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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
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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
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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?
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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
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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
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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
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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
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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)
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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
=′=
⊥
⊥
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Rotational Motion and Centripetal Force
rvmmaF rc
2
==
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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
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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)