Eunil Won Dept. of Physics, Korea University 1

Ch 03 Force

Movement of massive

object

Source of the move

Force Velocity, acceleration

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1 평(坪) = ? m2

~ 3.305 m2

~ 1.82 m x 1.82 m

Eunil Won Dept. of Physics, Korea University 3

Newton’s First Law

Law of Inertia (mass)Any body will remain at rest of in motion in a straight line with a constant velocity unless

acted upon by an outside force

If the net force is zero, there is a reference frame with zero acceleration

Eunil Won Dept. of Physics, Korea University 4

Newton’s Second Law

Law of MotionIf a force is applied, the state of motion is changed

avvpF mdtd

m)m(dtd

dtd =⎥⎦

⎤⎢⎣⎡ === ]m/skg[Nnewton 2⋅=⇒

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Newton’s Third Law

Action-Reaction LawWhenever one body exerts a force on a second body, the second body exerts a force back on the first that is equal in magnitude and opposite in direction

BAAB FF −=

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Example 3.1

N66.0m/skg0.66)m/skg)(1.5440.0(

,

22

,

=⋅==

=

extnet

extnet

F

maF

A 440-g can of food on a frictionless level surface is observed to accelerate at a rate of 1.5 m/sec2. What is the force on the can?

Eunil Won Dept. of Physics, Korea University 7

Example 3.2

100N50kg

20kg85kg

=

=

==

a

p

g

o

F

m

mm Acceleration

2m/s645.0502085

100 =++

=

++==

pgo

a

t

t

mmmF

mFa

An orderly

Gurney

The orderly exerts a backward force of 100 N on the floor. What acceleration is produced, assuming the friction in the wheels is negligible?

(question on the system I)

Force that the orderly exerts

An orderly

Patient

Eunil Won Dept. of Physics, Korea University 8

Example 3.3

100N50kg

20kg85kg

=

=

==

a

p

g

o

F

m

mm

Force

N2.45kgm/s2.45

)645.0)(5020(

)(

2 ==

+=

+== ammmaF pgo

An orderly

Gurney

Patient

Force that the orderly exerts

Calculate the force the orderly exerts on the gurney in Example 3.2

(Question on system 2)

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weight

gwaF

mm

==

Normal force

wNwN =⇒=−

=

0

0Ft

Weight and normal force

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Newton’s Universal Law of Gravitation

2rGMg =

2rmMGmgw ⇔=

2rmMGF =

: there is a force of attraction between any two masses that is proportional to the product of the masses and inversely proportional to the distance between their centers of mass

Eunil Won Dept. of Physics, Korea University 11

Eunil Won Dept. of Physics, Korea University 12

Static Friction

Nff kk µ== )(

Nf ss µ=

sff ≤

Friction

sµ

sµ

Kinetic Friction

: proportional to the normal force

kµ: coefficient of static friction

: coefficient of kinetic friction

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Friction

Coefficients of friction

Kinetic StaticRubber on dry concrete 0.7 1.0Rubber on wet concrete 0.5 0.7Wood on Wood 0.3 0.5Waxed wood on wet snow 0.1 0.14Metal on wood 0.3 0.5Steel on steel (dry) 0.3 0.6Steel on steel (oiled) 0.03 0.05Teflon on steel 0.04 0.04Bone lubricated with synovial fluid

(활액)0.015 0.016

Shoes on ice 0.05 0.1Ice on ice 0.03 0.1

Shoes on wood 0.7 0.9

Steel on ice 0.02 0.4

Eunil Won Dept. of Physics, Korea University 14

Example 3.4

N686)m/skg)(9.870( 2 === mgw

N96N)686)(14.0( === Nf sµ

00, =−⇒= fFF extnet

N68.6N)686)(1.0( ==== NfF kµ

A 70-kg cross-country skier is on level ground in wet snow wearing waxed wooden skis. (a) What is the maximum force that can be exerted on the skier in a horizontal direction without causing her to move? (b) Once the skier is moving, what horizontal force is necessary to keep her moving at a steady velocity?

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Tension

A tension is any force carried by a flexible string, rope, cable, chain etc

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Example 3.5

0, =−= wTF extnet

N98)m/skg)(9.810( 2 ==== mgwT

A child and basket with a total mass of 10 kg are suspended from a scale by a cord. Calculate the tension in the cord.

