Force Pressure

8/13/2019 Force Pressure

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LECTURE 19

Force and Pressure

Introduction

By now you are no doubt aware that I like toopen a lecture with some demonstration that hasresults that you would not predict unless youknow some physics. The demonstration for thislecture is set up by stringing a long cable acrossthe lecture platform. One end is tied to a rigidsupport point. The other is tied to a sturdy pieceof wood. You can see that this piece of wood issturdy from my swinging from it with all myweight.

CABLE

BEAM OF PROJECTION SCREEN

SUPPORTCLAMP

I then go to the center of the cable and pull itdownward. With very little effort there is a loudcrack as the wood breaks.

WOOD

The question is from where did this forcecome? It was certainly the force of my hand thatcaused it but the force of my hand was clearlymuch less than the force it caused. How couldthis be?

This amplification of force by geometry is avery important phenomenon in nature. It is ofparticular importance in the invention of tools by

humans. In medicine one of the routineapplications of the phenomenon is inphysiotherapy where devices have to be arrangedto put specific forces on parts of the human bodyto correct disorders. This lecture is anintroduction to such manipulation of forces. Itwill include the concept of pressure and how itrelates to such forces.

Physics Definitions of Force and Pressure

The words "force" and "pressure" have manyshades of meaning in English. However, they aregenerally associated with agents of change. Thusthings try to force you to change, or applypressure on you to change. Alternatively, they areoften thought of as things to be resisted if you donot want to change. You withstand the force, oryou cope with the pressure.

In physics, we again take a very restrictedmeaning for these words. In the case of force itis that of an agent which causes a mass to move.The most common such force is that of theweight of objects. This weight is the force thatcauses an object to fall to Earth.

To prevent an object from falling, the force ofits weight must be resisted. When you areholding an object this force is the force you mustapply by your hand. For example, if you areholding a mass in the palm of your hand, thatforce is the upward thrust of your hand againstthe bottom of the mass.

UPWARDFORCE OFMY HAND

UPWARDFORCE OFMY HAND

m m

DOWNWARDFORCE OFTHE WEIGHT

DOWNWARDFORCE OFTHE WEIGHT

To hold the object steady you must, of course,apply exactly the same force as the weight of theobject. For example, if the weight wasone kilogram and you applied an upward force of1100 grams, the mass would rise. If you appliedonly a force of 900 grams, the mass would fall.You must apply exactly 1000 grams. In fact, it is

one of the remarkable things about the humannervous system that it can generate exactly theright force to keep things like the mass in yourhand exactly where you want it to be.

However, suppose you had several different1-kilogram objects with different diameterbottoms. Suppose one had a nice flat diameter of6 cm while another had a stub on its bottom with


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Physics 101A - Physics for the life sciences 2

a diameter of only 1 cm and a third had a sharptip of only 1 mm diameter.

1 kg

6 cm1 cm 1 mm

1 kg 1 kg

Each of these masses would cause your handto apply an upward force of exactly one kilogram,but they would certainly feel different. Thediscomfort caused by the middle object, and thesevere pain caused by the pointy-bottomed objectwould be due to thepressurecaused by theweight of the objects where the pressure is relatedto the way the weight is spread out over yourhand. If it is spread out over a large area then the

pressure is said to be low. If the weight isconcentrated in a small area the pressure is said tobe high.

Again, in physics we must have exactdefinitions so that we can get numbers frommeasurements that can be studied for possiblemathematical relationships. Here the definition isstraightforward. It is simply that the pressure isthe force divided by the area over which that forceis spread:

p F

A=

In the example given here, the pressure of aone-kilogram force spread over a 6-cm diametercircle on your hand is 1 kg 28.3 cm2or 0.035kilograms per square centimeter (abbreviated tokg/cm2). The pressure from the kilogram withthe 1-cm bottom would be 1 kg 0.785 cm2or1.127 kg/cm2, while the pressure due to thestiletto-bottomed mass would be 112.7 kg/cm2.

(For those of you who have difficultyextracting yourselves from the US culture, the

unit of pressure there is, as you may well know,the pound per square inch, or "psi", often calledsimply "pounds pressure". If you are everunfortunate enough to have to deal with apressure in this unit it can be converted to therest-of-the world unit by dividing by 2.93, orabout 3. Thus a pressure of 30 psi, a commonpressure for automobile tires, is actually apressure of about 10 kg/cm2.)

A pressure of 112.7 kg/cm2is about 10 timesthat of the air pressure in an automobile tire andwill come close to piercing your skin, dependingon how tough your skin actually is. The weightwould certainly pierce your skin if the mass wassitting on a thumbtack pointing downward toward

your hand.So it is not the weight of the object that willdamage your skin but the pressure. This isgenerally true for all materials and is one of themain reasons why the concept of pressure isimportant. An architectural example arose in themid-sixties in North America when high stilettoheels suddenly came into women's fashion. Theseheels eventually became ridiculously small; insome cases to about 5-mm diameter at their tips.This resulted in a pressure from a 50 kg womanresting most of her weight on just one of theseheels (a common poise for a woman suffering

from these torture devices) of about 150 kg/cm2.This pressure not only dented linoleum floors; iteven broke the edges of ceramic tiles.

