7.1 interacting systems p. 183-185 7.2 action/reaction ... · pdf file1...

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7.3 Newton’s Third Law p. 189-194

Chapter 7: Newton's 3rd law

7.4 Ropes and Pulleys p. 194-198

7.5 Interacting-system Problems p. 198-202

7.1 Interacting Systems p. 183-185

7.2 Action/reaction pairs p. 185-189

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7.1 Interacting Systems p. 183--- 7.2 Action/reaction pairs p. ---189

Chapter 7: Newton's 3rd lawPh

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One of the most important examples of (almost) circularmotion is the motion of planets around the sun

None of these orbits is really circular (especially Pluto's),but in most cases it is a very good approximation

Many celestial bodiesare in a kind of circularmotion:moons around planetsplanets around starsstars around the centre of the galaxy

To understand this motion we first have to introduce anew concept: action/reaction pairs of forces


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As we said to understand planets motion we first have tointroduce a new concept: action/reaction pairs of forces

Any object A that exerts a force on another object Bexperiences itself a force that is caused by object B

Example: if you are sitting on achair, you (A) feel the chairpressing against you, and thechair (B) may break if you are tooheavy

FB on A is here the normal force,

FA on B is related to the weight of A

A

B

FB on A

FA on B

→

→→

→


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Two objects with an action/reaction pairs of forces have amutual influence on each other, We say that the objectsinteract with each other.

Sometimes object B is much larger (usually heavier) thanobject A. Then A's influence on B can be neglected sincethe state of B is hardly changed by FA on B

Example: a train (B) moving at5 km/h can easily push you (A)aside without noticeable changein its velocity. But if you walkinto the train, it will (usually)hardly move.

→

http://www.ananova.com/news/story/sm_1417228.html


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In such a situation we simply can consider object A alone.Object B is then part of the environment whichexerts forces on A

This is what we implicitly did whendrawing free-body diagrams forsingle-particle dynamics

In general we both have forcesfrom the environment (like weight)and action/reaction pairs


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We will also encounter situationswhere more than two systems areinteracting with each other.

We then will consider action/reactionpairs for different pairs of systems(see figure)

Note that Newton's 2nd lawapplies separately to eachsystem:


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Identifying action/reaction pairs (Tactic box 7.1):

Draw each object separately Identify all forces and draw them on the object on which they act Identify action/reaction pairs: FA on B goes with FB on A Identify objects that are systems of interest (i.e., which are not part of the environment) Draw a free-body diagram for each object of interest

→ →


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Example: a person pushesa crate with(action) force FP on C .The reaction force is FC on P

The person uses surface fric-tion fS on P to be able to push.

There is a reaction force fP on S, but surface and Earth can beconsidered as part of the en-vironment

Other forces: weight, normal f.

→→

→

→


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Action/reaction pairs are also behindany mechanism for propulsion(friction, thrust)

Ropes can also be considered as asystem. The forces acting on a ropeare usually tension forces


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What, if anything, is wrong with the force diagram for abicycle that is accelerating toward the right?

1. It does not draw each objectseparately.

2. It draws interactions between onebody and itself.

3. It fails to identify all relevantforces.

4. Nothing is wrong.

Chapter 7: teasersPhy221/R. Ouyed

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What, if anything, is wrong with the force diagram for abicycle that is accelerating toward the right?

1. It does not draw each objectseparately.

2. It draws interactions betweenone body and itself.

3. It fails to identify all relevantforces.

4. Nothing is wrong.


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What, if anything, is wrong with this forcediagram for a bicycle that is accelerating to theright?

A. Nothing is wrong.

B. One or more forces have the wrong length.

C. One of more forces have the wrong direction.

D. One or more action/reaction pairs are wrong.

E. Both C and D.


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What, if anything, is wrong with this forcediagram for a bicycle that is accelerating to theright?

A. Nothing is wrong.

B. One or more forces have the wrong length.

C. One or more forces have the wrong direction.

D. One or more action/reaction pairs are wrong.

E. Both C and D.

X

X


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End of Week 7


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7.3 Newton’s Third Law p. 189-194

Newton’s third law (or action/reaction) recognizes force as an

INTERACTION BETWEEN OBJECTS rather than as some “THING”

with an independent existence of its own!


