friction static friction sliding (kinetic) friction circular...

Physics,

Chapter 6. Force and motion II

• Friction– Static friction– Sliding (Kinetic) friction

• Circular motion

Physics,

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,

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,

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,

Static Kinetic

fs ≤ μsn , μs : coefficient of static frictionfk = μkn , μk : coefficient of kinetic friction

Physics,

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,

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,

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,

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,

Example 4

a b

Which case requires the lesser force to overcome static friction?a) ab) bc) the same

Physics,

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,

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,

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,

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,

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,

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,

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,

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,

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,

Example: Carnival Ride

g

Find minimum coefficient of static friction so that you don’t fall.

v

R

Answer: μS > gR/v2

Physics,

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,

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,

Terminal speed (종단속력)Vterminal when a = 0

vbR −= : 마찰력 (Assumption)

amvbgmF =−=

mabvmg =−

0=−= vmbga

Terminal speed

gbmv t =

Physics,

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

friction static friction sliding (kinetic) friction circular...

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