Physics 111 - Valparaiso University 111 Lecture 11 • Mid-term survey results • Ch 5: Newton’s 3rd Law • Ch 6: Tension Examples Help this week: Wednesday, 8 - 9 pm in NSC 118/119

Physics 111

Title page

Thursday,September 30, 2004

Physics 111 Lecture 11

• Mid-term survey results

• Ch 5: Newton’s 3rd Law

• Ch 6: TensionExamples

Help this week:Wednesday, 8 - 9 pm in NSC 118/119

Sunday, 6:30 - 8 pm in CCLIR 468

Help sessions

AnnouncementsThursSept

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Don’t forget to read over the labwrite-up and be ready for the quiz.

labs


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Response rate: 23 out of 33

Several students provided little to nofeedback on the free-response page

I need to hear from you if you havesuggestions or comments!

Help sessions


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Most questions had mean of 3.0

Least favorite activities:PhysletsMastering Physics

Favorite activities:Interactive Exercises

Average time on class: ~7 hours/week

Help sessions


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What can you do?Lots of recommended problems…including Mastering PhysicsOne exam problem will be MPOne exam problem will be Walker

Help sessions


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What I will do in the future:

• Make sure we cover what you need beforeMastering Physics assignment is due.

• However, Physlets will cover material beforewe cover it in class.

We’ve looked at a lot of problems with but asingle “system” of interest.

Interacting Systems

Ch 5: Newton’s LawsThursSept

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Let’s now formalize the way to handle problemswith multiple systems.

In the process, we’ll encounter Newton’s 3rd Law.

Example of me pushing on the wall.

Just because neither I nor the wall seem to beaccelerating, however, does NOT mean thatthere are no forces acting. In fact, there areseveral relevant forces involved in this process.

I can huff and puff andpush on the wall, but itdoesn’t seem to beaccelerating. Andneither am I!!!

Big Bad Wolf


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Let’s just look at theforces acting on me.

A STATICSystem: all

the forces arein balance.

Nothingaccelerates!

All forces on hero - picture


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N = normal forceof wall onme

N = normalforce of flooron me

f = frictional forceof floor on me

W = force ofEarth’s gravityon me

Let’s just look at theforces acting on me.

friction

weight

Normal (floor)

All forces on hero -FBD

Normal (wall)GOODGOOD

free-bodyfree-bodydiagramdiagram


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Let’s now look at theforces acting on the wall.

A STATICSystem: all

the forces arein balance.

Nothingaccelerates!

All forces on wall - picture


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C = contact forceof wall onme

f = frictional forceof floor on wall

N = normalforce of flooron wall

W = force ofEarth’s gravityon wall

All forces on wall -FBD

friction of flooron wall

Contact Forceof the mepushingon the wall.

weightof wall

Normal forceof floor on wall

GOODGOODfree-bodyfree-bodydiagramdiagram


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Let’s now look atthe forces actingon the wall.

Let’s just look at theforces acting on the floor.

All forces on floor - picture


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N = normal forceof the Earth onthe floor

W = force ofEarth’s gravityon the floor

fmf = frictional forceof me on floor

Cwf = contactforce of wallon floor

Cmf = contact force of me onfloor

fwf = frictional forceof wall on floor

Let’s just look atthe forces acting

on the floor.

Contactof wall onthe floor

Friction (me)

Normal (Earth)

All forces on floor - FBD

Contactof me onthe floor

Friction (wall)

weightof floor



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Finally, let’s look at the forces actingon the Earth in this problem.

EARTH

All forces on Earth - picture


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C = Contactof floor onthe Earth

Fghero = GravitationalForce of Meon the Earth

Fgf = GravitationalForce of flooron the Earth

Fgw = GravitationalForce of wallon the Earth

Finally, let’s look at the forces actingon the Earth in this problem.

GravitationalForce of Meon the Earth

EARTH

Contactof floor onthe Earth

GravitationalForce of wallon the Earth

GravitationalForce of flooron the Earth

All forces on Earth - FBD


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If one object exerts a force on a second object,the second object necessarily exerts an equalbut oppositely directed force on the first.

We’re talking aboutTWO DIFFERENTFORCES HERE!!!

