chapter 4 dynamics: newton’s laws of motion · title: microsoft powerpoint - chapter 4.ppt...

11/03/2012

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Lecture PowerPoints

Chapter 4

Physics: for Scientists &Engineers, with Modern

Physics, 4th edition

Giancoli


Chapter 4

Dynamics: Newton’s Lawsof Motion

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Units of Chapter 4

• Force

• Newton’s First Law of Motion

• Mass

• Newton’s Second Law of Motion

• Newton’s Third Law of Motion

• Weight—the Force of Gravity; and theNormal Force

• Solving Problems with Newton’s Laws:Free-Body Diagrams

• Problem Solving—A General Approach


4-1 Force

A force is a push or pull. An objectat rest needs a force to get itmoving; a moving object needs aforce to change its velocity.

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4-1 Force

Force is a vector, having both magnitudeand direction.

The magnitude of a force can bemeasured using a spring scale.


4-2 Newton’s First Law of Motion

It may seem as though it takes a force to keepan object moving. Push your book across atable—when you stop pushing, it stops moving.

But now, throw a ball across the room. The ballkeeps moving after you let it go, even thoughyou are not pushing it any more. Why?

It doesn’t take a force to keep an object movingin a straight line—it takes a force to change itsmotion. Your book stops because the force offriction stops it.

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This is Newton’s first law, which is oftencalled the law of inertia:

Every object continues in its state of rest, or ofuniform velocity in a straight line, as long as no netforce acts on it.



Conceptual Example 4-1: Newton’s first law.

A school bus comes to a sudden stop, and allof the backpacks on the floor start to slideforward. What force causes them to do that?

There is no force. By Newton's first law thebackpacks continue their state of motion,maintaining their velocity. They will bebrought to rest by friction or collision.

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Inertial reference frames:

Newton’s first law does not hold in everyreference frame, such as a reference frame thatis accelerating or rotating.

An inertial reference frame is one in whichNewton’s first law is valid. This excludesrotating and accelerating frames.

How can we tell if we are in an inertialreference frame? By checking to see ifNewton’s first law holds!


4-3 Mass

Mass is the measure of inertia of an object,sometimes understood as the quantity ofmatter in the object. In the SI system, mass ismeasured in kilogram (kg).

Mass is not weight.

Mass is a property of an object. Weight is theforce exerted on that object by gravity.

If you go to the Moon, whose gravitationalacceleration is about 1/6 g, you will weigh muchless. Your mass, however, will be the same.

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4-4 Newton’s Second Law of Motion

Newton’s second law is the relation betweenacceleration and force. Acceleration isproportional to force and inversely proportionalto mass.

It takes a force to changeeither the direction or thespeed of an object. Moreforce means moreacceleration; the sameforce exerted on a moremassive object will yieldless acceleration.


Newton’s Second Law – A Correction

Newton expressed his 2nd law in terms ofmomentum which is not discussed in Giancoliuntil chapter 9.

Momentum is a vector symbolized by thesymbol , and is defined as

Unit: kg.m/s

The correct statement of Newton’s 2nd law is:

The time rate of change of momentum of abody is proportional to the resultant externalforce acting on the body and takes place in thedirection of the force.

p

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Newton’s Second Law:

dt

d pF

In SI Units the constant is defined as 1 so we canwrite:

dt

dm

dt

dm

dt

md

dt

dv

vvpF

)(

Sincedt

d va

and in the special casewhere the mass of thebody is constant

0

dt

dm

Do not try to solve variable mass problems, eg. Rocketmotion without using the full form of Newton’s Second Law.

We can use aF

m



Force is a vector, so is true along eachcoordinate axis.

kjiF ˆˆˆwhere

ΣΣΣ

zyx

zzyyxx

FFF

maF,maF,maF

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The unit of force in the SIsystem is the newton (N).

Note that the pound is aunit of force, not of mass,and can therefore beequated to newtons butnot to kilograms.

Since

1 NL1 kg.m/s2



Example 4-2: Force to accelerate a fast car.

Estimate the net force needed to accelerate(a) a 1000 kg car at ½ g(b) a 200 g apple at the same rate.

2m/s4.92

9.82

kg1000 ga

m

2m/s4.92

9.82

g200 ga

m

kN 4.9N 49004.91000Σ aF

m

N 98.04.90.2Σ aF

m

(a)

(b)

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Example 4-3: Force to stop a car.

