Question

The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}^{3}\) and the typical density of seawater is about \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) a) Draw a free-body diagram of a human body floating in seawater and determine what percentage of the body's volume is submerged. b) The average density of the human body, after maximum inhalation of air, changes to \(945 \mathrm{~kg} / \mathrm{m}^{3}\). As a person floating in seawater inhales and exhales slowly, what percentage of his volume moves up out of and down into the water? c) The Dead Sea (a saltwater lake between Israel and Jordan ) is the world's saltiest large body of water. Its average salt content is more than six times that of typical seawater, which explains why there is no plant and animal life in it. Two-thirds of the volume of the body of a person floating in the Dead Sea is observed to be submerged. Determine the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the seawater in the Dead Sea.

Short answer

What will be the percentage of volume moving up and down if their body density decreases to \(945 \mathrm{~kg}/\mathrm{m}^{3}\) upon inhalation? What is the density of the seawater in the Dead Sea if two-thirds of the person's volume is submerged? Answer: The percentage submerged is approximately \(96.57\%\). The percentage of volume moving up and down upon inhalation is approximately \(3.92\%\). The density of the seawater in the Dead Sea is approximately \(1477.5 \mathrm{~kg}/\mathrm{m}^{3}\).

Step-by-step solution

Draw the Free-Body Diagram

To draw a free-body diagram (FBD), we must identify all the forces acting on the human body floating in seawater. These forces are the buoyancy force (\(F_b\)) acting upward and the gravitational force, or weight of the person (\(W\)), acting downward.

Calculate the Percentage Submerged using Archimedes' Principle

Since the person is floating, the buoyancy force must equal their weight: \(F_b = W\) The buoyancy force is given by \(F_b = \rho_{water} \cdot V_{submerged} \cdot g\) and the weight of the person is given by \(W = \rho_{body} \cdot V_{total} \cdot g\), where \(\rho_{water}\) is the density of seawater, \(\rho_{body}\) is the density of the human body, \(V_{submerged}\) is the submerged volume, \(V_{total}\) is the total volume of the person, and \(g\) is the acceleration due to gravity. To find the percentage submerged, set these forces equal: \(\rho_{water}\cdot V_{submerged}\cdot g = \rho_{body}\cdot V_{total}\cdot g\) Since we are looking for the percentage submerged, we can define a ratio \(R = \frac{V_{submerged}}{V_{total}}\). By substituting this into the previous equation and solving for \(R\), we get: \(R = \frac{\rho_{body}}{\rho_{water}} = \frac{985 \mathrm{~kg} / \mathrm{m}^{3}}{1020 \mathrm{~kg} / \mathrm{m}^{3}} \approx 0.9657\) The percentage submerged is therefore approximately 96.57%. #b) Inhalation and Exhalation Percentage#

Calculate the Percentage Submerged after Inhalation

Repeating the same process as before, we can calculate the new percentage submerged after the person inhales air, with the new density of the human body being \(945 \mathrm{~kg} / \mathrm{m}^{3}\). \(R_{inhalation} = \frac{\rho_{body(inhalation)}}{\rho_{water}} = \frac{945 \mathrm{~kg} / \mathrm{m}^{3}}{1020 \mathrm{~kg} / \mathrm{m}^{3}} \approx 0.9265\) The percentage submerged after inhalation is approximately 92.65%.

Calculate the Percentage of Volume Moving Up and Down

The difference in the percentage of the body submerged before and after inhalation will give us the percentage of volume moving up and down. \(\Delta R = R_{inhalation} - R = 0.9265 - 0.9657 = -0.0392\) Therefore, approximately 3.92% of the volume moves up and down as the person inhales and exhales. #c) Density of Dead Sea Seawater#

Determine the Ratio of Submerged Volume

The problem states that two-thirds of the volume of the body is submerged in the Dead Sea, so the current ratio of submerged volume is: \(R_{DeadSea} = \frac{2}{3}\)

Calculate the Density of the Dead Sea Seawater

Using the same relationships as before, we can set up the equation: \(R_{DeadSea} = \frac{\rho_{body}}{\rho_{DeadSea}} \rightarrow \rho_{DeadSea} = \frac{\rho_{body}}{R_{DeadSea}} = \frac{985 \mathrm{~kg} / \mathrm{m}^{3}}{2/3} \approx 1477.5 \mathrm{~kg} / \mathrm{m}^{3}\) The density of the seawater in the Dead Sea is approximately \(\boxed{1477.5 ~\mathrm{kg} / \mathrm{m}^{3}}\).

Key concepts

Buoyancy

Buoyancy is the upward force exerted by a fluid on an object submerged in it. This force allows objects to float. When an object, like a human body, is placed in water, it experiences both a downward gravitational force and an upward buoyant force. For an object to float, these forces must balance each other out. In other words, the buoyant force should equal the weight of the object.

Archimedes' Principle plays a crucial role in determining the buoyant force. It states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. This means that the volume of water pushed aside by the floating object helps us know the upward force acting on it. By applying this principle, we can calculate how much of an object, like a human body, will be submerged when it floats in a fluid, such as seawater.

• Buoyant Force: The upward force exerted by the fluid.
• Displacement: The volume of fluid displaced by the object.
• Floating Condition: Occurs when buoyant force equals weight.

Density Calculations

Density is defined as mass per unit volume and is a key factor in understanding buoyancy. It's given by the formula: \[ \text{Density} (\rho) = \frac{\text{mass}}{\text{volume}} \] To solve the problem of the body floating in seawater, we compare the density of the human body with that of seawater. If an object's density is less than the fluid's density, it will float; otherwise, it will sink.

For example, using the given densities, the human body has a lower density (985 kg/m³) compared to seawater (1020 kg/m³). This explains why a person can float in seawater. To find the submerged fraction or percentage, we use the ratio of body density to water density: \[ R = \frac{\rho_{body}}{\rho_{water}} \approx 0.9657 \] This calculation shows that approximately 96.57% of the body is submerged in seawater. Similarly, by manipulating the density values provided, we can calculate how much a person's volume moves when inhaling and exhaling in water. Here, inhalation decreases body density, meaning only 92.65% of the body will be submerged.

Free-Body Diagram

A free-body diagram (FBD) is an essential tool for visualizing forces acting on an object. For a floating body, like a human in water, two primary forces are involved: the buoyancy force and gravitational force (weight).

In our scenario, when drawing a free-body diagram, we start by marking the mass as a downward force due to gravity. This represents the weight of the person and is calculated using mass times gravitational acceleration. The buoyancy force is depicted as an upward force, which is equal to the weight of the displaced water. For a body floating at rest, these forces must balance out.

• Gravitational Force: Depicted as an arrow pointing downward.
• Buoyancy Force: Depicted as an arrow pointing upward.
• Force Equilibrium: When forces balance, the object floats.

By using free-body diagrams, one can better understand the equilibrium condition where the weight equals the buoyancy force, leading to the floating phenomenon. It helps in visualizing the interactions and calculations necessary for solving buoyancy-related problems.