Chapter 5, 6. Force and Motion
Transcript of Chapter 5, 6. Force and Motion
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..
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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!
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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
≡
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Example of Bad Thinking
• Since FFm,b = - FFb,m why isn’t FFnet = 0, and aa = 0 ?
a ??a ??FFb,m FFm,b
ice
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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
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(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...
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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
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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
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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
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Example
Find acceleration and normal force
Answers: a= g sin(θ) FN = mg cos (θ)
θ
am
mg
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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
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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
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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
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M
m
예) 차원 비교? T < mg ?
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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
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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