In general, the ability of a material to resistforces is related to the pressure that the materialcan withstand in resisting these forces. Thispressure is related, in a very complicated way, tothe intermolecular forces within the material aswell as the type of deformation that the pressureis tending to cause. For example, a dent is easierto create in a soft material which can flow then itis in a harder material that does not, even thoughthe harder material might break at a lowerpressure than the softer material. In technicalterms, the softer material is more "resilient". Infigurative sports terms, it "moves with the punch".The result of this complication is that materialshave to be tested under various circumstances tosee what their strengths actually are for aparticular application. The following figures arerepresentative results for straight compression.(See Hecht p. 349 Table 10.1.)

Material Compressive Strength(kg/cm2)

Concrete 350Wood (soft) 500Bone 1500

These figures may be at first surprising,particularly that soft wood is stronger thanconcrete and that bone is three times as strong aswood. This comes from the ability of wood toabsorb pressure by deforming before breaking,


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Lecture 19 - Force and Pressure 3

while concrete is brittle and breaks. Thus woodwill dent but not break while concrete will notdent but will break instead. The strength of boneis a result of its marvelous complexity (it is theultimate "composite material") that gives acombination of strength and ability to deform

without breaking .While pressure is easy to visualize as asqueeze, i.e. a compression, it can also be a pullas the various molecular bonds are stretched.Such an action can be thought of as a negativepressure. However, while it is still usuallymeasured in kg/cm2it is normally referred to as a"tensile stress" rather than a "pressure". (Asqueezing pressure is likewise often referred to asa "compressive stress".) In this case there is verylittle complication in the nature of the deformationof the material. The material just stretches until itsmolecular bonds break. Materials designed for

this sort of resistance to movement tend to belong and fibrous. Some representative values fortensile strength of material are given in the tablebelow. (For a more extensive list see Hecht p.349 Table 10.1).

Material Tensile Strength(kg/cm2)

Muscle 1Bladder wall 2Skin 100Tendon 800

Hair 1900Spider web 2400Concrete 40Wood (soft) 1000Bone 1100Ordinary steel 4000Piano wire 31000

Again some of these numbers may besurprising, in particular the extreme strength ofspider web (it is over half as strong as steel).Also the tensile strength of concrete might seem

low until you realize that concrete is used tosupport weights that press against it rather thanhang from it. In this respect, it may be surprisingthat bone material is so strong under bothcompression and tension but then bone hasevolved to take a great variety of forces withoutbreaking, some that try to stretch it while otherstry to crush it. A final number of greatimportance in the life sciences, particularly insports medicine, is the low tensile strength of

muscle. This comes about not by a neglect ofnature but rather by the fact that nature has foundno other fiber that can be activated by an electricalsignal to contract on command. (Note that it hasfound a fiber, the material of the tendon, whichdoes have a much higher strength but this is an

inert material.) The low tensile strength ofmuscle fiber requires that a great deal of musclemass is required for the sort of forces that areexerted in sports and until this muscle mass isachieved by gradual exercise it is very easy tocreate a muscle tear.

All of these stresses come about because ofthe requirement that the body create forces toresist forces that are applied to it from externalagents. In other word, to balance these outsideforces. I will therefore return to a more detailedconsideration of this requirement of the balancingof forces.

The Balancing of Forces

The requirement that any force on a stationarybody must be balanced by another force is one ofthe most important principles in physics. Yet,despite its importance, in the history of physics itwas also one of the most difficult principles todiscover, so don't feel inadequate if it is not atfirst obvious to you. To help, I will try to make itas believable as possible.

When you are yourself creating the upwardforce to resist the downward weight of an objectthe idea seems perhaps obvious. You are creating

that force and you know it. It may not be soobvious when the force is being created by aninsensate object such as a spring scale used forweighing amounts of material in the kitchen.However, if you look carefully at the reaction ofsuch a scale when you put an object on it you willsee that the pan of the scale moves down until thespring has built up sufficient upward force tostop the mass from falling any further. You canexperience this building up of an upward force bypushing down on the scale yourself. The moreyou push the scale downward, the move it resistsyour push by pushing upward. In fact, at anyinstant the dial on the scale will read the forcewith which you are pushing down on it.

It is probably not at all obvious that anydevice that keeps an object from falling creates anupward force that must balance the downwardweight of the object. Thus even if a book isresting on a table, that table must be creating anupward force exactly equal to the weight of thebook.


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DOWNWARD WEIGHT OF BOOK

UPWARD FORCE OF TABLE

BOOK

TABLE

It is here that, for most people, physics hasgone too far. Tables are inanimate things that donot create forces all on their own! Yet that isexactly what physics tells you is happening.

Since all objects have weight all objects onEarth must be balanced by forces if they are notto move, and the vast majority of them do notmove (they are, after all, called "inanimate").

