Friday, November 6, 1998

Chapter 9: shear modulus pressure Pascal’s Principle Archimedes’ Principle

Let’s look at a side view of what happens toour copper cylinder under the influence ofshear stress.

Fs

Fsx

l

Shear stressesproduce

shear strains

“shear force”

shear stressF

As

where A is the area of thetop of the cylinder.

Fs

Fsx

l shear strainx

l

Note: the units of shearstress and shear strainare the same as ourtensile and compressiveforms of stress and strain.

The manner in which an object respondsto shear stresses is also a characteristicof the material from which it’s made.

Here, we define the ratio of the shear stressto the shear strain as the shear modulus:

Mssshear stress

shear strain

F A

x

/

/ l

This quantity has the same units asYoung’s modulus: N/m2.

A child slides across a floor in a pairof rubber-soled shoes. The frictionalforce acting on each foot is 20 N, thecross-sectional area of each foot is14 cm2, and the thickness of the solesis 5 mm. Find the horizontal displace-ment of the bottom of the soles relativeto the top of the soles. The shearmodulus for rubber is 3.0 X 106 N/m2.

Ff

xl Ms

sF A

x

/

/ l

A child slides across a floor in a pairof rubber-soled shoes. The frictionalforce acting on each foot is 20 N, thecross-sectional area of each foot is14 cm2, and the thickness of the solesis 5 mm. Find the horizontal displace-ment of the bottom of the soles relativeto the top of the soles. The shearmodulus for rubber is 3.0 X 106 N/m2.

Ff

xl

3 1020 14 10

106

4

3

N / m N m

m2

2/ ( )

/ (5 )x

A child slides across a floor in a pairof rubber-soled shoes. The frictionalforce acting on each foot is 20 N, thecross-sectional area of each foot is14 cm2, and the thickness of the solesis 5 mm. Find the horizontal displace-ment of the bottom of the soles relativeto the top of the soles. The shearmodulus for rubber is 3.0 X 106 N/m2.

Ff

xl

x 2 4 10 5. m

When a force is applied over an area, wesay that the object feels pressure.

Usually, we talk about pressure of a fluidor a gas (like the atmosphere).

PF

A

[ ][ ]

[ ]P

F

A

N

mPa

2

“pascal”

At the Earth’s surface, we sit at the bottomof an entire column of molecules in theatmosphere. These molecules exert apressure on us at the surface of about101.325 kPa. We define this pressure tobe 1 atmosphere (atm).

As we go up through the atmosphere,what happens to the pressure we feel?

WHY?

Let’s back track for just a minute.

First, we note that fluids are unable to supportshear stresses. That means that when weexert a shear stress on the surface of a pondof water, the surface water slips across thewater below it.

surface water

deep water

wind

Our green fish is completely submerged. Thewater exerts a pressure on the fish. From whichdirection does the fish feel the pressure?

It’s all around you!

In fact, the pressure is exertedin the direction normal to thebody of the fish all over the fish.

It’s all around you!

Think about your own experience walkingaround outside (when there’s NOT a hardwind blowing). You don’t notice any differencein pressure over the surface of your body.

What would happen to you if there was apressure difference across your hand?

P1

P2

P P1 2

P1

P2

P P1 2

Your hand would feel a net force actingto the RIGHT in this case. Therefore, yourhand would start to accelerate to the right,or you would have to exert a force throughyour arm to counteract this force knownas a “pressure gradient” force.

Fnet

So, now let’s look at how pressure changeswith altitude. We know it’s a lot harder tobreathe at the top of a mountain than atsea level--there are fewer oxygen moleculesand fewer molecules in general up there.

Let’s look at the forcebalance for a little layerof atmosphere as wehead up the mountain.

P1

P2

Fg

The pressure ofall the moleculesabove our layer.

The pressure fromall the moleculesbelow our layer.

The gravitationalforce acting on thelayer itself. P P1 2

A = surface area

P1

P2

Fg

A = surface area

The pressure P1 on the top

surface of area A results ina force downward of F1 = P1 A.

The pressure P2 on the bottom

surface of area A results ina force upward of F2 = P2 A.

P1

P2

Fg

A = surface area

If our system is inequilibrium, the netforce must be 0. So...

F F Fg2 1 P A P A mg2 1

But what is the mass ofour little layer of atmosphere?

h

m V A h ( ) is the density of air.

P A P A mg2 1

P A P A A h g2 1 ( )

P P g h2 1 ( )

P g h ( )

If our system is inequilibrium, the netforce must be 0. So...

NOTE: our derivation here assumes a uniformdensity of molecules at a given layer in theatmosphere. In the real atmosphere, densitydecreases with altitude. Nevertheless, ourpressure and force balance diagram appliesso long as our layer is sufficiently thin so thatwithin it, the density is approximately constant.

If the atmosphere is in equilibrium (whichwould imply a uniform temperature andno winds blowing), the pressure at a givenheight above the surface would be the samearound the Earth.

The same arguments can be made forpressure under water. All other things beingequal, the pressure at a given depth belowthe surface is the same.

A scuba diver explores a reef 10 mbelow the surface. Assuming thedensity of water is 1 g/cc, what isthe pressure the diver feels?

We know the pressure at the surface of thewater is 1 atm = 101.3 kPa. The change inpressure as the diver drops under a 10 mcolumn of water is given by

P g h ( ) ( )( . )( )10 9 8 103 kg / m m / s m3 2

P 9 8 10 9 8 104 4. . kg / m / s N / m2 2

P P P 0 1013 98 199 3 ( . ) . kPa kPa

In solving the last problem, we applied aprinciple that we haven’t even defined yet,but that probably made good sense to us.

We said that the surface pressure at thebottom of the atmosphere equaled thepressure in the surface layer of water.

If this weren’t true, the water would fly outof the oceans or sink rapidly toward theocean floor!

In fact, Pascal’s Principle guarantees this willbe true. It states:

The pressure applied to an enclosed liquid istransmitted undiminished to every point inthe fluid and to the walls of the container.

Which means, that the pressure below thesurface of the water is equal to the surfacepressure + the pressure due to the columnof water above a given level.

What happens to a cork when we try tosubmerge it in water?

It shoots right back up to the surface.

What is responsible for the motion ofthe cork?

There must be a force acting upward onthe cork greater in magnitude than gravity.

But what happens to the cork when itgets to the surface?

It floats.

So what must be the net force on thecork as it’s floating on the surface?

ZERO!

What’s changed?

What have we noticed about our strangeunderwater force on the cork?

It’s greater than gravity when the corkis completely submerged.

It’s equal to gravity when the cork floatson the surface, only partially submerged.

Our new force relates to the volume ofthe cork that’s underwater!

Archimedes had this whole process figuredout some 2000 years ago! He said,

A body wholly or partially submerged ina fluid is buoyed up by a force equal to theweight of the displaced fluid.

So, the cork naturally float with just theright portion of its volume under the water’ssurface so that the buoyant force upwardfrom the water equals the gravitational force.