Physics, Page 1

Chapter 5, 6. Force and Motion

• Newton’s Laws– Concepts of Mass and Force– Newton’s Three Laws– Types of forces

But first, let’s review the last lecture..

Physics, Page 2

Summary of the last lecture1. Projectile Motion

2. Circular motion

3. Relative Motion

• x = x0 + v0t• v = v0x

• y = y0 + v0yt - 1/2 gt2

• vy = v0y- gt

• vy2 = v0y

2 - 2g Δy

PB PA BA PB PA BAr r r v v v= − → = −$ $ $ $ $ $

• Centripetal acceration for uniform circular motion :

a = v2 / r

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Isaac Newton (1642 ~ 1727)

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Question You are driving a car up a hill with constant velocity. How many forces are acting on the car?

12345

W

f

FNV

correct

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Question The net force on the car is,

1. Zero 2. Pointing up the hill 3. Pointing down the hill 4. Pointing vertically downward 5. Pointing vertically upward

W

f

FN V

W

f

FN

correct

ΣF = ma = 0

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QuestionYou are driving a car up a hill with constant acceleration.

How many forces are acting on the car?

12345

W

f

FN

acorrect

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2) Compare the magnitudes of the acceleration you experience, aA, to the magnitude of the acceleration of the spacecraft, aS, while you are pushing:

1. aA = aS2. aA > aS3. aA < aS

QuestionSuppose you are an astronaut in outer space giving a brief push to a spacecraft whose mass is bigger than your own .

1) Compare the magnitude of the force you exert on the spacecraft, FS, to the magnitude of the force exerted by the spacecraft on you, FA, while you are pushing:

1. FA = FS2. FA > FS3. FA < FS

correct

correct a = F/mF same ⇒ lower mass give larger a

Second Law!

Third Law!

Physics, Page 8

Force :- (loosely speaking) a push or pull on an object- Causes an object to accelerate

- unit : [N]- Vector quantity

<Principle of superposition for force>When two or more forces act on a body,

Net force (Resultant force) vector sum of all forces acting on the body

≡

Physics, Page 9

Newton’s First LawThe motion of an object does not change unless it is acted upon by a net force.

• If v=0, it remains 0.• If v is some value, it stays at that value.

Another way to say the same thing:• No net force (Fnet = 0) ⇔

• velocity is constant. • acceleration is zero.

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Mass or InertiaMass (m) is the property of an object that

measures how hard it is to change its motion.

Units: [M] = kg

Examples of 1st Law

• braking car vs. train• car vs. bus going around curve• push on light or heavy object

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Newton’s Second Law

netFam

=ur

rnetF F ma= =∑

ur ur r

Cause(원인)

Result(결과)

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QuestionAn airplane is flying from Seoul airport to Jegu. Many forces act on the plane, including weight (gravity), drag(air resistance), the trust of the engine, and the lift of the wings. At some point during its trip the velocity of the plane is measured to be constant (which means its altitude is also constant). At this time, the totaltotal force on the plane:

1. is pointing upward2. is pointing downward3. is pointing forward4. is pointing backward5. is zero

lift

weight

drag thrust

correct

Physics, Page 13

gravity would be working against the airplane flying at a level altitude, so the total force would need to be upward.

Gravity never goes away, so the force on the plane will be pointing downwards.

The force has to be pointing forward since the velocity is moving forward.

the up and down forces are zero because the plane is not going up or down, and the plane is pushing back against the air propelling it forward.

something to think about…the plane is constantly pointing down because the earth is round and constant altitude would mean its distance to center of earth may be changed to keep a constant distance from the ground

True, but the effect is small.. On a cross-country trip (300 miles), a “flat” trip would have distance to center of earth varying by 0.5%

Questions(common misconceptions)

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Example 1

MF1 M=10 kg, F1=200 NFind a.

a = Fnet/M = 200N/10kg = 20 m/s2

M=10 kg F1=200 N F2 = 100 N

Find a.

MF1 F2

a = Fnet/M = (200N-100N)/10kg = 10 m/s2

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Example 2• A force F acting on a mass m1 results in an acceleration a1.

The same force acting on a different mass m2 results in an acceleration a2 = 2a1. What is the mass m2?

(a)(a) 2m1 (b)(b) m1 (c)(c) 1/2 m1

F a1

m1 F a2 = 2a1

m2

• F=ma • F= m1a1 = m2a2 = m2(2a1)• Therefore, m2 = m1/2

• Or, in words… twice the acceleration means half the mass.

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Examples of Force1. Gravity (중력, 혹은 무게)

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12Fr

mgR

mGMFE

E ≡= 2

Gravitational Force m1

m221Fr

2112 FFrr

−= : Gravitational force between massive objects

12212

1212 r

rmGmF =

r

21221

2121 r

rmGmF =

r 2112 rr −=

On the Earth

REarth

m ME = 6 ×1024 kgRE = 6.4×106 m

( )2

26

2411

2 sec/8.9104.6

1061067.6 mR

GMgE

E =×

×⋅×==

−

G = universal gravitation constant = 6.67 x 10-11 N-m2/kg2

Physics, Page 18

Examples of Force2. Normal force (수직힘, 수직항력, 법선력)

book at rest on table:What are forces on book?