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Tensions of a patient

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Vector

{ } aa =aa ,,ρ

ba =

a−

01ˆˆ =⇒ ee

1. Vectors have direction &

magnitude

2. Two vectors are identical if

both direction and magnitude

are same

3. Negative vector.

4. Zero vector.

5. Unit vector.

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Vector analysis

sba =+

dba =−

a b

b−

s

d

a0a0a)(a

cb)(ac)(baabba

=+=−+

++=+++=+

a

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Components of Vectors

a

x

y

xy

xa

ya

)0,1,0(ˆ)0,0,1(ˆ

====

jyix

zyxaaaa zyx

ˆˆˆ zyx aaa ++=

++=

jijijiba

)()(

)()(

yyxx

yxyx

baba

bbaa

+++=

+++=+

yyy

xxx

yx

basbas

ss

+=+=

+==+ jisba

⎟⎟⎠

⎞⎜⎜⎝

⎛=

==

−

x

y

yx

aa

aaaa

1tan

sin,cos

θ

θθ

Eunil Won Dept. of Physics, Korea University 21

Example 3.6

jiFjiF

FFF

1

21t

yx

yx

FF

FF

222

11

+=

+=+=

94.23)342.0(7020sin78.65)939.0(7020cos

11

11

==°===°=

FFFF

y

x

53.15)259.0(6015sin95.57)966.0(6015cos

22

22

==°===°=

FFFF

y

x

jiFt )()( 2221 yyxx FFFF +++=

41.873.123

22

21

=−==+=

yyty

xxtx

FFFFFF

°=⎟⎠⎞

⎜⎝⎛=

=+=

− 89.3tan

01.124

1

22

tx

ty

tytxt

FF

FFF

θ

Calculate the total force exerted on a gurney

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Example 3.7/3.8

Acceleration of the cart

0=+++=

gNFFFF

v

vht

m

3.150.23.17 =−=−= fFF hn

2m/s61.025

3.15 ===mFa n

Normal Force

0=−−

=

vFwN

0Fy

N25510)8.9)(25( =+=

+=+= vv FmgFwN

Calculate the acceleration (cart+stuff=25 kg, frictional force=2.0 N)

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Torque and Rotation

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Torque

FrrFrFt ⊥=== φτ sin ωτ,τ //Fr ×=

The effectiveness of a force in producing a rotation is called torque

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Equilibrium Condition

0τ,0 == ∑∑F

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Example 3.9

Bta

netccwnetcw

Fww )cm0.4())(cm40())(cm15(,,

=+

=ττ

N4840.4

N1568N5.3670.4

)m/s8.9)(kg0.4)(40()m/s8.9)(kg5.2)(15(cm0.4

))()(cm40())()(cm15(

22

=+=

+=

+= gmgmF taB

N7.63=+=+=

wgmgmwww tata

N63.7 N484 ⟩⟩

(a) Calculate the force that the biceps muscle exerts to hold the book(b) Compare the force exerted by the biceps muscle with the weight of the forearm and book. (clockwise) (counterclockwise)

Eunil Won Dept. of Physics, Korea University 27

Example 3.10

TllFFlTl MM

netccwnetw

)/(,,

⊥⊥⊥⊥ ′=⇒′=

= ττ

N1715N)98cm/2.0cm)(35(

mg)l/l(FM

==

′= ⊥⊥

Calculate the force exerted by the extensor muscle in the upper leg when lifting a weight as shown below

Note: cm0.2lcm35l

=′=

⊥

⊥

Eunil Won Dept. of Physics, Korea University 28

Rotational Motion and Centripetal Force

rvmmaF rc

2

==

Eunil Won Dept. of Physics, Korea University 29

vr

vpqt )2( θ≈=∆

0coscos

=∆−

=∆−

=tvv

tvv

a pqpxqxx

θθ

⎟⎠⎞

⎜⎝⎛−=

−−=

∆−

=θ

θθ

θθ sin/)2(

sinsin 2

rv

vrvv

tvv

a pqpyqyy

jijiv

θθ sincos vv

vv pypxp

+=

+=

jijiv

θθ sincos vv

vv qyqxq

−=

+=

ja ya−=rv

rvaa yy

2

0

2

0

sinlimlim =⎟⎟⎠

⎞⎜⎜⎝

⎛==

→→ θθ

θθ

Eunil Won Dept. of Physics, Korea University 30

Example 3.11

m/s1062.2s60

)m05.0(250000 2×=×== πtsv

26222

m/s101.370.05m

m/s)10(2.62 ×=×==rvac

gaga

c

c

)1040.1(

1040.1m/s 9.8

m/s101.37

5

52

26

×=

×=×=

An ultracentrifuge spins at 50,000 revolutions per minute and the material being centrifuged is 5.0 cm from the pivot point. Calculate the centripetal acceleration and express it as a multiple of the acceleration of gravity (in g’s)