AP Chapter 4

7/25/2019 AP Chapter 4

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FORCE

A forceis any influence that can change the velocity of a

body. Forces can act either through the physical contactof two objects (contact forces: push or pull or at adistance (field forces: !agnetic force" gravitationalforce.


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FREE #O$% $&A'RA)

&n all but the si!plestproble!s that involve forces" itis helpful to draw a free bodydiagra! (F#$of the situation.

*his is a vector diagramthatshows all the forces that actonthe body whose !otion isbeing studied. Forces that the

body e+erts on anything elseshould not be included" sincesuch forces do not affect thebody,s !otion.


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F*force due to tension: a rope or

cordF

-force nor!al: acting perpendicular

to a surface

Ffforce of friction: opposes !otion

Fggravitational force or weight"

alwaysdownward

Faapplied force: push or pull

./Co!plete the free body diagra!showing all of the forces acting on the!ass M. #e sure to show the direction of

each force as an arrow and label each forceclearly0
http://images.google.com/imgres?imgurl=www.science.widener.edu/~dunne/Web/lecs/5pics/9.GIF&imgrefurl=http://www.science.widener.edu/~dunne/Web/l05.html&h=242&w=320&prev=/images%3Fq%3Dfree%2Bbody%2Bdiagrams%26svnum%3D10%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26sa%3DG


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Fg

FfF

a

F-

E+a!ple:


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Fg

Fs


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Fg/ Fg1

F-/

F-1

Ff

F*

F*

Fa


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Fa

Fg

F-

Ff


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Fg

Fa

F-

Ff


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Fg/F

g/2F

g1

F-/

Fa

F-1

F* FfFf


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F&R)* 3A4 OF O*&O-

According to -ewton,s

First Law of Motion:

5&f no net force acts on it" abody at rest re!ains at rest anda body in !otion re!ains in!otion at constant speed in a

straight line.5

Isaac Newton

(1642-1727)


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A))*he property a body has of resisting any change in itsstate of rest or of unifor! !otion is called inertia.*heinertia of a body is related to the a!ount of !atter itcontains. A 6uantitative !easure of inertia is mass.

*he )& unit of !ass is the 7ilogra! (7g.


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Motion tends to continue

unchanged.

The elephant at rest tends

to remain at rest.

Tablecloth trick:

Too little orce! too little time to

o"ercome #inertia# o tableware.


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)ECO-$ 3A4 OF O*&O-According to -ewton,s Second Law of Motion,the net

forceacting on a body e6uals the product of the !assand the accelerationof the body. *he direction of theforce is thesa!e as that of the acceleration. &n e6uation for!:

F = ma&n the )& syste!" the unit for force is the newton(-: Anewton is that net force which" when applied to a /87g!ass" gives it an acceleration of / !9s1.


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Net forceis so!eti!es designated F.*he second law of !otion is the 7ey to understandingthe behavior of !oving bodies since it lin7s cause(force and effect(acceleration in a definite way.


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.1A force of ;;; - is applied to a / ;;; -

m> / ; !9s

F = ma

m

Fa=

1500

3000= > 1 !9s1

b.4hat will its velocity be < s later=

vf= v

o+ at

> 1( /; !9s


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.A /;;; 7g car goes fro! /; to 1; !9s in < s. 4hat force is actingon it=

m> /;;; 7gv

o> /; !9s

vf> 1; !9s

t> < s

F = mat

vm

=

=

5

10201000 > 1;;; -


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.A ?;8g tennis ball approaches a rac7et at /< !9s" is in contactwith the rac7et for ;.;;< s" and then rebounds at 1; !9s. Find theaverage force e+erted by the rac7et.

m> ;.;? 7gv

o> /< !9s

t> ;.;;< sv

f> 8 1; !9s

=

005.0

152006.0

tvm=F = ma > 8 1; -


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. /;;; 7g

F> 8;;; -v

o> ; !9s

vf> ; !9s

mFa =

10003000= > 8 !9s1

a

vvt

of

=

3

300

= > /; s

b.@ow far will the car travel during this ti!e=

x = vot+at2

> ;(/;2 (8(/;1 > /


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*@&R$ 3A4 OF O*&O-According to -ewton,s third law of motion"when onebody e+erts a force on another body" the second body

e+erts on the first an e6ual force in opposite direction.