W

• Weight is downward• System is “in equilibrium” (acceleration = 0 ⇒ net force = 0)• Therefore, weight balanced by another force

FN

• FN = “normal force” = force exerted by surface on object• FN is always perpendicular to surface and outward• For this example FN = W

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Another Example:• Book at rest on table:• Push down with Fhand• What is FN?

W

Fhand

FN

• Equilibrium: • FN + W + Fhand = 0 • ⇒ FN = W+Fhand

What is direction of FN?

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Examples of Force3. Friction (쓸림, 쓸림힘, 마찰력)

(접촉마찰력)

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Static Friction (정지쓸림힘)

• Static Friction: a force between two surfaces that prevents motion• fs < μsFN =Fs,max (size is whatever is needed to prevent motion)

μs = coefficient of static frictiona property of the two surfaces1 ≥ μs ≥ 0

• direction is whatever direction needed to prevent motion

W

FN

Ffs

Physics, Page 22

Kinetic Friction (운동쓸림힘)

• Kinetic (sliding) Friction: a force between two surfaces that opposes motion

• fk = μkFN• μk = coefficient of sliding friction

a property of the two surfaces1 ≥ μk ≥ 0

• direction is opposite to motion

W

FN

Ffk

direction of motion

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Static Kinetic

fs ≤ μsn , μs : coefficient of static frictionfk = μkn , μk : coefficient of kinetic friction

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Example 1

v0 =8 m/sμk = 0.2

Find stopping distance

Answer: 16.3 m

F = μk FN = μk (mg), F = ma, V2 = Vo2 - 2as

mg

FN

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Example 5-1

Initial velocity v0 = 20 m/sec. Stop at l = 115 mFind the friction coef. : μk

nF kμ−= mamgk =μ−=ga kμ−=

gvt

tgvtavv

k

k

μ=

=⋅μ−=⋅+=

0

00 0

( )1770

11589220

2

2

2

220

202

21

02

21

0

.msec/m.

sec/mgl

v

gvgttvattvl

k

kk

=××

==μ

μ=μ−=+=

Example 1-2

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Example 2

T=50 N

M = 5 kg

μk = 0.2

Find acceleration of block

Answer: 8.04 m/s2

F = T- μkFN = ma

Physics, Page 27

Example 3

T=50 N

M = 5 kg

μk = 0.2

Find acceleration of block

θ=500

Answer: 6.0 m/s2

F = T cosθ - μkFN = ma

Physics, Page 28

Example 4

a b

Which case requires the lesser force to overcome static friction?a) ab) bc) the same

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Examples of Force4. Tension (장력)

M M : T = Ma

H : T = F

FH

M

m

M

F

M : T = Ma

m : T – mg = -ma

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F1 = T - μkm1g = m1aF2 = m2g - T = m2a+)

m2g - μkm1g = m1a +m2ag

mmmma k

21

12

+μ−

=

Example 5-2

=μkm1g

Example 5 : Find a.

m1

m2

m1

m2

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Terminal speed (종단속력)Vterminal when a = 0

vbD rr−= : 마찰력 (Assumption)

amvbgmF rrrr=−=

mabvmg =−

0=−= vmbga

Terminal speed

gbmv t =

Examples of Force5. Drag force (끌림힘)

유체와 물체사이에 상대적인 운동이 존재할 때, 물체의 운동을 방해하는 힘

D

Physics, Page 32

221 AvCD ρ=

Drag Coefficient

Air density Cross section

maAvCmgDmgF =−=−= 221 ρ

2

2v

mACga ρ

−= 0= : For terminal velocity

ACmgv t ρ

2=

3r∝

2r∝

rv t ∝

Terminal speed (종단속력) in air

Table 6-1 Some terminal speed in air

D

D

Physics, Page 33

Examples of Force# Centripetal force (구심력)

물체의 속력 변화는 없이 방향만을 바꾸어주는 힘

-

- 속도에 수직한 방향 (원운동의 경우, 원의 중심을 향하는 방향)

- 새로운 종류의 힘이 아니며, 쓸림힘, 중력, 장력 등 모든 힘이 구심력이 될 수 있음 !

rvmF

2

=Centripetal acceleration (구심가속도)

x

y

θv

Physics, Page 34

Question Consider the following situation: You are driving a car with constant speed around a horizontal circular track. How many forces are acting on the car?

1 2 3 4 5

Wf

FNcorrect

Physics, Page 35

The net force on the car is

1. Zero2. Pointing radially inward3. Pointing radially outward W

f

FN

ΣF = ma = mv2/R

a=v2/R Rcorrect

The car has a constant speed so acceleration is zeroA car that is driving in a circle is accelerating towards the center of the circle. According to Newton's second law the net force must be pointing toward the center of the circle.

when I sit in this car I can feel the FORCE...pulling me to the dark side...away from the center of the circle

Question

Physics, Page 36

Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude of the normal force FN exerted on you by the car seat as you drive past the bottom of the hill?