What average net force is required to bring a1500 kg car to rest from a speed of 100 km/hwithin a distance of 55 m?

m/s27.83600

1000100km/h100



N101.17.01500

m/s7.00552

27.80

2

2

4

2

0

20

2

020

2

maF

xx

vva

xxavv m = 1500 kgv0 = 27.8 m/sv = 0x0 = 0x = 55 m

Note that the force (and acceleration) is negativewhich shows that the force is in the oppositedirection to the initial velocity.

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4-5 Newton’s Third Law of Motion

Any time a force is exerted on an object, thatforce is caused by another object.

Newton’s third law:

Whenever one object exerts a force on a second object,the second exerts an equal force in the oppositedirection on the first.

Often stated as for everyaction there is an equal andopposite reaction.

Action and reaction forcesact on different objects.



A key to the correctapplication of the thirdlaw is that the forcesare exerted on differentobjects. Make sure youdon’t use them as ifthey were acting on thesame object.

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Rocket propulsion can also be explained usingNewton’s third law: hot gases from combustionspew out of the tail of the rocket at high speeds.The reaction force is what propels the rocket.

Note that therocket does notneed anything to“push” against.



Conceptual Example 4-4: What exerts theforce to move a car?

Response: A common answer is that theengine makes the car move forward. But it isnot so simple. The engine makes the wheelsgo around. But if the tires are on slick ice ordeep mud, they just spin. Friction is needed.On firm ground, the tires push backwardagainst the ground because of friction. ByNewton’s third law, the ground pushes on thetires in the opposite direction, acceleratingthe car forward.

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Helpful notation: the first subscript is the objectthat the force is being exerted on; the second isthe source.


Michelangelo’s assistant has been assigned the task ofmoving a block of marble using a sled. He says to his boss,“When I exert a forward force on the sled, the sled exerts anequal and opposite force backward. So how can I ever start itmoving? No matter how hard I pull, the backward reactionforce always equals my forward force, so the net force must bezero. I’ll never be able to move this load.” Is he correct?

Conceptual Example 4-5: Third law clarification.

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Conceptual Example 4-5: Third law clarification.

Action–reaction forces do not cancel because theyact on different objects.

Answer: No – in order to see whether the sled willaccelerate, we need to consider only the forces onthe sled. The force that the sled exerts on theassistant is irrelevant to the sled’s acceleration.


4-6 Weight—the Force of Gravity;and the Normal Force

Weight is the force exerted on an object bygravity. Close to the surface of the Earth, wherethe gravitational force is nearly constant, theweight of an object of mass m is:

where

Note that g varies with position on the earth’s surface and withaltitude.g varies from about 9.78 m/s2 at the Equator to about9.83 m/s2 at the poles, so an object will weigh about 0.5%more at the poles than at the Equator.

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An object at rest must have no net force on it. Ifit is sitting on a table, the force of gravity is stillthere; what other force is there?

The force exerted perpendicular to a surface is

called the normal force. It isexactly as large as neededto balance the force fromthe object. (If the requiredforce gets too big,something breaks!)



(a) (b)

(a) The net force on an object at rest is zero according toNewton’s second law. Therefore the downward force ofgravity ( ) on an object at rest must be balanced by anupward force (the normal force ) exerted by the table inthis case.

(b) is the force exerted on the table by the statue and isthe reaction force to by Newton’s third law.

GF

NF

'NF

NF

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A friend has given you a specialgift, a box of mass 10.0 kg with amystery surprise inside. The box isresting on the smooth (frictionless)horizontal surface of a table.

(a) Determine the weight of the boxand the normal force exertedon it by the table.

upwardsN98.00

N98.09.8010.0:boxofWeight

mgFmgFF

m

NNy

gW



A friend has given you a special gift,a box of mass 10.0 kg with amystery surprise inside. The box isresting on the smooth (frictionless)horizontal surface of a table.

(b) Now your friend pushes downon the box with a force of 40.0 N.Again determine the normalforce exerted on the box by thetable.

upwards.N138.040.098.040.0

00.40

mgF

mgFF

N

Ny

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A friend has given you a special gift,a box of mass 10.0 kg with a mysterysurprise inside. The box is resting onthe smooth (frictionless) horizontalsurface of a table.