*he *hird 3aw of otion appliestotwo different forceson two

different objects:5*he actionforceone object e+erts on theother" and the e6ual butopposite reaction forcethe

second object e+erts on thefirst.5

Action and reaction forces neverbalance outbecause they act on

different objects.


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$ction-%eaction &air 'amples


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.?A boo7 rests on a table.a.)how the forces acting on the table and the corresponding

reaction forces.

b.4hy do the forces acting on the table not cause it to !ove=

AC*&O-:F*#

REAC*&O-:F#*

*he forces F-and Fg

are balanced.


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4E&'@**he weightof a body is the gravitational force withwhich the Earth attracts the body. Weight(a vector

6uantity is different fro! mass(a scalar 6uantity. *heweight of a body varies with its location near the Earth(or other astrono!ical body" whereas its !ass is thesa!e everywhere in the universe. *he weightof a body

is the force that causes it to be accelerated downwardwith theacceleration of gravity g

Fro! the )econd 3aw of otion:

W > mg Bnits: -ewtons (-


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*@E -ORA3 FORCEA nor!al forceis a force e+erted by one surface onanother in a direction perpendicular to the surface of

contact.-ote: *he gravitational force and the nor!al force arenotan action8reaction pair.


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.A net horiDontal force of ;;; - is applied to a car at rest whoseweight is /;";;; -. 4hat will the car,s speed be after s=

Fa> ;;; -

Fg> /;";;; -t> s g

Fm g

=

8.910000= > /;1;. 7g

m

F

a = 4.1020

4000

= > .1 !9s1

vf= v

o+ at

> ; 2.1(

> ./ !9s


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.A < 7ga> ;. !9s1

Fg= mg

> ./ -

!F"= F

#$ F

g= ma

F#= ma + F

g

>


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.A 1;;8- wagon is to be pulled up a ;incline at constantspeed. @ow large a force parallel to the incline is needed if frictionis negligible=

F-

Fa

Fg

Fgy

Fg

+

G

Fg

> 1;; -

!Fx

= Fa% F

gx= ma = &

Fa= F

gx= F

gsin '&

> 1;;(sin;

> /;; -


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./;A cord passing over a frictionless pulley has a .; 7g !asshanging fro! one end and a .;87g !ass hanging fro! the other.(*his arrange!ent is called(twood)s machine.a.Find the acceleration of the !asses.

F*

F*

Fg/

Fg1

a (2 a (2

m*> 7gm

2> 7g


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F* F

*

Fg/ Fg1

a (2

a (2

!F= F

#% F

g*$F

#+ F

g2= m

#a

T

gg

m

FFa

12

=79

)8.9(7)8.9(9

+

= > /.11 !9s1


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F* F

*

Fg/ Fg1

a (2

a (2

b.Find the tension of thecord

Bse either side of thepulley:you get the sa!e answer0F

#% F

g*= m

*a

F#= m

*a+ F

g*

= m*a+g

> (/.112. > ./ -


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F-

F*

Fg

Fgy

Fg

+

G

.//A crate" which has a !ass of m>


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F-

F*

Fg

Fgy

Fg

+

G

b.4hat will be the !agnitude of the tension force (F# in the rope=

!Fx = F#% Fgx= &

F#= F

gx=F

gsin'-/

> /(sin 1


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FR&C*&O-: )*A*&C A-$ J&-E*&C FR&C*&O-Frictional forces act to oppose relative !otion between

surfaces that are in contact. )uch forces act parallel tothe surfaces.Static frictionoccurs between surfaces at rest relative to

each other. 4hen an increasing force is applied to aboo7 resting on a table" for instance" the force of staticfriction at first increases as well to prevent !otion. &n agiven situation" static friction has a certain !a+i!u!

value called starting friction. 4hen the force applied tothe boo7 is greater than the starting friction" the boo7begins to !ove across the table.


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FR&C*&O-: )*A*&C A-$ J&-E*&C FR&C*&O-*he 0inetic friction(or sliding frictionthat occurs

afterward is usually less than the starting friction" soless force is needed to 7eep the boo7 !oving than tostart it !oving.


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COEFF&C&E-* OF FR&C*&O-*he frictional force between two surfaces depends on

the normal(perpendicular force N pressing the!together and on the natures of the surfaces. *he latterfactor is e+pressed 6uantitatively in the coefficient offriction (mu whose value depends on the !aterials in

contact. *he frictional force is e+peri!entally found tobe:

)tatic friction

Jinetic friction

F Ff s N

F Ff k N=


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./1A horiDontal force of /; - is needed to pull a ?;.; 7g bo+across the horiDontal floor at constant speed. 4hat is thecoefficient of friction between floor and bo+=

Fa> /; -m> ?; 7g

!F"= F

N% F

g = &

FN = Fg=mg > ?;(.