Law 3


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NormalForce

Force of thewall pushingon me.

ContactForce

Force of theme pushingon the wall.

Interaction Pair -- Newton’s Third Lawalways act on DIFFERENT objects

Super Hero: normal & contact

NOT a free-body diagram


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NormalForce

Force of thewall pushingon me.

ContactForce

Force of theme pushingon the wall.

Interaction Pair -- Newton’s Third Lawalways appear in DIFFERENT FBDs.

Super Hero: normal & contact

NOT a free-body diagram


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Several notes:

• The NORMAL force acts perpendicular tosome surface. It is NOT NECESSARILYequal to mg!

• Each and every object in a problem has itsown free body diagram! Draw each oneseparately.

• Third Law Force pairs act on DIFFERENTOBJECTS. They NEVER act on the same object!

Summary notes


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King Henry and the Earth bothpossess “gravitational massgravitational mass”and exert equal but oppositelydirected forces on one another.

F

Earth on Henry

F

Henry on Earth

W1: Henry & Earth

1)

2)

3)

F

Earth on Henry=F

Henry on Earth

F

Earth on Henry<F

Henry on Earth

Worksheet Problem #1


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F

Earth on Henry>F

Henry on Earth

I’m going to jump off a chair.

Demo: my gravity on Earth


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Watch as the Earth rushesup to meet me!

Do you want to seethat again?

What’s going on here?

I need two student volunteers

Demo: students and 3rd Law


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Exp. #1: Each of you take onescale. Push on each otherthrough the scales and call outthe readings.

Exp. #2: One of you sit in the rolling chair. Theother push through the scale. Both call outreadings on the scales.

Our avant-guarde socialite pulls on the ropethat’s wrapped around the tree. Nothing happens.

Tension

Ch 6: Applying Newton’s LawsTuesSept

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skip

What must be true about the forces acting...

here here

here

tension along rope


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Let’s examine this piece more carefully...

Tension Tension

The forces balance -- The rope does NOT accelerate.

piece of rope


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In fact, no matter which little segment of therope I examine in this case, the tension forcesbalance in either direction, and the rope remainsstationary.

Okay, let’s look at tension in a rope that resultsin the acceleration of an object...

balance


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Remember this one?

What exactly is it that causesthe green block to accelerate?

block & ice


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Force Meter

Frictionless pond of ice

Let’s look at the free-bodydiagram for the green block.What forces are acting onthe green block?

Weight Normalforce

Contact

The contact force of therope on the block resultsin the block accelerating.

block accelerates


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What is the acceleration ofthe green block?

Weight Normalforce

Contact

block accelerates


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a

block=F

contact

mblock

What if we look at a pieceof the rope in this case?

Some mass mr

a =F

net

mr

=T

2−T

1

mr

Some mass mbRemember, thewhole systemis accelerating

at the same rate, a.

a

Tension 1 Tension 2

rope accelerates?

Ch 6: Applying Newton’s LawsThursSept

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Some mass mb

And clearly the block’s acceleration will bedictated by the magnitude of the contact force at theend of the rope (which in magnitude is equal to thetension in the rope at that end of the rope)connected to the block...

a =F

net

mb

=F

contact

mb

=T

end

mb

Contact

block again


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Back to the ropefor a minute...

TensionrightTensionleft

T

left>T

right

unequal tensions?


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However, what happensif mr = 0?

F

net=T

2−T

1= m

r

a = 0

a = 0

T

2=T

1

Tension 1 Tension 2

Massless string


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T

left=T

right

equal tensions


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TensionrightTensionleft



CQ2 Tension


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CQ2 – Tension the ?

In the 17th century, Otto von Güricke, a physicist in Mag-deburg, fitted two hollow bronze hemispheres together andremoved the air from the resulting sphere with a pump.Two 8-horse teams could not pull the halves apart even though thehemispheres fell apart when air was readmitted. Suppose vonGüricke had tied both teams of horses to one side and bolted theother side to a heavy tree trunk. In this case, the tension on thehemispheres would be

PI, Mazur (1997)

≥≥

1. twice2. exactly the same as3. half what it was before.