It is natural then to consider a force as theweight that it would balance. Consider a springscale, as used for weighing the fish you justcaught. It registers the weight by a pointer thatmoves against a scale. This scale is calibrated so

that when the device is used to suspend a fishfrom its hook the pointer is brought to the pointon its scale that corresponds to the weight of thefish.

gg

SPRINGRELAXED

STRETCHED

OF FISH

SPRING

9.4 kg

(9.4 kg)

. g

A very important concept here is that when aspring scale registers, say, 9.4 kg it is beingpulled on from both ends by 9.4 kg. One way tothink about this is to imagine the spring scale inthe diagram turned upside down. It would stillregister 9.4 kg if a 9.4 kg fish was hung on it.

Of course, such a scale does not have to beused just for weighing fish. It can be used formeasuring any force. For instance, if youconnect a spring scale to a wall and pull on thescale it will read the force with which you arepulling against the wall.

0

100

kg

50

85 kgPULL OF HAND

= 85 kg

PULL ON WALL= 85 kg(100 kg range)

Likewise, you can use it to measure thetension in a rope being used by two teamsplaying tug-of-war.

Tonne

1.0

0.5 0

g

g range

= 650 kg

= 650 kg

You could even use a set of these scales in acomplicated tug-of-war between three teams,where one of the teams is much bigger than theothers and so competes against the two smallerteams who help each other. The arrangement inthe diagram below would allow you to tell exactlyhow the force is shared between the two smaller

teams.

PULL OF LEFT TEAM 1 = 350 kgPULL OF RIGHT TEAM

= 800 kg

50 kg

450 kg

1

2

PULL OF LEFT TEAM 2 = 450 kg

00 kg50 kg

50 kg

Three tug-of-war teams do not have to line upto play the game. They could arrange themselves

in a triangular formation as shown in the diagrambelow, where now there are two bigger teams andone small team.

Here you would find that the one small teamactually balances the two larger teams. Theexplanation for this is in the fact that force is avector. Again, this is because force has bothmagnitude and direction. The force vectorsinvolved in this bizarre tug-of-war are shown.


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800k

g

450 kg

2

1

1

3

800kg

O

PULL OF TEAM 1 = 450 kgPULL OF TEAM 2 = 800 kg

PULL OF TEAM 3 = 800 kg

2

1

3

This method of having a small force producea very large reaction force has been known fromantiquity. A modern version of it can sometimesbe seen in off-road expeditions where a vehiclehas become stuck in rough terrain. (It does notappear as often as in the past because of electricwinches and/or cellular telephones to call foremergency assistance.) The trick is to attach asteel cable to a draw-bar point on the vehicle andstretch it to a distant very large tree. The otherend of the cable is then taken around the tree and,drawn tight, is wound around the tree several

times. A helper then pulls on this cable so that itdoesn't slip on the tree.

You then go to the middle of this cable andpull it sideways.

25METERS

2 METERS

25METERS

As an example, suppose the cable is50 meters long and you pull it sideways with a

force of 50 kg that actually pulls the rope adistance of 2 meters sideways. The force vectordiagram for this is shown below.

= 50 kg

. . = .-

REACTIONFORCEOFVEHICLE=T

REACTIONFORCEOFT

REE

=T

Here the transverse force you have appliedcreates forces in the cable that balance your forcewhen they are working in the directions shown.The components of these forces that are in thedirection from the vehicle to the tree actuallybalance each other. It is the perpendicular

components in the direction opposite to yourforce that work together to balance your force.Since the forces are symmetric they are equal inmagnitude. Letting this magnitude be T, thecomponents that balance your force are each

T Ty = sin .4 57o

Since there are two of them, the force beingexerted by the cable on your hands is

2 4 57 50T kgsin . o =

Solving this equations for T gives

T kgsin .

=

=

50

2 4 57313

o

Thus your 50 kg force has created a 313 kgreaction force in the cable. This is over a quarterof a Tonne, which is usually sufficient to movethe vehicle. If it is not then you could ask a helperto add to your force so as to get over half aTonne.

It is interesting to calculate how far your force

would have moved the vehicle. The distance ofthe vehicle before you pulled on the cable was 50meters. The distance is now

2 25 4 59 49 84 =cos . .o m

So you have ended up only moving thevehicle by 16 cm. However that is usuallyenough to backfill stones against the tires toprevent the vehicle from rolling backwards. Yourthen release your pull and your helper takes upthe slack in the cable by pulling on the end of the

cable wrapped around the tree. The cable is thenagain pulled sideways at its center and theexercise repeated until the vehicle has been pulledout of difficulty.

There are at least three distinct points of viewfrom which to consider this operation. First,consider it from the point of view of the vehicle(assuming vehicles have points of view, and fromthe behaviour of SUVs it certainly seems that


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they have). The forces on it are the force of thecable, being resisted, unsuccessfully you hope, bythe force of the mud.

313 kgCABLE

MUD

313 kg

Then there is the point of view of the tree.This is experiencing the pull of the cable towardthe vehicle, which is resisted, successfully youhope, by the force of its roots. It is also, ofcourse, resisting the pull of your helper so that itsroots have to resist both pulls but we will assumethat it is even strong enough for that.