> ;.1


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./A ;;8g bloc7 with an initial speed of ; c!9s slides along atabletop against a friction force of ;.; -.a.@ow far will it slide before stopping=

m> ;;+/;8

7gv

o> ;. !9s

Ff> ;. -

!Fx = % Ff= ma

Ff

FN

Fg

m

Fa

f

=310400

7.0

=

x> 8 /.< !9s1


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Ff

FN

Fg

axvv of 222+=

a

v

x o

2

2

=)75.1(2

)8.0( 2

= > ;./ !


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b.4hat is the coefficient of friction between the bloc7 and thetabletop=

!F"= F

N% F

g = &

FN = Fg=mg > ;;+/;8(. > .1 -

!Fx = % Ff= ma = & F

f= 1F

N

N

f

F

F=

92.3

7.0= > ;./


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./A ?;;87g go8cart is !oving on a level road at ; !9s.a.@ow large a retarding force is re6uired to stop it in a distance of; !=

m> ?;; 7g

vo> ; !9sx> ; !v

f> ; !9s x

vva

of

2

22

=

)70(2

)30(0 2

= > 8 ?.1 !9s

1

F = ma > ?;;(8?.1 > 8


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b.4hat is the !ini!u! coefficient of friction between the tires andthe road=

!F"= F

N% F

g = &

FN = Fg=mg > ?;;(. > ;.?


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./ ;; -

1 > ;. ;. ;; cos ;> ?.

-Fa"

> ;; sin ;> 1;; -

!F"= F

N+ F

a"$ F

g = &

FN

= Fg$ F

a"

> ;(. 8 1;; > ? -

Ff= 1F

N

> (;. 1 -


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F

g

FaF

f

FN

!Fx

= Fax

% Ff= ma

m

FFa

fax

=

70

2434.346 = > /. !9s1

/ A f f ;; - h 1< 7 b t l f


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./A force of ;; - pushes on a 1 ; !9s

vf> 1 !9s

t> s

t

vva f 0= 4

02 = > ;.

Fax

> ;; cos 1


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Fa

FN

Fg

Ff

!F"= F

N$ F

a"$ F

g = &

FN

= Fa"

+Fg

> ;?. 8 1


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F-

Ff

Fg

Fgy

Fg

+

G

./A 1;87g bo+ sits on an incline that !a7es an angle of ; withthe horiDontal. Find the acceleration of the bo+ down the incline ifthe coefficient of friction is ;.;.m> 1; 7g

> ;1 > ;.

Fg> 1;(.

>/? -

Fgx> /? cos ;> -F

g"> /? sin ;> /?. -

F


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F- F

f

Fg Fgy

Fg

+

G

!F"= F

N$ F

g"= &

FN

= Fg

> /?.

-F

f= 1F

N

> (;.(/?.

> 1.< !9s1

/ *wo bloc7s m (;; g and m (


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./*wo bloc7s m*(;; g and m

2(


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b.@ow large a force does m*then e+ert on m

2=

!1

alone

!Fx= m

2a

F*2

$ Ff2= m

2a

F*2= Ff2+ m2a> ;.< (. (;. 2 ;.< (1

> 1.? -

1; An object m > 1< 7g rests on a tabletop A rope attached to it


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.1;An object m(

> 1< 7g rests on a tabletop. A rope attached to it

passes over a light frictionless pulley and is attached to a !assm

3> /< 7g. &f the coefficient of friction is ;.1; between the table and

bloc7(" how far will bloc7 3drop in the first .; s after the syste!is released=

m(

> 1< 7g

m3> /< 7g1 > ;.1t> s

FN= F

g(

> 1 1< -

Ff(

= 1FN

> ;.1 (1 -

FN

Fg(

Ff F

#

F#

Fg3


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FN

Fg(

Ff F

#

F#

Fg3

!F = F#$ F

f(+ F

g3$ F

#= m

#a

1525

49)8.9(15

+

=T

fAgB

m

FFa

=

> 1.< !9s1

" = at2

> (1. // !


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AP Chapter 4

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