(c) If your friend pulls upward on thebox with a force of 40.0 N, whatnow is the normal force exertedon the box by the table?

upwards.N58.040.098.040.0

00.40

mgF

mgFF

N

Ny



Example 4-7: Acceleratingthe box.

What happens when aperson pulls upward on thebox in the previousexample with a forcegreater than the box’sweight, say 100.0 N?

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upwardsm/s0.2010.0

98.0100.0 2

m

mgFa

mamgFF

Py

yPy



Example 4-8: Apparent weight loss.

A 65 kg woman descends in anelevator that briefly accelerates at0.20g downward. She stands on ascale that reads in kg.

(a) During this acceleration, what isher weight and what does thescale read?

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0.8mg0.2g)m(ga)m(gmamgF

maFmgF

N

N

Weight is always = mg and will notchange when the elevator accelerates

W = mg = 65G9.8 = 640 N

Strictly scales measure force but arecalibrated to show the mass of theperson would read 65 kg if theelevator was not moving.

When the elevator is acceleratingdownwards (take as positive):

Scale reading = 0.8G65= 52 kg



Example 4-8: Apparent weight loss.

A 65 kg woman descends in anelevator that briefly accelerates at0.20g downward. She stands on ascale that reads in kg.

(b) What does the scale read whenthe elevator descends at aconstant speed of 2.0 m/s?

Without acceleration

kg65ofmasscorrectthereadsscaletheand

mgFFmgF NN 0

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4-7 Solving Problems with Newton’s Laws:Free-Body Diagrams

1. Draw a sketch.

2. For one object, draw a free-bodydiagram, showing all the forces actingon the object. Make the magnitudesand directions as accurate as youcan. Label each force. If there aremultiple objects, draw a separatediagram for each one.

3. Resolve vectors into components.

4. Apply Newton’s second law to eachcomponent.

5. Solve.


4-7 Solving Problems with Newton’s Laws: Free-Body Diagrams

0.113.52

2.10

F

F

N53.32.103.52

N2.011.8128.3

N52.30.4228.3

1180.60230.00.37

N0.420.799.0300.37

N28.30.70740.00.45

N28.30.70740.00.45

1

Rx

Ry1-

2222

tantan

FFF

FFF

FFF

N.sinFF

cosFF

sinFF

cosFF

RyRxR

ByAyRy

BxAxRx

BBy

BBx

AAy

AAx

Net force = 53.3 N at 11.0 to x axis

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y

x

A

B

C

A

B

C

By convention angles are measured relative to the x axis:Anti-clockwise (acw) is +veClockwise (cw) is –ve

A = 34

B = -56 = +304

C =144



(a) (b) (c)

Conceptual Example 4-10: The hockey puck.

A hockey puck is sliding at constant velocity across a flathorizontal ice surface that is assumed to be frictionless.Which of these sketches is the correct free-body diagramfor this puck? What would your answer be if the puckslowed down?

Note that if there was friction (c) would be the correct answer.

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Example 4-11: Pulling the mystery box.Suppose a friend asks to examine the10.0 kg box you were given previously,hoping to guess what is inside; and yourespond, “Sure, pull the box over toyou.” She then pulls the box by theattached cord along the smooth surfaceof the table. The magnitude of the forceexerted by the person is FP = 40.0 N, andit is exerted at a 30.0° angle as shown.Calculate(a) the acceleration of the box, and(b) the magnitude of the upward force FN

exerted by the table on the box.



(a) the acceleration of the box

righttom/s.463

0.10

6.34

N34.6866.00.4030

2

m

FamaF

cosFF

PxxxPx

PPx

N78.020.09.800.01

0

N0.02500.00.4030

PyN

PyN

PPy

FmgF

mgFF

sinFF

(b) the magnitude of the upward force FN

exerted by the table on the box.

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Example 4-12: Two boxes connectedby a cord.

Two boxes, A and B, are connectedby a lightweight cord and are restingon a smooth table. The boxes havemasses of 12.0 kg and 10.0 kg. Ahorizontal force of 40.0 N is appliedto the 10.0-kg box. Find (a) theacceleration of each box, and (b) thetension in the cord connecting theboxes.



(a) the acceleration of each box

Vertical forces cancel so only need toconsider horizontal forces.

Both boxes have the same acceleration.

2m/s1.8212.010.0

40.0

eliminatetoequationsAdd

:BBox

Σ :ABox

BA

P

BABATTP

T

BTB

ATPA

mm

Fa

mmaamamFFF

F

amFΣF

amFFF

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(b) the tension in the cordconnecting the boxes.