313kg

??BUT PROBABLY ABUT 75 kg

ABOUT

390kg

CABLE

HELPER

ROOTS

TREE

Finally there is your point of view, which hasalready been illustrated in the diagrams above.

The use of such diagrams, called "free-bodydiagrams", is essential in understanding theinterplay of forces in systems. I will follow withsome important examples of their application.

The Balancing of Forces on a Rope

Perhaps the easiest free-body diagram to

understand is that of a single piece of rope. Herethe forces are simply those at the ends of therope, each force directly counterbalancing theother. These forces are equal in magnitude butopposite in direction.

T T

In this case the magnitude of the force iscalled the "tension" in the rope. (That is why Ihave labeled it T.) The magnitude of this forcedoes not depend on the length of the rope. Forexample, you could cut the rope in two and have

the free body diagrams for the two halves

T TROPE

TROPE ROPE

TT

Likewise you could tie another piece of ropeto the one you already have and extend the actionof the tension the new end.

T TROPE ROPE

Thus the rope acts as a transmitter of theforce. In the case of the vehicle being pulled out

of the mud, the cable acts as a transmitter of thepulling force of the tree roots to the vehicle.

This action of a rope is one of the mostimportant examples in mechanics. It has beenused from antiquity to transmit forces overdistances. It is also extremely important in thehuman body where the force of muscles istransmitted to bones by the tension in thetendons.

T

MUSCLE

PULL OFTENDON

Compression can also be used to transmit aforce over a distance. Examples of this are theuse of a rod to push out a blockage in a sewerpipe or the hose of your household vacuumcleaner. Another example is the use of a poker topush a burning log in a fireplace.

LOGC

MUSCLE PUSH OF

POKER

Here the free-body diagram of the transmitterhas inward pointing forces on its ends.

C CPOKER

However, the forces that can be transmitted inthis fashion are usually much less than those thatcan be transmitted by tension. This is because,while tension keeps the transmitter straight,compression tends to cause it to buckle.

Once buckled, the device can support very littleforce, a flexible object like a rope being able tosupport virtually no compresssion whatsoever.(This is the basis of the cruel statement I firstheard as an engineering student to express thecontempt we had for what were then called"commerce" students but who are now moreeuphemistically referred to as "management"


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students: "They are so dumb they don't knowthat you can't push on a rope".)

The Balancing of Forces on a Pulley

One of the most important uses of tension asa force transmitter is in the pulley. While thisdevice is of no direct interest in the life sciences,since the pulley is not found in nature but, like thewheel, is completely a man-made artifact, itshould be of interest because of its importance inthe history of human development. (Evenmedical school admission committees look withfavour on any indication that a candidate is not acomplete medical nerd.) Besides it is a very clearexample of the use of free-body diagrams tounderstand what is going on in a system, andsuch diagrams are veryimportant inunderstanding the internal forces within thehuman body.

The simplest and earliest application of thepulley was as a simple lifting device. It seems tohave been invented shortly after the wheel, i.e. inprehistoric times. Many historians claim it to bethe first true machine invented by man. Aschematic of it is shown below.

BEAM

AXLEPULLEY

W

This device used the ability of a rope totransmit a tension force, in this case over andaround the pulley. In this way the upward forcerequired to lift a weight was converted into adownward force. The importance of this lay inthe easing of the forces involved in such lifting.When you lift an object in the ordinary fashion,the weight of the object being lifted adds to yourweight on your knees. Thus a person with an

upper body weight of 70 kg lifting a 50 kg weightwill experience 120 kg of force on his or herknees.

This can cause considerable strain on theknees and, as will be considered later, on themajor back muscles.

On the other hand, lifting the object bypulling downward on a rope lightens the load onthe knees, in the above example to 20 kg. Thiscan actually be beneficial to your knees and back.Of course, this method does not allow you to liftmore than your weight. But then in general youshould not be lifting more than your weight,particularly if you are to repeat this often in aday's work.

I said that the pulley is a good example forthe use of free-body diagrams. For the singlepulley the use is almost trivial, as in the diagrambelow where the body under consideration is the

pulley itself.

W W

2W

This diagram shows that there are twodownward forces on the pulley, each equal to thetension in the rope. This tension is, of course, thedownward pull you have applied which is, in turn,just the weight of the body. In the diagram this isindicated as W. These two applications of theweight W are balanced by the one upward forcecoming from the rope suspending the pulley from

the ceiling by its axle. The tension in this rope isthen twice the weight that is being lifted, animportant point in designing the beam supportingthe pulley.

Very shortly after the invention of the simplepulley, combinations of pulleys were used toenhance the lifting effect. Certainly these deviceswere well known to the ancient Greeks; the theoryof pulley systems having been thoroughly


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worked out by Archimedes. An example is thetwo-pulley system shown below.

BEAM

W

F = ?

Here the free-body diagrams for the pulleysare still relatively simple. The diagram for thelower pulley, which is lifting the weight, is shown

below.

It shows that now the tension in the rope (i.ethe force with which you have to pull to balancethe weight), is only half the weight. This isbecause there are two tensions acting on thepulley to balance the weight.

The diagram for the upper pulley is shownbelow.

T T

2T

This shows that the rope supporting thispulley, as in the case of the single pulley, must

have a tension that is twice that of your pull.However, your pull is now only half the weight sothis support cable will only have to withstand theweight itself.

Finally consider the support beam.

BEAM

2T

T

It shows that the roof support must nowprovide three times the force you have applied.This comes about again because of the necessityof balancing both your force and weight being

lifted.The free-body diagram technique can beapplied in any combinations of pulleys. The finalexample that I will present is shown below forwhat is called a "three-pulley block" system.Normally such combinations of pulleys have allthree pulleys of the same diameter and on thesame axle but they are drawn here of different


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diameters and on different axis to make the actionclear.

W

T

BEAM

The free-body diagram for the lifting pulley -block now shows that the force you have to apply,i.e. the tension in the rope, is only one-sixth ofthe weight. In other words, a 50 kg force will lifta 300 kg weight.

TT

T

T

T

T

W

The diagram for the upper pulley blockshows that the roof support must now supplyseven times your force, or again, your force plusthe weight of the object.

TT

T

T

T

T

T

7T

The Balancing of Pressures

As I have already said, in the real world thebalancing of forces always come down to thebalancing of pressures. This is because realobjects have finite size and the force is spreadover the area of contact. Since they share this areaof contact, the two bodies will have the samecompressive pressure.

1

2

C

C

AREA

However, there is another whole class ofinteractions involving liquids and gases where thebalancing of pressure is more subtle. Again I willtry to gradually introduce the subtleties of this


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sort of interaction in the hope they can then bemade more understandable.

First consider the application of a force to apiston in a chamber containing liquid.

F

AREA

F

REACTIONFORCE

The reaction pressure of the liquid due to thisforce is easy to calculate. This pressure suppliesthe reaction force required to balance the force

applied to the piston. The force supplied by thepressure is just the pressure times the area of thepiston. The pressure is therefore the forceapplied divided by the area.

F pA

p F

A

=

=

For example, if that force was 10 kg and thepiston had a diameter of 1 cm, i.e. an area of 0.78cm2, the pressure would be 12.7 kg/cm2. This isjust the same as the pressure of a solid object in

contact with another solid object.Now consider the first subtlety for liquids.Consider the volume of liquid just below thepiston and draw a free-body diagram of theforces on that volume.

F

REACTIONFORCE

FORCE OFPISTON

SECTION OF LIQUIDUNDER CONSIDERATION

The piston applies a downward force on theliquid that, for the liquid to be stationary, must beresisted by an upward force on the liquid. Thismust be provided by the liquid just beneath the

section of liquid shown in the diagram. Tobalance, this pressure must be the same as thepressure downward from the piston. Therefore,just as the tension force is transmitted along arope, the pressure is transmitted along the tube ofliquid.

However, now consider the very subtle pointconcerning what happens if the tube expands to alarger diameter. (This point is so subtle that itwas not fully understood until it was studied bythe scientist Pascal in the 17thcentury in Europe.)As a specific example, suppose the tube expandsfrom a diameter of 1 cm to a diameter of 10 cm.Taking the same force of 10 kg as in the previouscase, resulting in a pressure of 12.7 kg/cm2, weget the situation shown below.

10 kg

DIAMETER= 1 CM

PRESSURE = ?

DIAMETER= 10 CM

PRESSURE = 12.7 kg/cm2

The important question is; what is thepressure of the liquid in this larger tube?

In the seemingly similar case of a rigid objectof this shape the calculation is simple. Theupward force must balance the downward and sothe upward pressure is reduced because of thelarger area over which it acts.

g

= 1 CM

PRESSURE = 12.7/100 = 0.127 kg/cm

= 10 CM

PRESSURE = 12.7 kg/cm

g


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F F

p A p A

p p A

A

up down

up up down down

up downdown

up

=

=

=

For the example shown in the diagram, if theupper surface had a diameter of 1 cm at apressure of 12.7 kg/cm2and the bottom had adiameter of 10 cm the pressure on the bottom ofthe object would be only 0.127 kg/cm2.

However, the case for the liquid is not sosimple. Unlike solids, liquids can flow. If putunder pressure they will flow to press against anysurface that restrains them, such as the walls oftheir containers. It is therefore the pressure thatis transmitted by a liquid, not the force,

Thus while the effect is subtle the result is

very simple. The pressure on the bottom of thevolume of liquid under consideration here is thesame as the pressure on the top!

This has enormous ramifications in liquidsystems, ranging from machinery for supplyingvery large forces to very heavy objects (where thesubject is called "hydraulics") to the heartpumping blood through your arteries to supplyyour brain. Because the machinery aspect issimpler I will consider that first.

Returning to the above example, the pressurein the liquid produced as a reaction to the appliedforce of 10 kg on the 1-cm diameter piston wouldalso act on a 10 cm diameter piston in the bottomtube.

