<< Chapter < Page Chapter >> Page >

This force is the weight of about a 680-g mass. A mass of 680 g resting on the eye (imagine 1.5 lb resting on your eye) would be sufficient to cause it damage. (A normal force here would be the weight of about 120 g, less than one-quarter of our initial value.)

People over 40 years of age are at greatest risk of developing glaucoma and should have their intraocular pressure tested routinely. Most measurements involve exerting a force on the (anesthetized) eye over some area (a pressure) and observing the eye’s response. A noncontact approach uses a puff of air and a measurement is made of the force needed to indent the eye ( [link] ). If the intraocular pressure is high, the eye will deform less and rebound more vigorously than normal. Excessive intraocular pressures can be detected reliably and sometimes controlled effectively.

The tonometer being used by an eye care professional to determine the fluid pressure inside the eye.
The intraocular eye pressure can be read with a tonometer. (credit: DevelopAll at the Wikipedia Project.)

Calculating gauge pressure and depth: damage to the eardrum

Suppose a 3.00-N force can rupture an eardrum. (a) If the eardrum has an area of 1 . 00 cm 2 size 12{1 "." "00"`"cm" rSup { size 8{2} } } {} , calculate the maximum tolerable gauge pressure on the eardrum in newtons per meter squared and convert it to millimeters of mercury. (b) At what depth in freshwater would this person’s eardrum rupture, assuming the gauge pressure in the middle ear is zero?

Strategy for (a)

The pressure can be found directly from its definition since we know the force and area. We are looking for the gauge pressure.

Solution for (a)

P g = F / A = 3 . 00 N / ( 1 . 00 × 10 4 m 2 ) = 3 . 00 × 10 4 N/m 2 . size 12{P rSub { size 8{g} } =F/A=3 "." "00"`N/ \( 1 "." "00" times "10" rSup { size 8{ - 4} } `m rSup { size 8{2} } \) =3 "." "00" times "10" rSup { size 8{4} } `"N/m" rSup { size 8{2} } } {}

We now need to convert this to units of mm Hg:

P g = 3 . 0 × 10 4 N/m 2 1.0 mm Hg 133 N/m 2 = 226 mm Hg.

Strategy for (b)

Here we will use the fact that the water pressure varies linearly with depth h size 12{h} {} below the surface.

Solution for (b)

P = hρg size 12{P=hρg} {} and therefore h = P / ρg size 12{h=P/ρg} {} . Using the value above for P size 12{P} {} , we have

h = 3.0 × 10 4 N/m 2 ( 1.00 × 10 3 kg/m 3 ) ( 9.80 m /s 2 ) = 3.06 m.

Discussion

Similarly, increased pressure exerted upon the eardrum from the middle ear can arise when an infection causes a fluid buildup.

Got questions? Get instant answers now!

Pressure associated with the lungs

The pressure inside the lungs increases and decreases with each breath. The pressure drops to below atmospheric pressure (negative gauge pressure) when you inhale, causing air to flow into the lungs. It increases above atmospheric pressure (positive gauge pressure) when you exhale, forcing air out.

Lung pressure is controlled by several mechanisms. Muscle action in the diaphragm and rib cage is necessary for inhalation; this muscle action increases the volume of the lungs thereby reducing the pressure within them [link] . Surface tension in the alveoli creates a positive pressure opposing inhalation. (See Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action .) You can exhale without muscle action by letting surface tension in the alveoli create its own positive pressure. Muscle action can add to this positive pressure to produce forced exhalation, such as when you blow up a balloon, blow out a candle, or cough.

The lungs, in fact, would collapse due to the surface tension in the alveoli, if they were not attached to the inside of the chest wall by liquid adhesion. The gauge pressure in the liquid attaching the lungs to the inside of the chest wall is thus negative, ranging from 4 size 12{ - 4} {} to 8 mm Hg size 12{ - 8`"mm"`"Hg"} {} during exhalation and inhalation, respectively. If air is allowed to enter the chest cavity, it breaks the attachment, and one or both lungs may collapse. Suction is applied to the chest cavity of surgery patients and trauma victims to reestablish negative pressure and inflate the lungs.

Questions & Answers

what's Thermochemistry
rhoda Reply
the study of the heat energy which is associated with chemical reactions
Kaddija
How was CH4 and o2 was able to produce (Co2)and (H2o
Edafe Reply
explain please
Victory
First twenty elements with their valences
Martine Reply
what is chemistry
asue Reply
what is atom
asue
what is the best way to define periodic table for jamb
Damilola Reply
what is the change of matter from one state to another
Elijah Reply
what is isolation of organic compounds
IKyernum Reply
what is atomic radius
ThankGod Reply
Read Chapter 6, section 5
Dr
Read Chapter 6, section 5
Kareem
Atomic radius is the radius of the atom and is also called the orbital radius
Kareem
atomic radius is the distance between the nucleus of an atom and its valence shell
Amos
Read Chapter 6, section 5
paulino
Bohr's model of the theory atom
Ayom Reply
is there a question?
Dr
when a gas is compressed why it becomes hot?
ATOMIC
It has no oxygen then
Goldyei
read the chapter on thermochemistry...the sections on "PV" work and the First Law of Thermodynamics should help..
Dr
Which element react with water
Mukthar Reply
Mgo
Ibeh
an increase in the pressure of a gas results in the decrease of its
Valentina Reply
definition of the periodic table
Cosmos Reply
What is the lkenes
Da Reply
what were atoms composed of?
Moses Reply
what is chemistry
Imoh Reply
what is chemistry
Damilola
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 5

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask