitational force (weight) and the upward buoyant force.
The net force on the tank is simply the difference between these two forces. At the beginning of a dive, when the tank is full of air, the total weight of the tank (calculated by multiplying mass by the acceleration due to gravity, which is 9.81 m/s²) is higher. This leads to a downward net force, indicating the tank will tend to sink unless counteracted.
Conversely, at the end of a dive, when the tank is empty, its weight is reduced. Since the buoyant force remains constant (157.54 N in this example), the net force becomes upward, making the tank more buoyant and likely to rise.
  • Initial weight calculation: \(W = (14.0 \, \text{kg} + 3.0 \, \text{kg}) \times 9.81 \, \text{m/s}^2\)
  • Buoyant force: \(F_b = \rho \cdot V \cdot g\)
  • Net force: \(F_{\text{net}} = F_b - W\)
This net force calculation is essential for determining stability and movement, guiding divers in adjusting their own buoyancy for proper navigation underwater.

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Most popular questions from this chapter

A two-component model used to determine percent body fat in a human body assumes that a fraction \(f(<1)\) of the body's total mass \(m\) is composed of fat with a density of \(0.90 \mathrm{~g} / \mathrm{cm}^{3},\) and that the remaining mass of the body is composed of fat-free tissue with a density of \(1.10 \mathrm{~g} / \mathrm{cm}^{3} .\) If the specific gravity of the entire body's density is \(X,\) show that the percent body fat \((=f \times 100)\) is given by $$ \% \text { Body fat }=\frac{495}{X}-450 . $$

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