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Chapter 14: Problem 9

\(\cdot 9 \Rightarrow\) Blood pressure in Argentinosaurus. (a) If this longnecked, gigantic sauropod had a head height of \(21 \mathrm{~m}\) and a heart height of \(9.0 \mathrm{~m},\) what (hydrostatic) gauge pressure in its blood was required at the heart such that the blood pressure at the brain was 80 torr (just enough to perfuse the brain with blood)? Assume the blood had a density of \(1.06 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} .\) (b) What was the blood pressure (in torr or \(\mathrm{mm} \mathrm{Hg}\) ) at the feet?

Short Answer

Expert verified
(a) Pressure at heart: 1015 torr. (b) Pressure at feet: 1718 torr.

Step by step solution

01

Understand the Given Problem

We need to calculate the hydrostatic pressure difference due to gravity between two points in the Argentinosaurus, namely the heart and the brain/feet. The terms given are brain pressure (80 torr), heart height (9 m), and head height (21 m). The density of blood is provided as \(1.06 \times 10^{3} \, \mathrm{kg/m}^3\).
02

Calculate Blood Pressure at Heart for Brain Perfusion

To maintain 80 torr at the brain, we need to find the pressure required at the heart. The pressure at the heart must be higher than at the brain due to the vertical height of 12 m (height of brain minus height of heart: \(21 - 9 = 12\, \text{m}\)). Using the hydrostatic pressure formula, we have:\[ P = \rho gh \]where \(\rho = 1.06 \times 10^3 \, \text{kg/m}^3\) (density of blood), \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity), and \(h = 12 \, \text{m}\).Calculate difference in pressure:\[ \Delta P_{heart ightarrow brain} = 1.06 \times 10^3 \times 9.81 \times 12 = 1.246 \times 10^5 \, \text{Pa} \]Convert \(1.246 \times 10^5\, \text{Pa}\) to torr:\[1\, \text{torr} = 133.322\, \text{Pa}\]\[ \Delta P_{heart\_torr} = \frac{1.246 \times 10^5}{133.322} \approx 935\, \text{torr}\]Thus, the pressure at the heart must be:\[P_{heart} = P_{brain} + \Delta P_{heart\rightarrow brain}\]\[P_{heart} = 80 \, \text{torr} + 935 \, \text{torr} = 1015 \, \text{torr}\]
03

Calculate Blood Pressure at the Feet

Using the heart pressure as a reference, compute the pressure at the feet. The feet are 9 m below the heart. Again applying the hydrostatic pressure formula, find:\[ \Delta P_{foot} = \rho gh \]\[ \Delta P_{foot} = 1.06 \times 10^3 \times 9.81 \times 9 = 9.367 \times 10^4 \, \text{Pa} \]Convert \(9.367 \times 10^4\, \text{Pa}\) to torr:\[ \Delta P_{foot\_torr} = \frac{9.367 \times 10^4}{133.322} \approx 703\, \text{torr}\]Therefore, total pressure at the feet is:\[P_{feet} = P_{heart} + \Delta P_{foot}\]\[P_{feet} = 1015 \, \text{torr} + 703 \, \text{torr} = 1718 \, \text{torr}\]

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Key Concepts

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

Blood Pressure
Blood pressure is the force exerted by circulating blood upon the walls of blood vessels. It's crucial for maintaining life as it ensures oxygen and nutrients are delivered throughout the body. Blood pressure is often measured in units called torr or mmHg.

Let's break down our task using fluid mechanics for something grand like a sauropod! Blood pressure in these massive creatures is particularly intriguing because of the enormous height difference from their heart to their brain and feet.

To keep the brain properly perfused with blood, the heart must create a pressure that not only pushes blood to a height but compensates for gravitational pressure loss. In this exercise, hydrostatic pressure formulas help calculate the necessary pressure at different points in a sauropod's body, similar to how human blood pressure is more at our feet than our heart due to gravity. However, these creatures needed much larger pressures due to their size! Utilizing the values of blood density and gravitational acceleration helps us compute these differences in pressure.
Sauropods
Sauropods were one of the largest groups of dinosaurs. With long necks and tails, they relied on strong hearts to pump blood to their elevated heads. When considering blood pressure in these magnificent creatures, their size presents unique challenges.

An Argentinosaurus, for example, could have a head located as high as 21 meters above its heart. Thus, its circulatory system had to be remarkably efficient and capable of producing extremely high blood pressures to supply blood to its brain. Modern-day sauropods would face a massive head height challenge, which means their hearts would be more powerful than any animal today.

