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Chapter 11: Problem 43

A person can change the volume of his body by taking air into his lungs. The amount of change can be determined by weighing the person under water. Suppose that under water a person weighs \(20.0 \mathrm{~N}\) with partially full lungs and \(40.0 \mathrm{~N}\) with empty lungs. Find the change in body volume.

Short Answer

Expert verified
The change in body volume is approximately \(0.00204 \ \mathrm{m^3}\).

Step by step solution

01

Identify the force difference

First, find the difference in buoyant force when the person has their lungs partially full versus empty. This is given by the difference in underwater weights: \[ F_{empty} = 40.0 \, \mathrm{N} \]\[ F_{partial} = 20.0 \, \mathrm{N} \]The difference \( \Delta F \) is:\[ \Delta F = F_{empty} - F_{partial} = 40.0 \, \mathrm{N} - 20.0 \, \mathrm{N} = 20.0 \, \mathrm{N} \]
02

Relate force difference to volume change

This difference in force corresponds to the change in buoyancy force, which is caused by the change in body volume. The buoyant force \( F_b \) is given by Archimedes' principle:\[ F_b = \rho \cdot V \cdot g \]where \( \rho \) is the density of water \( (1000 \, \mathrm{kg/m^3}) \), \( V \) is the volume of water displaced, and \( g \) is the acceleration due to gravity \( (9.81 \, \mathrm{m/s^2}) \). Therefore, the change in body volume \( \Delta V \) can be found from:\[ \Delta F = \rho \cdot \Delta V \cdot g \]
03

Solve for the volume change

Rearrange the equation from Step 2 to solve for volume change \( \Delta V \):\[ \Delta V = \frac{\Delta F}{\rho \cdot g} = \frac{20.0 \, \mathrm{N}}{1000 \, \mathrm{kg/m^3} \cdot 9.81 \, \mathrm{m/s^2}} \].Calculate \( \Delta V \):\[ \Delta V = \frac{20.0}{1000 \cdot 9.81} \approx 0.00204 \, \mathrm{m^3} \]

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Key Concepts

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

Buoyant Force
The concept of buoyant force is central to understanding how objects behave in water. Buoyancy is the upward force that water exerts on an object immersed in it. This force is what makes objects feel lighter in water.

According to Archimedes' principle, the magnitude of the buoyant force is equal to the weight of the water displaced by the object. In our example, the difference in underwater weights shows how buoyant force changes depending on lung volume. The greater the volume of air in the lungs, the more water is displaced, and the larger the buoyant force.

  • Buoyant Force = Weight of Displaced Water
  • Increased body volume = Increased displaced water
  • Larger buoyant force leads to feeling lighter underwater
Body Volume
Body volume plays a crucial role in determining the buoyant force. When you inhale, your body expands, increasing the amount of water displaced. This expansion is directly linked to the buoyancy you experience.

The step-by-step solution demonstrated how to calculate the change in body volume based on the difference in buoyant force. The volume of water displaced translates to how much space your body takes up in the water.

Water displacement depends on:
  • The anatomical shape and size
  • The amount of air held in the lungs
Changing your body volume by breathing in and out illustrates how our bodies can dynamically interact with buoyant forces.
Density of Water
The density of water is vital in calculating buoyant force. Density is defined as mass per unit volume. For water, this density is approximately 1000 kg/m鲁.

Knowing the density of water allows us to relate it to buoyant force using Archimedes' principle. The more dense a fluid, the greater the buoyant force on an object submerged in it. This principle is why denser fluids like saltwater provide more buoyancy than freshwater.

  • Higher density = Stronger buoyant force
  • Vital for accurate calculations of volume and force
The consistency of water's density enables precise calculations in various applications, especially in exercises involving changes in body volume as shown in this problem.
Gravity
Gravity is the force that pulls objects toward the Earth and plays an essential role in the buoyant force. The standard acceleration due to gravity is typically approximated as 9.81 m/s虏.

The effect of gravity on buoyant force comes into play through its relationship with the weight of the displaced fluid. In the formula for buoyant force \[ F_b = \rho \cdot V \cdot g \]gravity contributes to determining how much weight of water is displaced, and therefore, how much buoyancy an object experiences.

  • Key factor in buoyancy calculations
  • Rules how buoyant force counters weight
Understanding gravity's interaction with fluid mechanics helps explain why objects rise or sink in water, based on their volume and density as explored in the given exercise.

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Most popular questions from this chapter

A dentist's chair with a patient in it weighs \(2100 \mathrm{~N}\). The output plunger of a hydraulic system begins to lift the chair when the dentist's foot applies a force of \(55 \mathrm{~N}\) to the input piston. Neglect any height difference between the plunger and the piston. What is the ratio of the radius of the plunger to the radius of the piston?

A suitcase (mass \(\mathrm{m}=16 \mathrm{~kg}\) ) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures \(0.50 \mathrm{~m}\) by \(0.15 \mathrm{~m}\). The elevator is moving upward, the magnitude of its acceleration being \(1.5 \mathrm{~m} / \mathrm{s}^{2}\). What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

A suitcase (mass \(m=16 \mathrm{~kg}\) ) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures \(0.50 \mathrm{~m}\) by \(0.15 \mathrm{~m} .\) The elevator is moving upward, the magnitude of its acceleration being \(1.5 \mathrm{~m} / \mathrm{s}^{2}\). What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is \(1.80 \times 10^{5} \mathrm{~Pa}\). Assuming that the top and bottom surfaces of the cap each have an area of \(4.10 \times 10^{-4} \mathrm{~m}^{2},\) obtain the magnitude of the force that the screw thread exerts on the cap in order to keep it on the bottle. The air pressure outside the bottle is one atmosphere.

One kilogram of glass \(\left(\rho=2.60 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)\) is shaped into a hollow spherical shell that just barely floats in water. What are the inner and outer radii of the shell? Do not assume the shell is thin.
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