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A worker drags a crate across a factory floor by pullingon a rope tied to the crate. The worker exerts a forceof 450 N on a rope that is inclined at 38o to thehorizontal, and the floor exerts a horizontal force of111 N that opposes the motion. (a) Calculate theacceleration of the crate if its mass is 310 kg.(b) Calculate the normal force of the floor on the crate.

W1: Worker & Crate

38o111 N

y

x

Problem Sheet #1


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a


W1: Worker & Crate

FBD


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Ffloor

n

θ

T

W


W1: Worker & Crate


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Knowns:θ= 380 , ay = 0Tx = (450 N) cos θ = 354.6 NTy = (450 N) sin θ = 277.0 NFfloor = -111 N xm = 310 kgg = 9.81 m/s2

Unknowns:ax, n

W1: Worker & Crate

(a)

Fx,net = Tx + Ffloor

Fx,net = 354.6 N – 111 N

Fx,net = m ax = (310 kg) ax

Fx,net = 254.6 N

ax= Fx,net / m

ax= 254.6 N/ 310 kg

ax= 0.821 m/s2


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W1: Worker & Crate

(b)

Fy,net = Ty + W + n = 0

0 = |T| sin 38o – mg + n

- n = 277 N – (310 kg) (9.81 m/s2)

Fy,net = m ay = (310 kg) (0 m/s2) = 0

n = 2760 N y



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A worker sits in a bosun’s chair that is supported by amassless rope that runs over a massless, frictionlesspulley and back down to the man’s hand. Thecombined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed? (b) How would theforce be different if it were exerted by a second manon the ground instead of the man in the chair?

This looks like a pretty complexproblem…And it can be tricky…so, let’s be careful and useNewton’s Laws explicitly foreach moving object.

? Worker, cart, pulley (W3)

Problem Sheet #2


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The combined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed?

First, what are the forces acting on the man?

Normal force ofchair on man.

n

Force ofgravityon man

Wman

Tension ofrope on chair. T

? Soln – F on man


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FBD: man


Next, what are the forces acting on the chair?

Force of gravityon chair

W

chair

Tension ofrope on chair. T

Contact force ofman on chair.

n '

? Soln – F on chair


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FBD: chair

A worker sits in a bosun’s chair that is supported by amassless rope that runs over a massless, frictionlesspulley and back down to the man’s hand. Thecombined mass of the man and the chair is 95.0 kg.(a) With what force must the man pull on the rope forhim to rise at a constant speed? (b) How would theforce be different if it were exerted by a second manon the ground instead of the man in the chair?

? Worker, cart, pulley (W3)


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Knowns:mman+chair = 95.0 kgg = 9.81m/s2

a = 0n = - n’

+y

Unknowns:

Tn


What’s Newton’s Law say about the forces on the man?

F

net= ma

But the acceleration of the man is ZERO!(i.e., constant speed means NO acceleration.)

F

net ,man=T + n −

W

man= ma = m(0) = 0

? Soln – 2N on man


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F

net ,chair=T − n '−

W

chair= ma = m(0) = 0

? Soln – 2N on chair


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But the acceleration of the chair is ZERO!(i.e., constant speed means NO acceleration.)

What’s Newton’s Law say about the forces on the chair?

F

net= ma


F

net ,cart=T − n '−

W

cart= 0

F

net ,man=T + n −

W

man= 0 Add these two

together.

2T + n − n '−

W

man−

Wcart

= 0

3rd Law PairSame magnitude

2T = m

man

g + m

cart

g

? Soln - combine


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T = (m

man+ m

cart)g / 2

T = =( . )( . ) /950 9 81 2 466kg Nm

s2

? Soln - tension


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2T = m

man

g + m

cart

g

(b) How would the force be different if it were exertedby a second man on the ground instead of the man inthe chair?

Now the man on the groundmust exert a force on the ropesuch that the tension in therope balances the force ofgravity on the man+chair.

T m m gman cart= + =( ) 932N

? Soln – ground


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Documents

Physics 111 - Valparaiso University 111 Lecture 11 • Mid-term survey results • Ch 5: Newton’s 3rd Law • Ch 6: Tension Examples Help this week: Wednesday, 8 - 9 pm in NSC 118/119