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AP Physics Chapter 6

Force and Motion – II

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AP Physics

Turn in Chapter 5 Homework, Worksheet & Lab Report

Take Quiz 6 Lecture Q&A

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Review on Chapter 5

Newton’s Second Law: F = ma Newton’s Third Law: Action and Reaction

Forces Forces

Weight Normal Force Tension Apparent Weight

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Four Fundamental (Basic) Forces in Nature

Gravitational Force Electromagnetic Force Strong (Nuclear) Force Weak (Nuclear) Force

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Gravitational Force

Gravity or Weight (Gravitational force Earth pulling on objects around it): W

W mg Gravitational force always exist, but value of g

can be different at different location. – On surface of earth, g = 9.81 m/s2. – Higher elevation, smaller g.– Higher latitude, larger g.

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Electromagnetic Force

Almost all other forces we encounter in daily life are electromagnetic force in nature.

Electric force: F Magnetic force: F Tension (pull, string): T Push: F Normal (support): N (not to be confused with

Newton, the unit of force) Friction: ƒ (ƒs or ƒk)

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Strong (Nuclear) Force

Holds protons and neutrons within the nucleus Exist in very short distance only ( 10-15 m) Stable nucleus

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Weak (Nuclear) Force

Also holds protons and neutrons within the nucleus

Also very short range ( 10-15 m) Not strong enough. Protons and neutrons can

escape nuclear reaction Unstable nucleus

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Relative Strength of Forces

Weakest Strongest

Gra

vita

tion

al

Ele

ctro

mag

neti

c

Wea

k

Str

ong

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Force at Distance

Contact force: Force giver and receiver must be in contact

Tension, push, normal force, friction, …

Field force / Distant force: Force giver and receiver do not have to be in contact

Gravity, electric force, magnetic force, …

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Unification of Forces

Electromagnetic and weak forces have been unified into (understood as) one force: Electroweak force

It is believed that all four forces are different aspects of a single force.– Grand Unification Theories (GUTs) and

Supersymmetric theories– Not yet successful

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Friction

Friction: force opposing relative motion or tendency of relative motion between two rough surfaces in contact

v

Smooth surface No friction No normal force No friction

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Two kinds of frictions

Static friction: ,maxs sf N

k kf N

s: coefficient of static friction Static friction force is not constant, it has a maximum.

Kinetic (or sliding) friction: k: coefficient of kinetic friction Kinetic friction force is constant.

is a constant depending on the properties of the two surfaces = 0 when one of the surfaces is smooth (frictionless.) s > k for same surfaces. has no unit.

s sf N

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Fapp

f

0

Table

not m

oving

yet.

Static

fricti

on in

creas

es as

appli

ed

force

incre

ases

. fs =

F appsN Once table is moving, a smaller

constant kinetic friction. fk < fs, max

But what if the applied force increases just slightly?

•

•

sN

•

• •

Example:What happens to the frictional force as you increase the force pushing on a table on the floor?

Fapp f

W

N

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Is friction always opposite to motion?

v

vf1 f2

v1

2 65

4

3

When you step on the pedal

v

v

7

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Example:

A trunk with a weight of 220N rests on the floor. The coefficient of static friction between the trunk and the floor is 0.41, while the coefficient of kinetic friction is 0.32.

a). What is the minimum magnitude for a horizontal force with which a person must push on the trunk to start in moving?

b) Once the trunk is moving, what magnitude of horizontal force must the person apply to keep it moving with constant velocity?

c) If the person continued to push with the force used to start the motion, what would be the acceleration of the trunk?

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Solution

NWks 220,32.0,41.0

min) 0, ?b a F

min) 0, ?a a F

Ff

W

N =

y

x

W

F

min ,maxsF f

minF

min ,maxsF f m a 0

sN sW

0.41 220 90N N

Similarly,

kf kN kW 0.32 220 70N N

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Solution

NWks 220,32.0,41.0

) ?c a

Ff

W

N =

y

x

W

F

kF fa

m

/

F f

W g

2

2

90 70

220 / 9.8 /

0.89 /

N N

N m s

m s

kF f ma

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Example: Pg135-57

The coefficient of kinetic friction in between the incline and the block is 0.20, and angle is 60o. What is the acceleration of the block if

a) It is sliding down the slope and

b) It has been given an upward shove and is still sliding up the slope?

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Solution

W

Nf

x

y) ?a a

cossin mmg mg a

a

xF

yF

0.20, 60.0o

xW f ma

sinmg maN

yN W 0 Why?

yN W cosmg

xW

Wx

Wy

sin cos sin cosg g g

2 29.8 sin 60 0.20cos 60 7.5o om m

s s

sinW

yW cosW

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Solution (2)

) ?b a

cossin mmg mg a

a

xF

yF

0.20, 60o

xW f ma

sinmg maN

yN W 0

yN W cosmg

sin

cosx

y

W W

W W

W

N

f

x

y

Wx

Wy

sin cos sin cosg g g

2 29.8 sin 60 0.20cos 60 9.5o om m

s s

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Practice:

An 11 kg block of steel is at rest on a horizontal table. The coefficient of static friction between block and table is 0.52.

a) What is the magnitude of the horizontal force that will just start the block moving?

b) What is the magnitude of a force acting upward 60o from the horizontal that will just start the block moving?

c) If the force acts down at 60o from the horizontal, how large can its magnitude be without causing the block to move?