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We have seen that the two forces of the gravitationalaction/reaction pair between two objects are equal inmagnitude but have opposite direction

This is true for any force and constitutes an example ofNewton's 3rd law:

Every force occurs as one member of an action/reactionpair of forces

The two forces in an action/reaction pair act on different objects The two forces are equal in magnitude but opposite in direction: FA on B = - FB on A

→ →


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It is important to note that Newton's 3rd law states theequality of the forces, not the acceleration of the twoobjects

The example of the satellite and the planet shows thateven if FS on P = FP on S the associated accelerations canbe very different in magnitude:

Hence aS = (mP/mS) aP >> aP for mP >> mS. Note thatthis is true independent of the type of force


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Car B is stopped for a red light. Car A, which has the samemass as car B, doesn’t see the red light and runs into theback of B. Which of the following statements is true?

1. B exerts a force on A but A doesn’t exert a force on B.2. B exerts a larger force on A than A exerts on B.3. B exerts the same amount of force on A as A exerts on B.4. A exerts a larger force on B than B exerts on A.5. A exerts a force on B but B doesn’t exert a force on A.

Cliction 7.1


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Sometimes two objects A and Bare connected and move together

In this case their accelerations arethe same, but not their net forces:

Acceleration Constraint


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Here is another example of anacceleration constraint.

In general, the accelerations neednot have equal direction. In thisexample, aAx = - aBy

Action/reaction pairs

x

y


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Boxes A and B are sliding to the right across a frictionlesstable. The hand H is slowing them down. The mass of Ais larger than the mass of B. Rank in order, from largest tosmallest, the horizontal forces on A, B, and H.

1. FB on H = FH on B = FA on B = FB on A 2. FB on H = FH on B < FA on B = FB on A3. FB on H = FH on B < FA on B = FB on A4. FH on B = FH on A > FA on B

Cliction 7.2


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7.4 Ropes and Pulleys p. 194-198


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Let's also take another look at strings with tension.

In equilibrium we would have TA on S = TB on S .However, if the string is accelerated this would not bethe case:

However, the string is often so light that this differencedoes not matter. In the future we will make the masslessstring approximation mS =0 so that TA on S = TB on S


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In the example below:the climber is not in contact with the knot. So, strictlyspeaking forces T and T are not an action/reactionpair! BUT if the ropes are massless then T and T actas if they are an action/reaction pair.

C on K

C on K

K on C

K on C


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In Phys 221 we will also makethe approximation that pulleysare massless and have no friction

Important principle: the magnitudeof the tension in a massless stringremains constant as it passes overa massless, frictionless pulley


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1. greater than2. less than3. equal to

All three 50 kg blocks are at rest. Is the tension in rope 2greater than, less than or equal to the tension in rope 1?

Cliction 7.3


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NOTE 1: the force on the axle of the pulley is shared equally by the two lines looping through the pulley.

NOTE 2: a single movable pulley allows a unit weight to be lifted with only half the force needed to lift the weight without assistance.

The weight lifted divided by the lifting force is

defined as the advantage

of the pulley system.


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In diagram 3, the addition of a fixed pulley yields a lifting advantage of 3. The tension in each line is ⅓ the unit weight, and the force on the axles of each pulley is ⅔ of a unit weight. As in the case of diagram 2a, another pulley may be added to reverse the direction of the lifting force, but with no increase in advantage. This situation is shown in diagram 3a.


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7.5 Interacting-system Problems p. 198-202


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Pushing a PackageProblem 1 :

The boy needs to push the packagewith an acceleration of at least 1 m/s in order for the package to make it to the top.

Can he give the package a big enoughshove/push to reach the top of the ramp?

i) The coefficient of static friction of theground is 0.25ii) The coefficient of kinetic friction of the ramp is 0.40.iii) The boy is in static equilibrium.

This is the ONLY

action/reaction pair in this problem.