10 kg

DIAMETER= 1 CM

DIAMETER= 10 CM



FORCE = 1000 kg

The result would be a force in proportion tothe area of the piston. In other words, the forceon this piston would be 1000 kg, or one Tonne!Since the liquid presses against any surface, nomatter how it is oriented, the geometry can berearranged so that the larger piston also facesdownward. The force of the liquid is then upward,

allowing it to support a weight of 1 Tonne. Thusa 10-kg force could raise a mass of 1 Tonne.

10 kg


FORCE = 1000 kg

1 CM 10 CM

This ability of pressure to produce very largeforces is very important in the construction ofautomobile tires, where a relatively modestpressure of 10 kg/cm2in four tires can supportover a Tonne of machinery hurtling along ahighway at 160 km/hr (at least in Germany where

such speeds are legal). It is even important in theconstruction of bicycle tires, where a pressure ofabout 20 kg/cm2can support the weight of a100 kg rider. In the life sciences it is extremelyimportant because of the forces produced on theheart walls by the very modest blood pressure(about 0.15 kg/cm2) required to push bloodthrough the arteries.

Pressure Due to Depth in a Liquid

So far I have considered the pressure in aliquid from a force applied to a piston. However,there is another very important source of pressurein a liquid; that due to its own weight. Thispressure is well known to deep-sea divers.

That this pressure is due to the weight of thewater can be seen from the free-body diagram ofthe water above you. For example, consider thefree-body diagram of a column of water onecentimeter square and extending from your ear tothe ocean surface. Suppose also that your ear is30 meters below the surface.

30 m

1 cm SQUARE

EAR


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The weight of this column of water, assumingthe density to be same as ordinary water, which isgood enough for this type of calculation, wouldbe 2.36 kg. The force on the bottom of thiscolumn to support this weight would thereforealso have to be 2.36 kg. The area of the bottom of

this column is /4 cm2

. The pressure at thebottom is therefore 3 kg/ cm2.

If the water is pressing against your ear, thenyour ear must provide a pressure of 3 kg/cm2tokeep it out. This is the source of the pain if youdive suddenly into deep water. You can onlysafely take deep dives if you allow the pressureon your ear drums to equalize by allowing high-pressure air to seep through the canal leading tothe interior behind the eardrum so that the ear-drum itself is not required to provide all thenecessary resistance to the water pressure.

From these numbers you can see that

3 kg/cm2at 30 m of depth corresponds to0.1kg/cm2per meter of depth. If you convert thepressure to kg/m2, by simply multiplying by10000, you get that the pressure is 1000 kg/m2

per meter, or 1000 kg/m3. This is, of course, justthe density of water. Thus pressure rises inproportion to the depth, the proportionalityconstant being just the water density. In equationform

p hkg m/ 2

=

where is the internationally approved symbol

for density and his the depth of water.This variation of pressure with height ofliquid that is supported is very important in manymammals, particularly those that have their headsfar above their feet. In humans it results in ablood pressure that is significantly higher at thefeet than at the brain, where the blood is mostneeded. This is why you often feel momentarilydizzy when you get up suddenly, before the hearthas livened up to provide the increased pressureneeded to get blood to the brain when you arestanding. It is also why putting up your feet asoften as you can is very beneficial to blood

circulation in your legs. (Not as much pressureis then needed to force the blood out of your feetand back to your heart.) As an example withnumbers, a person whose brain is 60 cm abovehis heart will have a blood pressure at his brainthat is 0.06 kg/cm2less than at his heart.

This problem is very severe for the giraffe,where the head can be several meters above theheart and a special blood circulation system is

needed to get blood up to its brain. This is alsoprobably one of reasons why the brontosaurusdinosaurs had such small brains. It was just toomuch effort to pump the blood to the brain whenthey reached to eat high leaves.

Buoyancy

A very important consequence of the variationof pressure with depth is the buoyancy of objectsin water. To make the arithmetic simple, considera rectangular block that is floating.

The water underneath this object must besupplying the force needed to prevent the objectfrom sinking any farther. Since water can onlysupply force as a pressure it is then, strictlyspeaking, the pressure of the water that is holdingup the block.

For a rectangular block floating with its edgesvertical, the pressure on the sides produces onlyhorizontal forces that have no components that

can resist the weight of the block. The weightmust therefore be resisted by the pressure on thebottom of the block.

The force produced by this pressure is thepressure times the bottom area of the block. Thispressure (in kg/m2) is the density of water timesthe height of the surface above the bottom. Theforce due to this pressure is therefore the densityof the water times the depth of the bottom of theblock times the area of the bottom of the block.This is just the density times the volume of water


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that the block displaces. Thus the block isbuoyed up by a force equal to the weight of thewater it displaces.

WATER

DISPLACED

FORCE = WEIGHT OFWATER DISPLACED= WEIGHT OF OBJECT

This is the famous theorem of Archimedeswhich, according to legend, he discovered while

taking a bath and which caused him to leap out ofhis bath and run naked down the street shoutingEureka!