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Solution

W

N

f F

)a

f

FN

W

y

x

)b

11 , 0.52m kg

yF

cosF

cos sinF F mg

cos sinF mg

20.52 11 9.8 /

59cos sin cos 60 0.52sin 60o o

mg kg m sF N

xF xF f 0 cosF N

yF N W 0 N sinmg F

sinmg F

,s maxf N 20.52 11 9.8 / 56mg kg m s N minF

Fx

Fy

xF

cosF

yF sinF

cosF f

sinF N W

sinmg F

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Solution (2)

c)

2

30.52 11 9.8 /1.1 10

cos sin cos 0.60 652sin 0o o

mg kg m sF N

Similarly to Part b), we have

Another approach is to use the same equation as in b) but change the angel to –60o.

F 2

30.52 11 9.8 /1.1 10

cos sin cos 60 0.52sin 60o o

mg kg m sN

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Practice: Pg133-21

Block B in the diagram weighs 711 N. The coefficient of static friction between block and horizontal surface is 0.25. Find the maximum weight of block A for which the system will be stationary?

30o

A

B

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Solution: Pg133-21

x

N

T

WB

f

y

B

WA

TTA

y

x

:P

P

pxF

pyF

:Bm

BxF

AW

cosAT T 0 AT

AW

tanT sincos

T

ByF BW

T N

N T

cos

T

sinA AT W 0

sinAT

T f 0 f

BN W 0 BW

tan 0.25 711 tan 30 103oBW N N

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Drag Force

Drag Force:

C: drag constant (not a true constant)

: density of fluid (air or liquid)

A: cross-sectional area of object moving in fluid

v: speed of object relative to fluid

Source: fluid in relative motion pushing on object

Direction: opposing relative motion of object relative to fluid, same direction as relative motion of fluid

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2D C Av

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Terminal velocity

0F

W

+

W D D 21

2C Av

2mgv

C A

D

21

2D C Av

v: increases

Eventually, D ___ W: 0

0 constant a =

Initially, v: small D:

Fnet:

small

downward downward

W _____ D>

Increases

a:

D:

Fnet = =

: v:

0 W mg

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21

2D C Av

10

5

km

km

D

D

2103

2

103

0.38

0.672

kgv

mvkg

m

210 10

1

2C Av

210 10

25 5

v

v

10 5 5 103 3

10.38 , 0.67 ,

2km km km km

kg kgv v

m m

25 5

1

2C Av

Practice: Pg133-33Calculate the ratio of the drag force on a jet flying with a speed of 1000 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at half the speed and half the altitude. The density of air is 0.38 kg/m3 at 10 km and 0.67 kg/m3 at 5.0 km. Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient C.

210

210

0.38

0.674

v

v

4 0.382.3

0.67

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Review: Uniform Circular Motion and Centripetal Acceleration

Direction of acceleration is always toward the center of circle (or circular arc)

Centripetal

2va

r

v: speed of particle

r: radius of circle or circular arc, where

a

a

a

v

v

v

for uniform circular motion at any time.a v

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Centripetal Force

Centripetal force is in general not a single physical force; rather, it is in general the net force.

Do not draw centripetal force on force diagram (Free Body Diagram)

c cF ma2v

mr

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Examples of centripetal forces

Rounding a curve in a car

max sv Rg

tanv Rg

Max velocity w/o skidding

– Flat curve:

– Banked curve:

Orbiting the Earth (Sun or other object)– Gravity

N

f

W

Static friction provides the centripetal force (when no skidding)

a component of Normal force (net force)

Rear View

N

W

No friction2

1tanv

Rg

33

W

f

NW

Yes, successful.

x

y

) ?a f

,max) ?sb f

F

,maxsf

m

2vmR

s N 3 30.35 10.7 10 3.75 10s W N N 33.21 10 N

2

3 313.4 /1.09 10 3.21 10

61.0

m skg N

m

mg 3

3

2

10.7 101.09 10

9.8

W Nkg

mgs

cF f

Example: Pg138-87A car weighing 10.7 kN and traveling at 13.4 m/s attempts to round an unbanked curve with a radius of 61.0 m. a) What force of friction is required to keep the car on its circular path?b) If the coefficient of static friction between the tires and road is 0.35, is the attempt at taking the curve successful?