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Pushing a PackageProblem 1 :

x-axis f - F cos(30) = 0y-axis n - m g - F sin(30) = 0S

B BPonB

PonB

x-axis -f + F - m gsin(30) = m ay-axis n - m g cos(30) = 0

k

P P

BonP P P x

BOY

PACKAGE

NEWTON’s third law F = F = FBonP PonB

From the package equations one can show that F = 139 N

From the boy’s x-equation we can calculate howmuch static friction he needs to push the box f = F cos(30) = 120 N

S

The maximum static friction is f = 0.25n = 115 NS,max B

STEP 1

STEP 2

STEP 3

CONCLUSION: He CANNOT shove hard enough without slipping!


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Cliction 7.4A small car is pushing alarger truck that has adead battery. The massof the truck is larger thanthe mass of the car. Which of the following statements is true?

1. The car exerts a force on the truck but the truck doesn’texert a force on the car.

2. The car exerts a larger force on the truck than the truckexerts on the car.

3. The car exerts the same amount of force on the truck asthe truck exerts on the car.

4. The truck exerts a larger force on the car than the carexerts on the truck.

5. The truck exerts a force on the car but the car doesn’texert a force on the truck.


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What is the acceleration of the man M hanging at the rope?

Here we have an acceleration constraint aSy = - aMy

Fceil→

Problem 2


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Fceil→

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Draw the free-body diagrams ofthe three systems (man, pulley,set):

M

wM

T1

→

→

P

Fceil

-T2

→

→

S

wS

T2

→

→

-T1

→

Because the tension is the same on both sides of thepulley we have T1 = T2

In principle we also need to consider the rope segmentsas objects, but since they are massless this is trivial

Fnet,M→ Fnet,S

→


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Fceil→

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Newton's 2nd law for M, P, S: weneed only to consider the y-com-ponent of each equation:

M

wM

T1

→

→

P

Fceil

-T2→

→

S

wS

T2

→

→

-T1→

Fnet,M→ Fnet,S

→

Tension remains thesame over pulley ⇒

T1y = T2y ⇒(Fceil)y = 2 T1y


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Fceil→

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Insert acceleration constraint:


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Let's a bit elaborate on pulleys. In combination they can create an ancient but still very useful machine: a tackle

The principle of a tackle isbased on multiple applicati-on of our previous example

http://de.wikipedia.org/wiki/Flaschenzug http://en.wikipedia.org/wiki/Pulley

Problem 3


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Let's analyze a compound pulley.Pictorial representation:

P1

P2

WFpull→

Fceil,1→

Fceil,2→

w→

Fpull→


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Free-body diagrams: we don't careabout the ceiling forces. In equilibrium we then have

P1

P2

W

Fpull→

Fceil,1→

Fceil,2→

w→

T3→

T2→

T1→

All tensions with a minus sign are partof a respective action/reaction pair:T2 between P1 and P2, andT3 between P2 and W

w→

T3→

Weight W

-T3→

T2→ T1

→Pulley P2

-T2→

Pulley P1

Fceil,1→

Fpull→


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Newton's 2nd law: Fnet = 0 for allthree systems ⇒

w→

T3→

Weight W

-T3→

T2→ T1

→Pulley P2

-T2→

Pulley P1

Fceil,1→

Pulley principle: same tension on both sides ⇒

Fpull→

→

Eq.(6) is only useful for findingFceil,1→


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(5) follows from inserting(3) and (1) into (4)

(5) allows to calculateT2y = mW g/2

from (2) we then find thepulling force

(Fpull)y = - mW g/2

We therefore need only half the weight force tokeep mW in the air


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Chapter 7Reading Quiz


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The propulsion force on a car is due to

1. static friction.2. kinetic friction. 3. the car engine.4. elastic energy.


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Is the tension in rope 2 greater than, less than, or equal tothe tension in rope 1?

1. greater than rope 22. less than than rope 23. equal to rope 2


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An acceleration constraint says that in some circumstances

1. the acceleration of an object has to be positive.2. two objects have to accelerate in the same direction.3. the magnitude of the accelerations of two objects have

to be equal.4. an object is prevented from accelerating.5. Acceleration constraints were not discussed in this chapter.


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Selected Problems

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End of Chapter 7

IMPORTANT:

Print a copy of the SUMMARY page (p. 203)and add it here to your lecture notes.

It will save you crucial time when trying to recall:Concepts, Symbols, and Strategies


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7.1 interacting systems p. 183-185 7.2 action/reaction ... · pdf file1...

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