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Physics, Page 1
Chapter 6. Force and motion II
• Friction– Static friction– Sliding (Kinetic) friction
• Circular motion
Physics, Page 2
Summary of last lecture• 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
Physics, Page 3
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
Physics, Page 4
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
Physics, Page 5
Static Kinetic
fs ≤ μsn , μs : coefficient of static frictionfk = μkn , μk : coefficient of kinetic friction
Physics, Page 6
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
Physics, Page 7
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
Physics, Page 8
Example 2
T=50 N
M = 5 kg
μk = 0.2
Find acceleration of block
Answer: 8.04 m/s2
F = T- μkFN = ma
Physics, Page 9
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
Physics, Page 10
Example 4
a b
Which case requires the lesser force to overcome static friction?a) ab) bc) the same
Physics, Page 11
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
Physics, Page 12
Uniform Circular Motion(circular motion with constant speed)
vv
RRa v
R=
2
centripetal acceleration
• Instantaneous velocity is tangent to circle
• Instantaneous acceleration is radially inward
• There must be a force to provide the acceleration
a
Physics, Page 13
Centripetal force (구심력) and Centrifugal force (원심력)
Angular velocity
dtd θ
≡ωω
π=
2T
x
y
θ
v
T = Fr = mrω2
rvm
2
=
Centripetal forceCentrifugal force?
Physics, Page 14
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
Physics, Page 15
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 R
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
correct
Physics, Page 16
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
Question
correct
Physics, Page 17
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
Physics, Page 18
Example: Circular motion and gravity
• Newton’s Law of Gravity: F = GMEm/R2
• Newton’s 2nd law: F=ma
• Centripetal acceleration: a=v2/R
• Thus…GMEm/R2 = mv2/R
• v = √GME/R
• T = time for one orbit = 2πR/v
ME
R m
E
3/2
GMR2 T π
=
Find the period for one orbit.
Physics, Page 19
Example: Circular motion and gravity
E
3/2
GMR2 T π
= G = 6.67 x 10-11 N-m2/kg2
ME = 5.58 x 1024 kg
1. Rmoon-earth = 3.85 x 108 m Plug in and calculate: T=27.5 days
2. Rshuttle = RE = 6.38 x 106 m Plug in and calculate: T= 84.5 minutes
3. Rsynchronous = 4.23 x 107 m Plug in and calculate: T=1 day
ME
R m
Physics, Page 20
Example: Carnival Ride
g
Find minimum coefficient of static friction so that you don’t fall.
v
R
Answer: μS > gR/v2
Physics, Page 21
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==
Physics, Page 22
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
=
An
Physics, Page 23
Terminal speed (종단속력)Vterminal when a = 0
vbR −= : 마찰력 (Assumption)
amvbgmF =−=
mabvmg =−
0=−= vmbga
Terminal speed
gbmv t =
Physics, Page 24
221 AvDR ρ=
Drag Coefficient
Air density Cross section
maAvDmgRmgF =ρ−=−= 221
2
2v
mADga ρ
−= 0= : For terminal velocity
ADmgv t ρ
=2
3r∝
2r∝
rv t ∝
Terminal speed (종단속력) in air