1. FN < mg2. FN = mg3. FN > mg

v

mg

FN

R

ΣF = ma = mv2/R

FN - mg = mv2/R

FN = mg + mv2/R

a=v2/R

You feel like you are being pulled down in your seat so the mg must be greater then the normal force -- wrong!

FN is always equal to mg --- wrong!

The net force must point towards the center

correct

Question

Physics, Page 37

Going over top of hill….

1. FN < mg 2. FN = mg 3. FN > mg

mg

FN

a=v2/RR

ΣF = ma = mv2/R

mg - FN = mv2/R

FN = mg - mv2/R

v

Question

correct

Physics, Page 38

Example : Find the angle θ.

Tsinθ

Tcosθ

θ

0=−θ= mgcosTF:y y

θ=

cosmgT

rvmmasinTF:x x

2

==θ=

rmvtanmg

2

=θ

θθ

sintan

22

gLv

grv

==

Example: Find the maximum velocity on the track.

rvmmafF s

2

===

mgf smax,s μ=

grm

rfv s

max,smax μ==

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Example : Find the normal forces at bottom, top, and at A-point.

rmvmgn bot

2

=−At Bottom: mgr

mvn bot +=2

rmvmgn top

2

=+ mgr

mvn top −=2

At Top:

At A-point: rmvn A

2

=

Anr

Physics, Page 40

Newton’s Third LawFor every action, there is an equal and opposite reaction.

• Finger pushes on box • Ffinger→box = force exerted on box by finger

Ffinger→box

Fbox→finger • Box pushes on finger• Fbox→finger = force exerted on finger by box

• Third Law: Fbox→finger = - Ffinger→box

(두 물체가 상호작용할 때 서로에게 작용하는 힘(짝힘)은 항상 크기가 같고 방향이 반대이다.)

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Newton's Third Law...

• FFA ,B = - FFB ,A. is true for all types of forces

FFw,m FFm,w

FFf,m

FFm,f

Physics, Page 42

Example of Bad Thinking

• Since FFm,b = - FFb,m why isn’t FFnet = 0, and aa = 0 ?

a ??a ??FFb,m FFm,b

ice

Physics, Page 43

Example of Good Thinking• Consider only the boxonly the box!

–– FFon box = maabox = FFm,b

– Free Body Diagram (which means a diagram including all forces existing)

What about forces on man?

aaboxbox

FFb,m FFm,b

ice

Physics, Page 44

(a) 참외에 작용하는 힘은?

(b) 참외와 지구 사이에 작용하는 제3법칙의 짝힘은?

FCE (중력), FCT (식탁이 주는 수직힘)

이 두 힘은 짝힘인가? 아니다!

FCE (지구 인력) = - FEC (참외 인력)

(c) 참외와 식탁 사이에 작용하는 제3법칙의 짝힘은?

FTC (참외가 식탁에 가한 수직힘) = - FCT (식탁이 참외에 가한 수직힘)

Newton's Third Law...

Physics, Page 45

Summary• Newton’s First Law:

The motion of an object does not change unless it is acted on by a net force

• Newton’s Second Law:Fnet = ma

• Newton’s Third Law:Fa,b = - Fb,a

• Types of Forces– Gravity – Normal Force– Friction– Tension– Drag force

Physics, Page 46

M1 = 2 kg

Find acceleration of blocks and tension T1

Answers: a = 4 m/s2 and T1= 16 N

M2 = 4 kg

F = 24 NT1

Application of Newton's Laws

Physics, Page 47

Example

M1 = 4 kg

Turn blocks around….Find acceleration of blocks and tension T1

Answers: a= 4 m/s2 (the same)and T1= 8 N (smaller)

M2 = 2 kg

T=24 NT1

Physics, Page 48

Example

Find acceleration and normal force

Answers: a= g sin(θ) FN = mg cos (θ)

θ

am

mg

Physics, Page 49

Example : Pulley Problem I

What is the tension in the string?

A) T<W

B) T=W

C) W<T<2W

D) T=2W

E) T>2W

WW

Pull withforce = W

W

Same answer

T

Physics, Page 50

Example : Pulley Problem II

What is the tension in the string?

A) T<W

B) T=W

C) W<T<2W

D) T=2W

E) T>2W W2Wa

a

TT

Physics, Page 51

Example : Pulley Problem III

• Identify forces• Set up axes• Write F net = ma for each axis• Solve!• and, Verify

Procedure for Solving Problems

GravityTensionNormal forceFriction

Physics, Page 52

M

m

예) 차원 비교? T < mg ?

Physics, Page 53

m2

m1m1g

T

m2g

T

amgmTF 111 =−=

amTgmF 222 =−=+)( ) ( )ammgmm 2112 +=−

gmmmma

21

12

+−

= , gmm

mmT21

212+

=

m1 m2F PP

amPF 22 ==amPFF 11 =−=+)

( )ammF 21 +=

21 mmFa+

= , Fmm

mP21

2

+=

Atwood’s Machine

Physics, Page 54

m

(1) a

T

mg

T

mg

(2) a

(1) F = T - mg = maT = m(g + a)

(2) F = mg - T = maT = m(g - a)

Example : Weighting in an Elevator