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Chapter 14: Problem 10

(a) Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.65 \(\mathrm{m}\) tall. (b) Consider a cylindrical segment of a blood vessel 2.00 \(\mathrm{cm}\) long and 1.50 \(\mathrm{mm}\) in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head?

Short Answer

Expert verified
(a) 17171.79 Pa; (b) 0.0303 N.

Step by step solution

01

Understand the Problem

We need to find the difference in blood pressure between two positions in the human body: the feet and the head. Blood pressure varies with height due to gravity and the fluid mechanics of blood. For a person 1.65 meters tall, we'll use the hydrostatic pressure formula to calculate this difference. We also need to determine the extra outward force a vessel must withstand between the feet and the head.
02

Calculate Blood Pressure Difference

The blood pressure difference due to being at different heights is given by \( \Delta P = \rho g h \), where \( \rho \) is the density of blood (approximately \( 1060\, \text{kg/m}^3 \)), \( g \) is the acceleration due to gravity (\( 9.81\, \text{m/s}^2 \)), and \( h \) is the height difference (\( 1.65\, \text{m} \)). Substitute these values: \[ \Delta P = 1060 \times 9.81 \times 1.65 = 17171.79\, \text{Pa} \] Thus, the difference in blood pressure is 17171.79 Pa.
03

Understand Force Requirement for the Vessel

We need to calculate the additional force due to the pressure difference on a cylindrical blood vessel. Use the formula for force due to pressure on the surface area: \( F = P \cdot A \). Here, we need the cross-sectional area \( A \) of the blood vessel.
04

Calculate Cross-sectional Area of the Vessel

The diameter of the blood vessel is 1.50 mm, so the radius \( r \) is 0.75 mm or \( 0.00075\, \text{m} \). The cross-sectional area \( A \) is given by \( A = \pi r^2 \). Calculate: \[ A = \pi \times (0.00075)^2 = 1.7671 \times 10^{-6} \text{ m}^2 \]
05

Calculate Additional Force

Using \( F = P \cdot A \), where \( P = 17171.79 \text{ Pa} \) is the pressure difference, and \( A = 1.7671 \times 10^{-6} \text{ m}^2 \), we find: \[ F = 17171.79 \times 1.7671 \times 10^{-6} = 0.0303 \text{ N} \] Thus, the additional outward force the vessel in the feet must withstand is 0.0303 N compared to the head.

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Key Concepts

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

Hydrostatic Pressure
Hydrostatic pressure is a fundamental concept in fluid mechanics, playing an essential role in how blood pressure varies within the human body. This kind of pressure is caused by the weight of a fluid above a certain point. According to the formula for hydrostatic pressure, \( P = \rho g h \), where \( \rho \) is the fluid's density, \( g \) is the gravitational acceleration, and \( h \) is the height difference, we can understand how pressure changes as we move from the head to the feet of a person.
When applied to blood pressure, hydrostatic pressure helps explain the additional pressure exerted by the blood at the feet compared to the head. Here, subtle variations in height lead to noticeable changes in pressure, largely influenced by gravity. The greater the height of the fluid column, the greater the hydrostatic pressure at its base.
For example, in the case of someone who is 1.65 meters tall, this results in a pressure difference of 17171.79 Pa between the head and the feet, showcasing how significant hydrostatic pressure can be in biological systems.
Fluid Mechanics
Fluid mechanics, the study of fluids and the forces on them, is crucial for comprehending how blood circulates through the human body. At its core, fluid mechanics involves understanding properties such as pressure and flow, which are essential for interpreting blood circulation in vessels.
Fluids, like blood, exhibit unique behaviors under various pressures and gravitational forces. In this case, understanding how blood flows vertically and horizontally through vessels, as well as how it adapts to changes in height, is key. Fluid mechanics principles explain why blood pressure is not uniform throughout the body and why it's generally higher in the feet than in the head.
The principles of fluid mechanics help to interpret how pressure differences from hydrostatic effects translate into forces that blood vessels need to withstand. This is particularly significant for understanding the functional requirements of blood vessels at different body parts. The knowledge gained from fluid mechanics principles can lead to the design of medical interventions and therapies for blood pressure management.
Cylindrical Blood Vessel
The structure of blood vessels, particularly their cylindrical shape, is a critical aspect in understanding how they withstand various pressures. Cylindrical blood vessels, due to their geometry, are regularly subjected to forces from the internal pressure of the blood they carry.
When examining a segment of a blood vessel with specific dimensions, the forces acting on it can be analyzed using the vessel's cross-sectional area. In the exercise, for example, a vessel with a 1.5 mm diameter and a length of 2 cm needs to resist additional forces corresponding to the pressure difference across different body parts.
The cross-sectional area \( A \) of the vessel can be calculated using \( A = \pi r^2 \), where \( r \) is the radius. This area plays an essential role in determining the force exerted on the vessel walls, calculated by \( F = P \cdot A \). The structural integrity of blood vessels is paramount because they need to withstand varying pressure levels, especially in vulnerable areas like the feet. This force calculation is crucial, as it warns of potential risks like vessel expansion or even rupture if the pressure becomes too great.

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Most popular questions from this chapter

The weight of a king's solid crown is w. When the crown is suspended by a light rope and completely immersed in water, the tension in the rope (the crown's apparent weight) is \(f w .\) (a) Prove that the crown's relative density (specific gravity) is \(1 /(1-f) .\) Discuss the meaning of the limits as \(f\) approaches 0 and \(1 .\) (b) If the crown is solid gold and weighs 129 \(\mathrm{N}\) in air, what is its apparent weight when completely immersed in water? (c) Repeat part (b) if the crown is solid lead with a very thin gold plating, but still has a weight in air of 12.9 \(\mathrm{N}\) .

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A\) . The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \(\omega\) . Gravitational forces are negligible. Consider a volume element of the fluid of area A and thickness \(d r^{\prime}\) adistance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .(\text { a) Apply }\) Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime} .\) (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0}\) , show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .(\mathrm{c})\) An object of volume \(V\) and density \(\rho_{o b}\) has its center of mass at a distance \(R_{\text { cmob }}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{cb}} R_{\mathrm{cmod}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}}(\mathrm{e})\) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 \(\mathrm{m}\) . (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth. \((b)\) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 \(\mathrm{cm}\) in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You can ignore the small variation of pressure over the surface of the window.)

A single ice cube with mass 9.70 g floats in a glass completely full of 420 \(\mathrm{cm}^{3}\) of water. You can ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 \(\mathrm{kg} / \mathrm{m}^{3}\) . What volume of salt water would the \(9.70-\mathrm{g}\) ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.

A cubical block of density \(\rho_{\mathrm{B}}\) and with sides of length \(L\) floats in a liquid of greater density \(\rho_{\mathrm{L}}\) (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \(\rho_{\mathrm{W}} )\) and does not with it If water is poured on the surface of the liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L, \rho_{\mathrm{B}}, \rho_{\mathrm{L}},\) and \(\rho_{\mathrm{w}}\) . (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and the side length is \(10.0 \mathrm{cm} .\)
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