The differences in pressure due to height are calculated using hydrostatic principles. This is similar to how tall plants need to move water from roots to leaves against gravity using capillary action, but in the case of sauropods, it was more about robust pumping action!
Fluid Mechanics
Fluid mechanics is the branch of physics that studies the behavior of fluids (liquids and gases) and the forces on them. It plays an essential role in understanding blood pressure variations in organisms.

Through the formulas of fluid mechanics, it becomes possible to calculate the pressure exerted by a fluid within the circulatory system. In our exercise, using the equation for hydrostatic pressure, we calculated the additional pressure required to compensate for the vertical distance between the heart and brain in a sauropod.

- **Hydrostatic Pressure Formula**: It states that pressure in a fluid at a given depth is determined by the fluid's density, gravity, and height. - This principle explains why the sauropod needed higher pressure at its heart to maintain a sufficient pressure at its high-altitude brain. - Similarly, pressure at the feet would be dramatically higher due to the gravitational effects pulling blood down.

Understanding such calculations allows us to gain insights into how massive creatures like sauropods managed their circulatory demands, reflecting the marvel of natural engineering in ancient times and compelling us to appreciate the underline physics in biological systems!

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Most popular questions from this chapter

The tension in a string holding a solid block below the surface of a liquid (of density greater than the block) is \(T_{0}\) when the container (Fig. \(14-57\) ) is at rest. When the container is given an upward acceleration of \(0.250 \mathrm{~g}\), what multiple of \(T_{0}\) gives the tension in the string?

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An armada of spaceships that is 1.00 ly long (as measured in its rest frame) moves with speed \(0.800 \mathrm{c}\) relative to a ground station in frame \(S\). A messenger travels from the rear of the armada to the front with a speed of \(0.950 \mathrm{c}\) relative to \(S .\) How long docs the trip take as measured (a) in the rest frame of the messenger, (b) in the rest frame of the armada, and (c) by an observer in the ground frame \(S ?\)

The car-in-the-garage problem. Carman has just purchased the world's longest stretch limo, which has a proper length of \(L_{c}=30.5 \mathrm{~m} .\) In Fig. \(37-32 a,\) it is shown parked in front of a garage with a proper length of \(L_{g}=6.00 \mathrm{~m}\). The garage has a front door (shown open) and a back door (shown closed). The limo is obviously longer than the garage. Still, Garageman, who owns the garage and knows something about relativistic length contraction, makes a bet with Carman that the limo can fit in the garage with both doors closed. Carman, who dropped his physics course before reaching special relativity, says such a thing, even in principle, is impossible. To analyze Garageman's scheme, an \(x_{c}\) axis is attached to the limo, with \(x_{c}=0\) at the rear bumper, and an \(x_{k}\) axis is attached to the garage, with \(x_{g}=0\) at the (now open) front door. Then Carman is to drive the limo directly toward the front door at a velocity of \(0.9980 c\) (which is, of course, both technically and financially impossible). Carman is stationary in the \(x_{c}\) reference frame; Garageman is stationary in the \(x_{z}\) reference frame. There are two events to consider. Event 1 : When the rear bumper clears the front door, the front door is closed. Let the time of this event be zero to both Carman and Garageman: \(t_{\mathrm{s} 1}=t_{c 1}=0\). The event occurs at \(x_{c}=x_{g}=0 .\) Figure \(37-32 b\) shows event 1 according to the \(x_{g}\) reference frame. Event 2 : When the front bumper reaches the back door, that door opens. Figure \(37-32 c\) shows event 2 according to the \(x_{g}\) reference frame. According to Garageman, (a) what is the length of the limo, and what are the spacetime coordinates (b) \(x_{R 2}\) and (c) \(t_{g 2}\) of event \(2 ?\) (d) For how long is the limo temporarily "trapped" inside the garage with both doors shut? Now consider the situation from the \(x_{c}\) reference frame, in which the garage comes racing past the limo at a velocity of \(-0.9980 \mathrm{c}\). According to Carman, (e) what is the length of the passing garage, what are the spacetime coordinates (f) \(x_{c 2}\) and \((g) t_{c 2}\) of event \(2,(h)\) is the limo ever in the garage with both doors shut, and (i) which event occurs first? (j) Sketch events 1 and 2 as seen by Carman. (k) Are the events causally related; that is, does one of them cause the other? (I) Finally, who wins the bet?
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