Only need to consider one box.

N21.81.8210.040.0

answer)samethe(givesABoxor

N21.81.8212.0:BBox

amFFamFF

amF

APTATP

BT



Example 4-13: Elevator andcounterweight (Atwood’s machine).

A system of two objects suspended overa pulley by a flexible cable is sometimesreferred to as an Atwood’s machine.Here, let the mass of the counterweightbe 1000 kg. Assume the mass of theempty elevator is 850 kg, and its masswhen carrying four passengers is 1150kg. For the latter case calculate (a) theacceleration of the elevator and (b) thetension in the cable.

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4-7 Solving Problems with Newton’s Laws: Free-BodyDiagrams

Example 4-13: Elevator and counterweight(Atwood’s machine).

(a) the acceleration of the elevator

2m/s0.68

10001150

100011509.80

:ghtCounterwei

:Elevator

CE

CE

CECE

CCEET

CCCT

EEET

mm

mmga

mmgmma

amgmamgmF

a

amgmF

amgmF

CECE aaaa


4-7 Solving Problems with Newton’s Laws: Free-BodyDiagrams

Example 4-13: Elevator andcounterweight (Atwood’s machine).

(b) the tension in the cable.

N10,5000.68)1000(9.80

answersamethegivesghCounterweiConsider

N10,5000.68)1150(9.80

:ElevatorConsider

agmF

amgmF

agmF

amgmF

CT

CCT

ET

EET

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Conceptual Example 4-14: Theadvantage of a pulley.

A mover is trying to lift a piano (slowly)up to a second-story apartment. He isusing a rope looped over two pulleys asshown. What force must he exert on therope to slowly lift the piano’s 2000 Nweight?

N10002

2000

22

0

0speedconstantatpianolift theTo

mgFmgF

mamgFFFy

TT

yTT

ya



Example 4-15: Accelerometer.

A small mass m hangs from athin string and can swing likea pendulum. You attach itabove the window of your caras shown. What angle doesthe string make (a) when thecar accelerates at a constanta = 1.20 m/s2, and (b) whenthe car moves at constantvelocity, v = 90 km/h?

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What angle does the string make (a)when the car accelerates at a constant a= 1.20 m/s2, and (b) when the car movesat constant velocity,v = 90 km/h?

00elocityconstant vAt(b)

07809

201

tan(2)by(1)divide

(2)(1)

0:Vertically

:lyHorizontal(a)

veas&Taking

11

θa

..

.tan

g

atanθ

g

aθ

cosθ

sinθ

F

mgcosθ

F

masinθ

mgcosθFF

masinθFF

TT

TTy

TTx



Example 4-16: Box slidesdown an incline.

A box of mass m is placed ona smooth incline that makesan angle θ with the horizontal. (a) Determine the normalforce on the box. (b)Determine the box’sacceleration. (c) Evaluate for amass m = 10 kg and an inclineof θ = 30°.

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(a) Normal force on the box.

(b) Box’s acceleration.

(c) Evaluate FN and a for m = 10 kgand θ = 30°.

θ

0θmgΣ :directiony

cosmgF

cosFF

N

Ny

θ

θΣ :directionx

singa

masinmgFx

2m/s4.9309.80a

N85309.8010.0

sin

cosFN


4-8 Problem Solving—A General Approach

1. Read the problem carefully; then read it again.

2. Draw a sketch, and then a free-body diagram.

3. Choose a convenient coordinate system.

4. List the known and unknown quantities; findrelationships between the knowns and theunknowns.

5. Estimate the answer.

6. Solve the problem without putting in any numbers(algebraically); once you are satisfied, put thenumbers in.

7. Keep track of dimensions.

8. Make sure your answer is reasonable.

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Summary of Chapter 4

• Newton’s first law: If the net force on an objectis zero, it will remain either at rest or moving ina straight line at constant speed.

•Newton’s second law:

• Newton’s third law:

• Weight is the gravitational force on an object.

• Free-body diagrams are essential for problem-solving. Do one object at a time, make sure youhave all the forces, pick a coordinate systemand find the force components, and applyNewton’s second law along each axis.

chapter 4 dynamics: newton’s laws of motion · title: microsoft powerpoint - chapter 4.ppt...

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