The reverse way of looking at this is that afloating object displaces an amount of water equalto its own weight. This is often demonstrated inhigh-school science courses by placing a floatingobject in a container for which any rise in thewater level results it water spilling out intoanother container. The water collected in theother container can then be compared with theweight of the object. For example, floating a 200gm object would result in 200 ml of water being

spilled into the collection container.

WATERDISPLACED

WEIGHT =WEIGHT OFBLOCK

What is not so obvious to most people is thateven an object that sinks will have a buoyant forceacting on it. Consider an object with the samerectangular shape as before but too heavy to float.Since the object sinks, the water pressure on itsbottom does not supply sufficient force to holdthe object in place. Some other force is needed,such as that from a rope tied to the object.

PRESSURE

T

W

But what is the size of the extra force that isneeded? Most people think that it is the same isif the water was not there. However, you caneasily prove this to be wrong. An easy

demonstration of this is to tie a rope to a concreteblock, which weighs about 10 kg, and lower itinto a garbage can almost filled with water. (Thehardest part of this demonstration is getting rid ofthe water afterwards, a problem that has oftencaused problems between physics teachers andschool custodians.)

You will find that lowering the block into thewater makes it seem surprisingly lighter. In fact,a typical 10 kg concrete block will weigh onlyabout 5.5 kg under water.

10 kg

. g

W

This effect is explained by the difference inpressure at the top of the block compared to the

bottom. This difference is just due to thedifference in the height of the top and the bottomof the block. This difference is, of course, justthe height of the block and the pressuredifference is this height times the density of thewater. Multiplying this pressure difference bythe area over which the pressures acts gives thedensity of the water times the volume of theblock. Thus the block is still buoyed up by the


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weight of water the block displaces. The onlydifference compared to the floating block is thatnow the block displaces an amount of water equalto its volume, not to its weight. The blockactually sinks because, due to its density beinghigher than that of water, this amount of water

does not weigh as much as the block itself. Inother words, the buoyant force is insufficient tosupport the block and so an extra support forcemust be supplied to make up the difference.

Pressure Due to The Atmosphere

The concept of a pressure change due todifferences in elevation is very important inatmospheric phenomena. In modern society it iswell known that changes in altitude result inchanges in atmospheric pressure, the higher yougo the lower the pressure. It is often experiencedas a feeling in your ears during take-off and

landing in modern air travel. (It is not nearly aspainful as it used to be before they hadpressurized cabins to counteract the effect.) It isalso well known in the difficulty of simplyboiling an egg in places like Denver, or as high-altitude sickness during strenuous athletics inplaces like Mexico City. These are due to thethinness of the atmosphere resulting from thereduced pressures at these elevations. Thislowers the temperature at which water boils andreduces the transfer of oxygen from the lungs tothe blood.

What is not so well know is exactly how

much the pressure changes with height and whatcauses it. From the consideration of the pressurechange with depth in water it must be just theresult of the weight of air between the twoelevations.

Most people do not even realize that air hasweight. However, at sea level it is about1.3 kg/m3. Perhaps an easy way for you tovisualize this is to consider having an extremelylow temperature refrigerator that can liquefy air.(They do exist and are widely used in medicine toproduce liquid oxygen and nitrogen.) If you thenfilled a balloon with one cubic meter of air (to adiameter of about 1.25 m) and fed all this air intothe refrigerator you would end up with 1.3 kg ofliquid.

Thus the pressure drop from sea-level to 1meter above sea-level is 1.3 kg/m2, the pressuredrop at 2 meters would be 2.6 kg/m2, etc. (Toobtain these pressures in the more usual gm/cm2

just divide by 10. The pressures then become0.13 gm/cm2and 0.26 gm/cm2respectively.)

However, unlike water, as you rise in air itgets thinner. It therefore weighs less. Thepressure drop per meter is therefore less. Inother words, unlike water, where the pressuredrops smoothly to exactly zero at the surface, thepressure of air fades away as you go into outer

space. There is no sharp point at which itbecomes zero. There is no "edge" to theatmosphere.

Yet in deep space there is no air and so itspressure must be zero. What then is the changein pressure from the Earth's surface to outerspace. This must be just the pressure of air at theEarth's surface.

But what is this?For a long time humans did not even realize

there was such a pressure. This is because airpervades everything on the Earth's surface andtherefore equalizes its pressure against all

surfaces. For example, the pressure of the air onthe outside of your eardrums is balanced by thepressure of the air on the insides of youreardrums and there is no net force on youreardrums.

It is only when the air on one side of asurface is removed, such as when you pump airout of a container, that the pressure of air can beseen. And it is enormous! This was firstdemonstrated in the Middle Ages in Europe whenthe air was pumped out of a chamber formed bytwo hemispheres of about 60 cm in diameter.The force of the atmospheric pressure on thesehemispheres was so great that two very stronghorses could not pull them apart. A more moderndemonstration of such forces, which is oftenperformed in high-school physics courses byremoving the air from an ordinary container, isillustrated by a photograph in Hecht.

Again, careful physics measurements, inwhich all the air has been removed from achamber and the resulting forces accuratelymeasured, has shown normal atmosphericpressure at sea-level to be about 10 Tonnes persquare meter! (That is about 1 kg/cm2.)