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W

Nf

100060 16.7 ,

3600o

km m h mv

h km s s

x

y

xF

sin cosN

2

sin cos cos sin

mv mg

R

2 cos sin sin cosv gR

2 2sin cos sin cosv gR gR v

222

22 2

9.8 / 200 sin8.1 11.1 / cos8.1sin cos0.079

sin cos 11.1 / sin8.1 9.8 / 200 cos8.1

o o

o o

m s m m sgR v

v gR m s m s m

yF

cos sinN W

40 11.1 , 200km m

v r mh s

sinN

2

sin cos

mvN

R

cosN

N

2 2cos sin sin cosv v gR gR

2 2sin cos sin cosv gR gR v

sin

cosx

y

N N

N N

cos

sinx

y

f f

f f

Practice: Pg135-50A banked circular highway curve is designed for traffic moving at 60 km/h. The radius of the curve is 200 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?

Nx

Ny

fx

fy

221 1

2

16.7 /tan tan 8.1

200 9.8 /oo m sv

Rg m m s

sin cosN f Cma cosN 2v

mR

cos sinN f W

2vmR

0 sinN W

cos sin

mg

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667 , 556topW N N N

) ?bottomb N

F

bottomN 667 111 778N N N

2vmR

F

Nbottom

W

Top:

Bottom:

topW N 2v

mR

bottomN W 2v

mR

2vW m

r

667 556 111topW N N N N

Ntop

W W

N

Practice: Pg134-45 (Modified wording)A student of weight 667 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the apparent weight of the student is 556 N. (a) Does the student feel “light’ or “heavy” there? (b) What is the apparent weight at the lowest point? If the wheel’s speed is doubled, what is the apparent weight at the (c) highest point and (d) lowest point?

a) Wapp < W

Feel “light”

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667 , 556 , ' 2topW N N N v v

) ' ?topc N

topF

'topN

667 4 111 223N N N

'topW N 2'v

mR

2

4v

W mR

22v

mR

2

4v

mR

Nbottom’

W

Ntop’

W

Practice: Pg134-45 (Modified wording)A student of weight 667 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the apparent weight of the student is 556 N. (a) Does the student feel “light’ or “heavy” there? (b) What is the apparent weight at the lowest point? If the wheel’s speed is doubled, what is the apparent weight at the (c) highest point and (d) lowest point?

) ' ?bottomd N

bottomF

'bottomN 667 4 111 1111N N N

'bottomN W 2'v

mR

2

4v

W mR

2

4v

mR

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a) In order to keep the same speed, and therefore the same centripetal force, the tension required is largest when the rock is at the bottom. •

•

•

Practice: Pg139-107A certain string can withstand a maximum tension of 40 N without breaking. A child ties a 0.37 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 m, slowly increasing the speed until the string breaks. a) Where is the stone on its path when the string break?b) What is the speed of the stone as the string breaks?

a

a

aW

W

W

T

T

) 40 , 0.37 , 0.91 , ?b T N m kg r m v

F T W ma 2v

mr T

2vW m

r

2vmg m

r

Tg

m

2v

r 2v T

r gm

v T

r gm

2

400.91 9.8 9.5

0.37

N m mm

kg s s

T

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Practice: Pg133-29Two blocks (m = 16 kg and M = 88 kg) are not attached to each other. The coefficient of static friction between the blocks is s = 0.38, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force F required to keep the smaller block from slipping down the larger block?

MmF

F

f

W

N N2

W2

N

x

y

m:xF

maF N

yF f W 0

M:xF MaN

F N maf W mg

N N mg

mgF ma

mgMa

a mg

M

mg mgF m

M 1

mg m

M

216 9.8 161

0.38 88

mkg kgs

kg

488N

mg

f

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Practice: A little girl of mass m1 = 10 kg sits on a slab of mass m2 = 20 kg on a frozen lake. Between girl and slab, the coefficient of static friction is 0.60, and the coefficient of static friction between the sled and lake is 010. The sled is pulled forward by a horizontal force F. What is the maximum magnitude of F such that the girl stays on the sled?

F

W1

W2

N1

N2

f1

N1

f1

f2

F

m1:

1 2 1 2 max10 , 20 , 0.60, 0.10, ?m kg m kg F

xF

yF

m2:

xF

yF

1 1m 1g m a

2N

1f 1m a 1 1 1N m a

1 1N W 0 1N 1W 1m g

a 1g

1 2F f f 2m a

F 1 2 2f f m a

02 2 1N W N

2 1W N 2 1 1 2m g m g m m g

1 1 2 2 2N N m a

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Practice (Continued)

1 1 2 2 2F N N m a

1 1 2 1 2 2 1m g m m g m g

1 2 2 1m m g

1 1 2 1 2 1 2m g m g m m g

1 1 2 2 1 2g m m m m g

20.10 0.60 10 20 9.8

210

mkg kg

s

N

F

W1

W2

N1

N2

f1

N1

f1

f2

F