As a concluding thought, this measurement ofthe atmospheric pressure at sea-level actuallydetermined the total mass of the Earth'satmosphere. Since the pressure on the surface isdue to the weight of all the atmosphere directlyabove that surface, then the total weight is theatmospheric pressure times the surface area of theEarth. The surface area was known in the 16thcentury to be about 5 1014m2. The total mass


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of the atmosphere is therefore about 5 1015

Tonnes, or about 5 1018kg.In many operations on Earth, such as

determining the pressure underwater, it is thedeviation of a pressure from normal atmosphericpressure that is important. This is obtained by

simply subtracting 1 kg/cm2

from the actualpressure. For example, when you pump air intoan automobile tire the pressure is actually1 kg/cm2when the tire is flat. What you thenwant to know is how much the pressure hasactually increased by pumping in the air. Thuswhen the pressure is actually 4 kg/cm2, thepressure gauge on your filling hose will read3 kg/cm2. To distinguish this pressure from the"Absolute Pressure", i.e. 4 kg/cm2, this pressureis often referred to as the "Gauge Pressure".(See Hecht.)

The Newton and The PascalBy now I can almost feel the wrath of hordesof physics teachers who are loudly protesting thatforce and pressure are not measured in kg andkg/cm2. My response to this is that 99% of thehuman race, who even understand what force andpressure are all about, do measure them and talkabout them this way. (Except, again, for thosepoor souls who have to contend with US units.) Itis only the physicists, and a few chemists andother physical scientists, who do not use theseunits. Certainly most of the engineers of theworld use them.

Why then do the physicists make such a bigfuss about them?

The problem arises for objects where theforce is the weight of a mass. This results in adouble meaning for the word "kilogram", often inthe same sentence. Thus the weight of a onekilogram object is one kilogram, a statement thatseems utterly redundant until one realizes theword kilogram in this sentence has referred toboth the amount of matter (i.e. its mass) and theforce on it.

To remove this confusion, which can be veryconfounding when it comes to the motion ofmasses (see lecture 21) physicists use,exclusively, Newtons (abbreviated N) to specifyforces. The reason for this will be a subject forLecture 21 but for now you only need to knowthat it is a force of about 100 grams, or one-tenthof a kilogram. More accurately, it is the weight of102 grams, so that one kilogram weighs 9.8Newtons.

You may have already suspected, even if youhaven't come across it before, that this is theacceleration of gravity on the surface of theEarth. The number 9.8 is so important inmechanics that, by International agreement, it isdesignated by the symbol g.

Thus a physicist will say that a 1 kg massweighs 9.8 N, thereby implying that the force ofgravity on the mass is 9.8 N.

Similarly for the units of pressure. Here theInternational unit for area is the square meter sothe International unit for pressure is the Newtonper square meter. This mouthful has beenredefined as the Pascal (abbreviated to Pa), inhonour of Pascal's studies of pressure. Aconversion, good enough for elementary physicscourses and other university work is just tomultiply kg/cm2by 100,000 to get Pascals. Thusnormal atmosphere is about 100 kPa.

In these SI units, the equation for the pressureversus height of liquid becomes

p gh=

and this is the way you will deal with it in the restof the course. In fact, to appease the souls ofdeceased physics teachers I promise not to useforce as a kilogram or pressure in kg/cm2 everagain (well almostnever again) in this course.

You might ask why did I use kg and kg/cm2

in the first place? I did so because I have found

that the difficult concepts of force and pressurebecome even more bewildering to beginningstudents if they are introduced using the totallyunfamiliar units of Newtons and Pascals.

Relevant material in Hecht

Section 4.2, solved problems 4.1 and 4.2, Section4.9 (Introduction to, but not including ParallelForce Systems, solved problems 4.11 and 4.12,Section 9.4, with solved problems 9.3 and 9.5,Section 9.5 with solved problem 9.6 and Section9.6 with solved problems 9.7 and 9.8

Solved Example Problems

1. The Figure shows two blocks, block 1 andblock 2, held in equilibrium by massless,unstretchable strings. The mass of block 1is 18.2 kg, the mass of block 2 is 3.8 kgand the angle is 25o. What are thetensions in the strings?


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The diagram, with numbers, is shown below.

The free body diagram of the point where thethree strings meet is

The tension T2 is just that resisting the weight ofthe 3.8 kg mass, i.e. 37.24 N.

The other forces can now be obtained bytaking the vector sum to be zero.

2. A dentist's chair of mass 233 kg issupported by a hydraulic lift having a largepiston of cross-sectional area 1410 squarecentimeters. The dentist has a foot pedalattached to a small piston of cross-sectional

area 74 square centimeters. What forcemust be applied to the small piston to raisethe chair?


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3. A cord runs around two pulleys as shownbelow. 1.67 kg bucket hangs from onepulley. If the mass of the pulleys and thefriction of the system are negligible, by howmuch must you pull on the rope if you areto lift the bucket?


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Force Pressure

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